Trend of Atomic Radius
Defined as half the internuclear distance between 2 similar atoms bonded only by a single bond.
Atomic radius decreases across the period
• Proton no./nuclear charge increases. Hence, drawing electrons closer to nucleus.
• Negligible increase in shielding effect as electrons are added to the valence shell
(no. of inner electron shells remain constant)
Trend of Ionic Radius
Defined as the radius of a spherical ion in an ionic compound.
Na, Mg, Al and Si
• Positive ions decreases from Na+, Mg2+, Al3+ to Si4+
• Nuclear charge increases as no. of proton increases
- Equal no. of electrons
- Negligible increase in screening effect as equal number of electron shells
- Hence, attraction force between the nucleus and the electrons in Si4+ ion is stronger than
that in Na+
P, S and Cl
Negative ions radius decreases from P, S to Cl
• Nuclear charge increases as no. of proton increases
- Equal no. of electrons
- Negligible increase in screening effect as equal no. of electron shells
- Therefore, the attraction force between the nucleus and the electrons in CL ion is
stronger than that in P ion.
Comparison between ionic radius and atomic radius
Size of positive ions : size of parent atoms
• Positive ions are smaller than parent atoms because:
- Cations are formed when a neutral atom loses electon/s. Thus cations have lesser
electrons than parent atoms.
- Cations and parent atoms have equal number of protons.
- nuclear charge is therefore shared among less no of electrons in cations.
- Hence the attraction force between the nucleus and the remaining electrons in cations
is stronger than that of parent atoms.
Size of negative ions : size of parent atoms
• Negative ions are larger than parent atoms
- Anions are formed when a neutral atom gains electron/s. Thus anions have more
electrons than parent atom.
- Anions and parent atoms have equal number of protons.
- Nuclear charge is therefore shared among more no of electrons in cations.
- Hence the attraction force between the nucleus and the remaining electrons in
cations is weaker than that of parent atoms.
Chemical formula of how these elements exist:
Na
Mg
Al
Si
P4
S8
Cl2
Thursday, April 16, 2009
Chemical periodicity part 3
Learning Outcome : Interpret the variation in melting point and electrical conductivity in terms of presence of simple molecular ,giant molecular and metallic bonding in elements
Bonding and structure of elements
Ø Metallic Bonds
Ø Giant Metallic Structure
>Na, Mg, Al
Ø Covalent bonds
Ø Giant covalent structure
>Si
Ø Intramolecular : Covalent bonds
Ø Intermolecular: Van Der Waals forces
Ø Simple molecular structure
>P, S, Cl
Trend of Melting Point
· Na, Mg ,Al
- High Melting Point
Since it is a :
>Giant Metallic structure
» High energy is required to overcome the strong attractive forces between the metal cations and delocalized electrons
>Melting point: Na< Mg
-decrease in cation radius
- increase in no of delocalized electrons
»increase in charge/size ratio, increase in the metallic bond strength
· Si
-Has the highest melting point
>Giant molecular structure
»large amount of energy required to break many strong covalent bonds in the giant molecular structure
· P, S, Cl
-Low melting point
>exists as simple molecular structures
>Intermolecular forces : Van der Waals forces
>Low thermal energy is required to overcome the weak Van der Waals forces
»The melting point of these elements is dependant on the size the electron cloud
»The higher the molecular mass(Mr), the larger the electron cloud,the stronger the VDW forces. Hence,the higher the melting point.
Trend of Electrical Conductitvity
· Na, Mg, Al
-Good electrical conductors
»they are metals with delocalized electrons
-Degree of electrical conductivity :
Al >Mg >Na
»As more valence electrons are contributed to the ‘sea of electrons’,electrical conductivity increases
· Si
-Low electrical conductivity under ordinary conditions
»Conductivity increases with an increase in temperature
»At a higher temperature ,electrons are excited to higher energy levels. Silicon is a semi- conductor .
· P ,S, Cl
- Non-conductor of electricity because absence of delocalized electrons or mobile ions.
Bonding and structure of elements
Ø Metallic Bonds
Ø Giant Metallic Structure
>Na, Mg, Al
Ø Covalent bonds
Ø Giant covalent structure
>Si
Ø Intramolecular : Covalent bonds
Ø Intermolecular: Van Der Waals forces
Ø Simple molecular structure
>P, S, Cl
Trend of Melting Point
· Na, Mg ,Al
- High Melting Point
Since it is a :
>Giant Metallic structure
» High energy is required to overcome the strong attractive forces between the metal cations and delocalized electrons
>Melting point: Na< Mg
-decrease in cation radius
- increase in no of delocalized electrons
»increase in charge/size ratio, increase in the metallic bond strength
· Si
-Has the highest melting point
>Giant molecular structure
»large amount of energy required to break many strong covalent bonds in the giant molecular structure
· P, S, Cl
-Low melting point
>exists as simple molecular structures
>Intermolecular forces : Van der Waals forces
>Low thermal energy is required to overcome the weak Van der Waals forces
»The melting point of these elements is dependant on the size the electron cloud
»The higher the molecular mass(Mr), the larger the electron cloud,the stronger the VDW forces. Hence,the higher the melting point.
Trend of Electrical Conductitvity
· Na, Mg, Al
-Good electrical conductors
»they are metals with delocalized electrons
-Degree of electrical conductivity :
Al >Mg >Na
»As more valence electrons are contributed to the ‘sea of electrons’,electrical conductivity increases
· Si
-Low electrical conductivity under ordinary conditions
»Conductivity increases with an increase in temperature
»At a higher temperature ,electrons are excited to higher energy levels. Silicon is a semi- conductor .
· P ,S, Cl
- Non-conductor of electricity because absence of delocalized electrons or mobile ions.
Chemical Equilibria
a.) A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as
aA + bB cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.Dynamic Equilibrium refers to a reversible reaction in which the rates of forward and reverse reactions become equal and there is no change in the concentration of the reactants and products.b.) Le Chatelier's Principle can be used to predict the effect of a change in conditions on a chemical equilibrium.
c.) If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.
Concentration
Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. The chemical system will attempt to partially oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent and yield of products will be altered corresponding to the impact on the system.
This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.
CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced. This observation is supported by the "collision theory". As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.
Temperature
Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favour the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.
Changes in pressure owing to changes in volume
The equilibrium concentrations of the products and reactants do not directly depend on the pressure subjected to the system. However, a change in pressure due to a change in volume of the system will shift the equilibrium.
Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. When the volume of the system is changed, the partial pressures of the gases change. Because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.
Thus, an increase in pressure due to decreasing volume causes the reaction to shift to the side with the fewer moles of gas.[2] A decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical system (or equation)
g.) The Haber synthesis was developed into an industrial process.Conditions:medium temperature (~500oC) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).Osmium is a much better catalyst for the reaction but is very expensive.
This process produces an ammonia, NH3(g), yield of approximately 10-20%.
The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
N2(g)nitrogen
+
3H2(g)hydrogen
heat, pressure, catalyst
2NH3(g) ammonia
H = -92.4 kJ mol-1
By Le Chetalier's Principle:
Increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation. Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules.
Decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as wellHowever, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia.
The equilibrium expression for this reaction is:
Keq =
[NH3]2
[N2][H2]3
h.) Chain Disproportionation occurs when two radicals meet, instead of coupling, they exchange a proton. That gives two terminated chains, one saturated and the other with a terminal double bond.
2R-(CH2CH2)n-CH2CH2• → R-(CH2CH2)n-CH=CH2 + R-(CH2CH2)n-CH2CH3
http://www.scribd.com/doc/6618373/InDepth-Overview-and-Comparison-of-2006-2007-AL-Chem-Syllabus
aA + bB cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.Dynamic Equilibrium refers to a reversible reaction in which the rates of forward and reverse reactions become equal and there is no change in the concentration of the reactants and products.b.) Le Chatelier's Principle can be used to predict the effect of a change in conditions on a chemical equilibrium.
c.) If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.
Concentration
Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. The chemical system will attempt to partially oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent and yield of products will be altered corresponding to the impact on the system.
This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.
CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced. This observation is supported by the "collision theory". As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.
Temperature
Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favour the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.
Changes in pressure owing to changes in volume
The equilibrium concentrations of the products and reactants do not directly depend on the pressure subjected to the system. However, a change in pressure due to a change in volume of the system will shift the equilibrium.
Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. When the volume of the system is changed, the partial pressures of the gases change. Because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.
Thus, an increase in pressure due to decreasing volume causes the reaction to shift to the side with the fewer moles of gas.[2] A decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical system (or equation)
g.) The Haber synthesis was developed into an industrial process.Conditions:medium temperature (~500oC) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).Osmium is a much better catalyst for the reaction but is very expensive.
This process produces an ammonia, NH3(g), yield of approximately 10-20%.
The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
N2(g)nitrogen
+
3H2(g)hydrogen
heat, pressure, catalyst
2NH3(g) ammonia
H = -92.4 kJ mol-1
By Le Chetalier's Principle:
Increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation. Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules.
Decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as wellHowever, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia.
The equilibrium expression for this reaction is:
Keq =
[NH3]2
[N2][H2]3
h.) Chain Disproportionation occurs when two radicals meet, instead of coupling, they exchange a proton. That gives two terminated chains, one saturated and the other with a terminal double bond.
2R-(CH2CH2)n-CH2CH2• → R-(CH2CH2)n-CH=CH2 + R-(CH2CH2)n-CH2CH3
http://www.scribd.com/doc/6618373/InDepth-Overview-and-Comparison-of-2006-2007-AL-Chem-Syllabus
Chemical periodicity part 2
Periodic Table
1st ionization energy is the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous cations.
It is a endothermic process.
Across Periods:
There are more protons in each nucleus so the nuclear charge in each element increases.
Therefore the force of attraction between the nucleus and outer electron is increased.
There is a negligible increase in shielding because each successive electron enters the same energy level.
More energy is needed to remove the outer electron.
1st I.E increases
First ionization energy generally increases going across Period 3.
The first ionization energy drops between magnesium and aluminium before increasing again.
The first ionization energy drops between phosphorus and sulphur before increasing again.
Irregularities:
1st I.E of Magnesium is greater than Aluminium.
Magnesium: 1s2 2s2 2p6 3s2
Aluminium: 1s2 2s2 2p6 3s2 3p1
3p1 electron is at a higher energy level. Thus it is much easier to be removed than 3s2 electrons.
1st I.E of Phosphorus is greater than Sulphur.
Phosphorus: 1s2 2s2 2p6 3s2 3p3
Sulphur: 1s2 2s2 2p6 3s2 3p4
The 3p electrons in phosphorus are all unpaired. In sulphur, there is one group of paired electrons and 2 unpaired ones.
There is some repulsion between paired electrons (interelectronic repulsion) in the same orbital.
Less energy needed to remove one electron from 3p4 than 3p3.
Down the Group:
There will be an additional shell of electrons
There is an increase in screening/shielding effect which outweighs the increase in effective nuclear charge.
Valence electrons is further away from the nucleus.
Less energy needed to remove electrons
1st I.E decreases
1st ionization energy is the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous cations.
It is a endothermic process.
Across Periods:
There are more protons in each nucleus so the nuclear charge in each element increases.
Therefore the force of attraction between the nucleus and outer electron is increased.
There is a negligible increase in shielding because each successive electron enters the same energy level.
More energy is needed to remove the outer electron.
1st I.E increases
First ionization energy generally increases going across Period 3.
The first ionization energy drops between magnesium and aluminium before increasing again.
The first ionization energy drops between phosphorus and sulphur before increasing again.
Irregularities:
1st I.E of Magnesium is greater than Aluminium.
Magnesium: 1s2 2s2 2p6 3s2
Aluminium: 1s2 2s2 2p6 3s2 3p1
3p1 electron is at a higher energy level. Thus it is much easier to be removed than 3s2 electrons.
1st I.E of Phosphorus is greater than Sulphur.
Phosphorus: 1s2 2s2 2p6 3s2 3p3
Sulphur: 1s2 2s2 2p6 3s2 3p4
The 3p electrons in phosphorus are all unpaired. In sulphur, there is one group of paired electrons and 2 unpaired ones.
There is some repulsion between paired electrons (interelectronic repulsion) in the same orbital.
Less energy needed to remove one electron from 3p4 than 3p3.
Down the Group:
There will be an additional shell of electrons
There is an increase in screening/shielding effect which outweighs the increase in effective nuclear charge.
Valence electrons is further away from the nucleus.
Less energy needed to remove electrons
1st I.E decreases
Chemcial Bonding part 3
Chemical Bonding
Learning outcomes:
(a) Describe ionic bonding, as in sodium chloride and magnesium oxide, including the use of ‘dot and cross’ diagrams.
Answer: NaCl
x
x
xx
x
x
x
xxNa + Cl x [ Na ] + [ x Cl ]-
x
x x
Sodium atom has one valence electron, it loses one to become a positive 1+ charged ion to achieve the stable octet arrangement.
Chlorine atom has seven valence electrons, it gains one electron from sodium atom to form a negative -1 charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attract each other and are bonded in a single ionic bond to form the lattice structure, NaCl.
Answer: MgO
xx
xx
xx
xx Mg + x O à [ Mg ] 2+ [ O ] 2-
Magnesium atom has two valence electrons, it loses two electrons to become a positive 2+ charged ion to achieve the stable octet arrangement.
Oxygen atom has six valence electrons, it gains two electrons from Magnesium atom to form a negative 2- charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attracted each other by electrostatic force and are bonded in a double ionic bond.
(i) Describe metallic bonding in terms of a lattice of positive ions surrounded by mobile electrons.
Answer: The valence electrons from metals are far away from the nucleus hence they are not strongly attracted to the positive charged of the protons in the nucleus. These valence electrons can easily leave the atom and becomes free mobile delocalized electrons. Metals are structures held up by metallic bonding which is the attraction force between the positive cations and the “sea of negative delocalized electrons”. The more the delocalized electrons, the stronger the metallic bonding, the higher the melting point of the metal.
(m) Describe, in simple terms , the lattice structure of a crystalline solid which is:
(i) ionic, as in sodium chloride, magnesium oxide.
Answer: NaCl
The lattice structure of sodium chloride consists of large number of Na+ and Cl- ions arranged in an orderly manner. One sodium ion is bonded three dimensionally to six chloride ions, each by a single ionic bond. The ions are arranged in straight rows and forms a large structures with flat sides, a crystal structure.The ions of Na+ and Cl- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence NaCl is hard but brittle.
Answer: MgO
The lattice structure of magnesium oxide consists of large number of Mg2+ and O2- ions arranged in an orderly manner. One magnesium ion is bonded three dimensionally with six oxygen ions, each by a double ionic bond. The ions of Mg2+ and O2- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence MgO is hard but brittle. Magnesium oxide also has a higher melting point than sodium chloride because the electrostatic force of attraction between the oppositely charged ions in magnesium oxide is twice that of sodium chloride’s.
(ii) Simple molecular, as in iodine.
Answer: Iodine is made up of small discrete iodine molecules. Each molecule is made up of two iodine atoms covalently bonded to one another. These iodine molecules attracted each other by weak Van Der Waal’s intermolecular forces to form a lattice structure which has very low melting point and boiling point.
(iii) giant molecular, as in graphite; diamond.
Answer: Graphite
Graphite is a carbon allotrope in which large number of carbon atoms are arranged in flat, hexagonal parallel layers to form a giant covalent structure. The carbon atoms in each layer are arranged in rings of six atoms with each carbon atom covalently bonded strongly to three other atoms in its layers. Weak Van Der Waal’s forces are present between the layers. The layers of atoms can slide over each other easily making graphite soft and slippery.
Answer: Diamond
Diamond is a carbon allotrope in which each carbon atom forms four covalent bonds with four other carbon atoms in a three dimensional tetrahedral arrangement. The structure is a giant network of carbon atoms held together by covalent bonds. Each carbon atom in diamond is at the centre of a tetrahedron.
(iv) Hydrogen-bonded, as in ice.
Answer: Ice has an open regular structure, with the O atom in H2O molecule forms two hydrogen bonds with two water molecules. The two H atoms form two more hydrogen bonds with two other water molecules. Each H2O molecule is therefore hydrogen bonded to four other H2O molecules in a tetrahedral arrangement.
(v) Metallic, as in copper.
Answer: Metals have giant structures. In metal like copper , the atoms are packed together closely in regular three-dimensional patterns to form giant lattice structures. In copper, it consisted of positive Cu2+ ions bonded to a sea of delocalized electrons.
(n) Outline the importance of hydrogen bonding to the physical properties of substances, including ice and water.
Answer: Each H2O molecule is hydrogen –bonded to four other H2O molecules in a tetrahedral arrangement. The open structure explains the fact that ice is less dense than liquid water at 0 oC . Hence, during winter, only the top layer of water is frozen. The bottom part of water is relatively warmer and keeps the aquatic life alive.
Learning outcomes:
(a) Describe ionic bonding, as in sodium chloride and magnesium oxide, including the use of ‘dot and cross’ diagrams.
Answer: NaCl
x
x
xx
x
x
x
xxNa + Cl x [ Na ] + [ x Cl ]-
x
x x
Sodium atom has one valence electron, it loses one to become a positive 1+ charged ion to achieve the stable octet arrangement.
Chlorine atom has seven valence electrons, it gains one electron from sodium atom to form a negative -1 charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attract each other and are bonded in a single ionic bond to form the lattice structure, NaCl.
Answer: MgO
xx
xx
xx
xx Mg + x O à [ Mg ] 2+ [ O ] 2-
Magnesium atom has two valence electrons, it loses two electrons to become a positive 2+ charged ion to achieve the stable octet arrangement.
Oxygen atom has six valence electrons, it gains two electrons from Magnesium atom to form a negative 2- charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attracted each other by electrostatic force and are bonded in a double ionic bond.
(i) Describe metallic bonding in terms of a lattice of positive ions surrounded by mobile electrons.
Answer: The valence electrons from metals are far away from the nucleus hence they are not strongly attracted to the positive charged of the protons in the nucleus. These valence electrons can easily leave the atom and becomes free mobile delocalized electrons. Metals are structures held up by metallic bonding which is the attraction force between the positive cations and the “sea of negative delocalized electrons”. The more the delocalized electrons, the stronger the metallic bonding, the higher the melting point of the metal.
(m) Describe, in simple terms , the lattice structure of a crystalline solid which is:
(i) ionic, as in sodium chloride, magnesium oxide.
Answer: NaCl
The lattice structure of sodium chloride consists of large number of Na+ and Cl- ions arranged in an orderly manner. One sodium ion is bonded three dimensionally to six chloride ions, each by a single ionic bond. The ions are arranged in straight rows and forms a large structures with flat sides, a crystal structure.The ions of Na+ and Cl- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence NaCl is hard but brittle.
Answer: MgO
The lattice structure of magnesium oxide consists of large number of Mg2+ and O2- ions arranged in an orderly manner. One magnesium ion is bonded three dimensionally with six oxygen ions, each by a double ionic bond. The ions of Mg2+ and O2- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence MgO is hard but brittle. Magnesium oxide also has a higher melting point than sodium chloride because the electrostatic force of attraction between the oppositely charged ions in magnesium oxide is twice that of sodium chloride’s.
(ii) Simple molecular, as in iodine.
Answer: Iodine is made up of small discrete iodine molecules. Each molecule is made up of two iodine atoms covalently bonded to one another. These iodine molecules attracted each other by weak Van Der Waal’s intermolecular forces to form a lattice structure which has very low melting point and boiling point.
(iii) giant molecular, as in graphite; diamond.
Answer: Graphite
Graphite is a carbon allotrope in which large number of carbon atoms are arranged in flat, hexagonal parallel layers to form a giant covalent structure. The carbon atoms in each layer are arranged in rings of six atoms with each carbon atom covalently bonded strongly to three other atoms in its layers. Weak Van Der Waal’s forces are present between the layers. The layers of atoms can slide over each other easily making graphite soft and slippery.
Answer: Diamond
Diamond is a carbon allotrope in which each carbon atom forms four covalent bonds with four other carbon atoms in a three dimensional tetrahedral arrangement. The structure is a giant network of carbon atoms held together by covalent bonds. Each carbon atom in diamond is at the centre of a tetrahedron.
(iv) Hydrogen-bonded, as in ice.
Answer: Ice has an open regular structure, with the O atom in H2O molecule forms two hydrogen bonds with two water molecules. The two H atoms form two more hydrogen bonds with two other water molecules. Each H2O molecule is therefore hydrogen bonded to four other H2O molecules in a tetrahedral arrangement.
(v) Metallic, as in copper.
Answer: Metals have giant structures. In metal like copper , the atoms are packed together closely in regular three-dimensional patterns to form giant lattice structures. In copper, it consisted of positive Cu2+ ions bonded to a sea of delocalized electrons.
(n) Outline the importance of hydrogen bonding to the physical properties of substances, including ice and water.
Answer: Each H2O molecule is hydrogen –bonded to four other H2O molecules in a tetrahedral arrangement. The open structure explains the fact that ice is less dense than liquid water at 0 oC . Hence, during winter, only the top layer of water is frozen. The bottom part of water is relatively warmer and keeps the aquatic life alive.
Chemical Bonding part 2
(f) Hydrogen bonding – a specialized form of permanent dipole –dipole interaction.
Hydrogen bonding is formed between the hydrogen atom of an ammonia or water molecule (with a small positive charge) in one molecule and the lone pair of electrons of the electronegative atom such as nitrogen, oxygen or fluorine in another neighbouring molecule.
(g) Bond length – the distance between the nuclei of the two atoms in the covalent bond or sum of covalent (atomic) radii.
The shorter the bond length, the stronger the covalent bond.
Bond energy – the amount of energy required to break one mole of covalent bond at gaseous state.
The greater the bond energy, the stronger the bond.
Bond polarity – the dipole-dipole intermolecular forces between the slightly positively-charged ends of one molecule to the negative end of another or the same molecule.
(h) Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Permanent dipole-dipole attraction – it is the attraction between molecules that have permanent dipole moments (polar molecules) in liquid and solid states.
(j) Ionic bonding – they are formed when electrons are transferred from the valence shell of atoms of one element to other atoms.
1. high melting and boiling point
2. soluble in water, insoluble in organic solvents (exceptions)
3. poor electrical conductors in solid state but good conductors in molten and aqueous state.
4. hard and brittle
Covalent bonding – the sharing (pairing) of electrons between atoms.
Hydrogen bonding – a specialized form of permanent dipole –dipole interaction.
Metallic bonding – the force of attraction between positive metal ions and the negative delocalized electrons.
1. high degree of electrical conductivity
2. high degree of thermal conductivity
3. high melting and boiling point
4. shiny metal lustre
5. insoluble in water
6. malleable
Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Hydrogen bonding is formed between the hydrogen atom of an ammonia or water molecule (with a small positive charge) in one molecule and the lone pair of electrons of the electronegative atom such as nitrogen, oxygen or fluorine in another neighbouring molecule.
(g) Bond length – the distance between the nuclei of the two atoms in the covalent bond or sum of covalent (atomic) radii.
The shorter the bond length, the stronger the covalent bond.
Bond energy – the amount of energy required to break one mole of covalent bond at gaseous state.
The greater the bond energy, the stronger the bond.
Bond polarity – the dipole-dipole intermolecular forces between the slightly positively-charged ends of one molecule to the negative end of another or the same molecule.
(h) Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Permanent dipole-dipole attraction – it is the attraction between molecules that have permanent dipole moments (polar molecules) in liquid and solid states.
(j) Ionic bonding – they are formed when electrons are transferred from the valence shell of atoms of one element to other atoms.
1. high melting and boiling point
2. soluble in water, insoluble in organic solvents (exceptions)
3. poor electrical conductors in solid state but good conductors in molten and aqueous state.
4. hard and brittle
Covalent bonding – the sharing (pairing) of electrons between atoms.
Hydrogen bonding – a specialized form of permanent dipole –dipole interaction.
Metallic bonding – the force of attraction between positive metal ions and the negative delocalized electrons.
1. high degree of electrical conductivity
2. high degree of thermal conductivity
3. high melting and boiling point
4. shiny metal lustre
5. insoluble in water
6. malleable
Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Chemical Bonding
b) Describe, including the use of ‘dot-and-cross’ diagrams,
i) covalent bonding in H2, O2, Cl2, HCl, CO2, CH4, C2H6
Covalent bond is formed from sharing of electrons between atoms. A covalent bond is the electrostatic force of attraction between the nuclei of 2 atoms for the shared pair of electron.
Molecule
‘dot-and-cross’ diagram
Description(to achieve stability)
H2
2 hydrogen atoms shared 1 pair of e- in the valence shell
O2
2 oxygen atoms shared 2 pairs of e- in the valence shell
Cl2
2 chlorine atoms shared 1 pair of e- in the valence shell
HCl
1 hydrogen & 1 chlorine atoms shared 1 pair of e- in the valence shell
CO2
1 carbon & 2 oxygen atoms shared 2 pairs of e- in the valence shell
CH4
1 carbon & 4 hydrogen atoms shared 4 pairs of e- in the valence shell
C2H6
2 carbon & 6 hydrogen atoms shared 7 pair of e- in the valence shell
ii) dative bond, in BF3->NH3
Dative bond is a type of covalent bond where the shared paired of electrons comes from one donor only. The donor must possess 1 lone pair of electron in its valence shell. The acceptor must have an empty orbital in its outer quantum shell to accommodate the lone pair.
NH3 + BF3----à BF3->NH3
NH3’s 1 lone pair donates to BF3. Now both N and B have 8e- in valence shell.
c) Explain shapes and bond angles of molecules by, using qualitative model of electron pair repulsion, using as simple examples: BF3 (trigonal), CO2 (linear), CH4 (tetrahedral), NH3 (pyramidal), H2O(V-shape), SF6(octahedral), by using VSEPR Theory.
Valence Shell Electron Pair Repulsion Theory can be use to predict the shape f simple covalent molecules.
I. These b.p & l.p. electrons around central atom determine the shapes of the molecule.
II. The electrons repel each other as far as possible.
III. Degree of repulsion
l.p.-l.p > l.p-b.p. > b.p.-b.p.
105’ 107’ 109’
Molecule
Structural formula
Kind of repulsion
Bond angle
Shape
Description
BF3
l.p.-l.p.
360’/3=120’
Trigonal planar
(because it is a triangle in 2D.)
B is in Grp III F is in Grp VII. There are 3F atoms. The strength of repulsion is equal because all are l.p.
CO2
l.p.-l.p.
360’/2=180’
Linear (because it is in a straight line)
There is repulsion between 2 O atoms, so the repulsion is the largest.
CH4
b.p-b.p.
109’
Tetrahedral (because there are 4 faces)
There is repulsion between H atoms.
NH3
l.p.-b.p.
107’
Trigonal pyramidal (because it is a triangle in 3D.
There is repulsion between N and H atom. So the angle will be pushed smaller.
H2O
l.p.-l.p.
105’
V-shape
There is repulsion causing the H atoms to be pushed closer to each other.
SF6
b.p-b.p
90’
Octahedral (because it has two 4 sided pyramid)
This is a molecule with more than 8e-.
Such case can be seen in period 3 and 4. There is no l.p. repulsion, so there is equal bond angle.
Factors affecting Bond Size:
I. Total number of e- pairs.
The more the number of e- pairs the central atom has, the smaller the bond angle.
II. Number of lone pairs.
L.p. electrons cause more repulsion than the b.p. electrons. Bond pairs are pushed closer to one another.
III. Size of central atom.
As size of central atom increases, the l.p. region increases, resulting in greater repulsion between the l.p.-l.p. & b.p-l.p.. The b.p. are pushed closer to one another, bond angle decrease.
IV. Electronegativity of central atom
The greater the electronegativity of a central atom, the closer the b.p. are towards the central atom, increasing b.p.-b.p. repulsion, hence bond angle increases.
d) Describe covalent bonding in terms of orbital overlap, giving σ and π bonds
A sigma bond is formed from the head-on overlapping of orbitals.
A Pi bond is formed from the side-on overlapping of p orbitals.
e) Predict the shapes and bond angles of molecules analogous to those specified in c).
1) Identify the central atom.
2) Determine the no. of electrons surrounding the central atom.
· For molecules with no charge,
No. of e- surrounding central atom = No. of valence e- of central atom + No. of surrounding atoms
· For charged particles,
No. surrounding atom = No. of valence e- of central atom ± magnitude of charge + No. of surrounding atoms
3) Determine the no. of electron pairs.
· No. of electron pairs = ½ x No. of e- surrounding central atom
4) Determine no. of b.p. electron and l.p. electron.
· No. of b.p. = No. of atoms around the central atoms
· No. of l.p. = No. of electron pairs – No. of b.p.
5) Identify shape of the molecule using the table below.
No. of Electron pairs
No. of Bond pairs
No. of Lone pairs
Geometrical shape
Bond angle
2
2
0
Linear
180’
3
3
0
Trigonal planar
equal or near 120’
2
1
V-shape/bent
4
4
0
Tetrahedral
equal or near 108’
3
1
Trigonal pyramidal
2
2
V-shape/bent
5
5
0
Trigonal bipyramidal
90’, 120’
4
1
Distorted tetrahedron
-
3
2
T-shape
90’
2
3
Linear
180’
6
6
0
Octahedral
90’
5
1
Square pyramidal
4
2
Square planar
i) covalent bonding in H2, O2, Cl2, HCl, CO2, CH4, C2H6
Covalent bond is formed from sharing of electrons between atoms. A covalent bond is the electrostatic force of attraction between the nuclei of 2 atoms for the shared pair of electron.
Molecule
‘dot-and-cross’ diagram
Description(to achieve stability)
H2
2 hydrogen atoms shared 1 pair of e- in the valence shell
O2
2 oxygen atoms shared 2 pairs of e- in the valence shell
Cl2
2 chlorine atoms shared 1 pair of e- in the valence shell
HCl
1 hydrogen & 1 chlorine atoms shared 1 pair of e- in the valence shell
CO2
1 carbon & 2 oxygen atoms shared 2 pairs of e- in the valence shell
CH4
1 carbon & 4 hydrogen atoms shared 4 pairs of e- in the valence shell
C2H6
2 carbon & 6 hydrogen atoms shared 7 pair of e- in the valence shell
ii) dative bond, in BF3->NH3
Dative bond is a type of covalent bond where the shared paired of electrons comes from one donor only. The donor must possess 1 lone pair of electron in its valence shell. The acceptor must have an empty orbital in its outer quantum shell to accommodate the lone pair.
NH3 + BF3----à BF3->NH3
NH3’s 1 lone pair donates to BF3. Now both N and B have 8e- in valence shell.
c) Explain shapes and bond angles of molecules by, using qualitative model of electron pair repulsion, using as simple examples: BF3 (trigonal), CO2 (linear), CH4 (tetrahedral), NH3 (pyramidal), H2O(V-shape), SF6(octahedral), by using VSEPR Theory.
Valence Shell Electron Pair Repulsion Theory can be use to predict the shape f simple covalent molecules.
I. These b.p & l.p. electrons around central atom determine the shapes of the molecule.
II. The electrons repel each other as far as possible.
III. Degree of repulsion
l.p.-l.p > l.p-b.p. > b.p.-b.p.
105’ 107’ 109’
Molecule
Structural formula
Kind of repulsion
Bond angle
Shape
Description
BF3
l.p.-l.p.
360’/3=120’
Trigonal planar
(because it is a triangle in 2D.)
B is in Grp III F is in Grp VII. There are 3F atoms. The strength of repulsion is equal because all are l.p.
CO2
l.p.-l.p.
360’/2=180’
Linear (because it is in a straight line)
There is repulsion between 2 O atoms, so the repulsion is the largest.
CH4
b.p-b.p.
109’
Tetrahedral (because there are 4 faces)
There is repulsion between H atoms.
NH3
l.p.-b.p.
107’
Trigonal pyramidal (because it is a triangle in 3D.
There is repulsion between N and H atom. So the angle will be pushed smaller.
H2O
l.p.-l.p.
105’
V-shape
There is repulsion causing the H atoms to be pushed closer to each other.
SF6
b.p-b.p
90’
Octahedral (because it has two 4 sided pyramid)
This is a molecule with more than 8e-.
Such case can be seen in period 3 and 4. There is no l.p. repulsion, so there is equal bond angle.
Factors affecting Bond Size:
I. Total number of e- pairs.
The more the number of e- pairs the central atom has, the smaller the bond angle.
II. Number of lone pairs.
L.p. electrons cause more repulsion than the b.p. electrons. Bond pairs are pushed closer to one another.
III. Size of central atom.
As size of central atom increases, the l.p. region increases, resulting in greater repulsion between the l.p.-l.p. & b.p-l.p.. The b.p. are pushed closer to one another, bond angle decrease.
IV. Electronegativity of central atom
The greater the electronegativity of a central atom, the closer the b.p. are towards the central atom, increasing b.p.-b.p. repulsion, hence bond angle increases.
d) Describe covalent bonding in terms of orbital overlap, giving σ and π bonds
A sigma bond is formed from the head-on overlapping of orbitals.
A Pi bond is formed from the side-on overlapping of p orbitals.
e) Predict the shapes and bond angles of molecules analogous to those specified in c).
1) Identify the central atom.
2) Determine the no. of electrons surrounding the central atom.
· For molecules with no charge,
No. of e- surrounding central atom = No. of valence e- of central atom + No. of surrounding atoms
· For charged particles,
No. surrounding atom = No. of valence e- of central atom ± magnitude of charge + No. of surrounding atoms
3) Determine the no. of electron pairs.
· No. of electron pairs = ½ x No. of e- surrounding central atom
4) Determine no. of b.p. electron and l.p. electron.
· No. of b.p. = No. of atoms around the central atoms
· No. of l.p. = No. of electron pairs – No. of b.p.
5) Identify shape of the molecule using the table below.
No. of Electron pairs
No. of Bond pairs
No. of Lone pairs
Geometrical shape
Bond angle
2
2
0
Linear
180’
3
3
0
Trigonal planar
equal or near 120’
2
1
V-shape/bent
4
4
0
Tetrahedral
equal or near 108’
3
1
Trigonal pyramidal
2
2
V-shape/bent
5
5
0
Trigonal bipyramidal
90’, 120’
4
1
Distorted tetrahedron
-
3
2
T-shape
90’
2
3
Linear
180’
6
6
0
Octahedral
90’
5
1
Square pyramidal
4
2
Square planar
Chem periodicity part 1
a. Describe and explain the acid/base behaviour of oxides and hydroxides, including, where relevant, amphoteric behaviour in reaction with sodium hydroxide (only) and acids
Na2O, MgO, Al2O3
All have giant structures in which the bonding is ionic. Therefore they have high melting points and boiling points and conduct electricity when molten.
Na2O, MgO: basic oxide (soluble, pH= 13)
Al2O3: amphoteric oxide, insoluble in water
SiO2
Silicon is a metalloid and forms the oxide SiO2 which has a giant covalently bonded structure. It is therefore a solid with high melting point and boiling point and does not conduct electricity when molten. Acidic oxide and is insoluble in water.
SiO2(s) + 2NaOH(aq) à Na2SiO3(aq) + H2O(l)
P4O10 , SO2 , SO3 , Cl2O7
The remaining oxides are formed by non-metals. These oxides are built up from molecules within which the bonding is covalent, but between the molecules the bonding is the weak Van der Waals type. These oxides therefore have low melting points and boiling points and do not conduct electricity in the liquid state. Acidic oxide and gives acidic solution (pH = 2)
P4O6(g) + 12NaOH(aq) à 4Na3PO3(aq) + 6H2O(l)
P4O10(g) + 12NaOH(aq)à 4Na3PO4(aq) + 6H2O(l)
SO2(g) + 2NaOH (aq) à Na2SO3(aq) + H2O(l)
SO3(l) + 2NaOH (aq) à Na2SO4(aq) + H2O(l)
Cl2O(g) + 2NaOH(aq) à 2NaClO(aq) + H2O
Cl2O7(g) + 2NaOH(aq) à 2NaClO4(aq) + H2O(l)
describe and explain the reactions of the chlorides with water
Na: NaCl dissolves in water, forming aqueous ions ( hydration), neutral pH=7
Mg: the high charge density in Mg2+ polarizes water molecules, giving H+ (hydration and hydrolysis) slightly acidic, pH = 6.5. MgCl2 undergoes hydration to form Mg2+ and Cl- ions. Mg2+ has a high charge density as it has a high charge and small ionic radius. (Charge density = charge / size of atom or ion). Hence, the Mg2+ forms a attraction force with the negative end of water molecules (polarises the water molecules). This weakens the O-H bond of water molecules and releases H+.
Al: Due to the high charge density of Al3+ ,it tends to form strong bond with oxygen and weakens the O-H bond of water molecules in the [Al(H2O)6]3+ ion. AlCl3 reacts with water to first form [Al(H2O)6]3+ ions . This Al-containing ion has a high charge density. (Charge density = charge / size of atom or ion). Hence, this ion forms a strong attraction force with the negative end of water molecules. (polarises the water molecules). This weakens the O-H bond of water molecules and releases H+.
Si and P: Si & P has vacant d orbitals available for dative bonding with water. Hence SiCl4 PCl3 and PCl5 can be hydrolysed by water.
interpret the variations and trends in (b), (c), (d), and (e) in terms of bonding and electronegativity
Oxides of elements (Bonding)
• Na2O, MgO, Al2O3: Exist as ionic structure. A large amount of energy is required to break the strong electrostatic attraction forces between the ions in the lattice
• SiO2: Exist as giant covalent structure where Si and O atoms are held by covalent bonds. A lot of energy is required to break the strong covalent bonds.
• P4O6 , P4O10, SO2, SO3: Exist as simple discrete molecules with Van der Waals’ forces and/or permanent dipole- permanent dipole between the molecules. Lower energy required to overcome these intermolecular forces.
Oxides of elements (Eletronegativity)
• Na2O, MgO, Al2O3: In molten state, the ions are mobile and therefore can carry charges around freely. Hence these oxides are able to conduct electricity.
• SiO2, P4O6 , P4O10, SO2, SO3: In giant covalent structure or simple discrete molecules, there are no mobile ions or electrons in all states. Hence they do not conduct electricity.
Reaction of elements with oxygen
4Na(s) + O2(g) -> 2Na2O(s)
On heating, burns in oxygen with a yellow flame to give a mixture of white sodium oxide and yellow sodium peroxide[2Na(s) + O2(g) ->Na2O2(s)]
2Mg(s) + O2(g) -> 2MgO(s)
On heating, burns in oxygen with brilliant white flame and a white magnesium oxide is produced.
4Al(s) + 3O2(g) -> 2Al2O3(s)
On heating, burns with white flame to give white aluminium oxide.
An oxide layer is formed to protect the metal against further oxidation under room temperature.
Si(s) + O2(g) -> SiO2(s)
On heating, burns to give white oxide. Does not undergo oxidation at room temperature.
Limited O2: P4(s) + 3O2(g)->P4O6(s)
Excess O2: P4(s) + 5O2(g)-> P4O10(s)
Red phosphorus: burns on heating. While white phosphorus burst into flame in oxygen to give a white oxide.
S(s) + O2(g) -> SO2(g)
On heating, burns with bright blue flame to give colourless gaseous oxide.
With Pt: 2SO2(g) + O2(g) -> 2SO3(g)
In the presence of a Pt catalyst, this oxide further reacts to give SO3.
Reaction of element with chlorine
2Na(s) +Cl2(g) ->NaCl(s)
On heating, reacts quite vigorously.
2Mg(s) +Cl2(g) -> MgCl2(s)
On heating, burns brightly.
3Al(s) +Cl2(g) -> Al2Cl6(s)
On heating under anhydrous conditions, forms solid chloride.
Si(s) +2Cl2(g) -> SiCl4(l)
On heating under anhydrous conditions, forms colourless covalent liquid.
P(s) +6Cl2(g) -> 4PCl3(l)
PCl3(l) + Cl2 (g)-> PCl5(s)
On heating, white phosphorus burns with pale green flame to give PCl3 and further reacts to give PCl5
Oxidation number in element of oxide and chloride
Na2O : +1 NaCl : +1 The variation between
MgO :+2 MgCl2 : +2 oxidation in element of oxide and
Al2O3 :+3 AlCl3 : +3 chloride in their oxidation number is in
SiO2 : +4 SiCl4 : +4accordance to their valance electron.As element Na Mg Al Si P AND S
P4O6 : +3 PCl3 : +3 are across the same period hence an electron is added to the previous
P4O10 : +5 PCl5 : +5 element electron number to be the next element electron number
SO2 : +4 (starting from Na electron number:11)
SO3 : +6
Reaction of element of oxide with water
Na2O + H2O -> 2NaOH
Basic oxides react with water to form alkaline solutions.
MgO + H2O -> Mg(OH)2
Basic oxides react with water to form alkaline solutions.
Al2O3 does not react with water due to high lattice energy.
SiO2 is insoluble in water due to its giant molecular structure.
P4O6 + 6H2O -> 4H3PO3
P4O10 + 6H2O -> 4H3PO4
Acidic oxides react with water to form acidic solutions.
SO2 + 2H2O -> H2SO3 + H2O
SO3 + 2H2O -> H2SO4 + H2O
Acidic oxides react with water to form acidic solutions.
Suggest types of chemical bonding present in chlorides and oxides from observations of their chemical and physical properties.
Elements- Na Mg Al Si P S Cl
O2 Na2O MgO Al2O3 SiO2 P4O10 SO2 /With ppt SO3
Cl2 NaCl MgCl2 Al2Cl6 SiCl4 PCL3/PCL5
Elements- Na2O MgO Al2O3 SiO2 P4O6 P4O10 SO2 SO3
Chem Bond ----------Giant Ionic------------- ---Giant Covalent----- ---------Simple Covalent----------------------
Elements- NaCl MgCl2 Al2Cl6 SiCl4 PCl3 PCL5
Chem Bond -----Ionic Bond----- -------------------Simple Covalent----------------------
Predict characteristic properties of an element in a given group by using knowledge of chemical periodicity.
Group 1 2 3 4 5 6 7 8
Elements- Na Mg Al Si P S Cl Ar
Structure -------Giant Metallic------ Giant Covalent --------------Simple Molecular------------
Bonding ------Metallic Bond------- ------Covalent---- ---------Van Der Waal's Forces----------
Mpt Pt. -----------HIGH-------------- ----- HIGH------- --------------------LOW------------------------
Electrical -----------GOOD------------ ------LOW-------- ---------NON-CONDUCTOR---------------
Electronegativity ------------Increases across a period & decreases down a group---------------------------
Deduce the nature, possible position in the Periodic Table and identity of unknown elements from given information of physical chemical properties.
Group 1 2 3 4 5 6 7 8
Elements- Na Mg Al Si P S Cl Ar
Physical ------------Metal------------ --Metalloid--- --------------Non-Metals--------------
Mpt Pt ---------HIGH--------------- -----HIGH----- ------------------LOW-----------------
Bonding ----Metallic Bond--------- ---Covalent-- ---------Van Der Waal's Forces---
Structure -----Giant Metallic-------- Giant Covalent -------Simple Molecular------------
Electrical ---------GOOD------------ -----LOW------- ----------NON-Conductor-----------
Na2O, MgO, Al2O3
All have giant structures in which the bonding is ionic. Therefore they have high melting points and boiling points and conduct electricity when molten.
Na2O, MgO: basic oxide (soluble, pH= 13)
Al2O3: amphoteric oxide, insoluble in water
SiO2
Silicon is a metalloid and forms the oxide SiO2 which has a giant covalently bonded structure. It is therefore a solid with high melting point and boiling point and does not conduct electricity when molten. Acidic oxide and is insoluble in water.
SiO2(s) + 2NaOH(aq) à Na2SiO3(aq) + H2O(l)
P4O10 , SO2 , SO3 , Cl2O7
The remaining oxides are formed by non-metals. These oxides are built up from molecules within which the bonding is covalent, but between the molecules the bonding is the weak Van der Waals type. These oxides therefore have low melting points and boiling points and do not conduct electricity in the liquid state. Acidic oxide and gives acidic solution (pH = 2)
P4O6(g) + 12NaOH(aq) à 4Na3PO3(aq) + 6H2O(l)
P4O10(g) + 12NaOH(aq)à 4Na3PO4(aq) + 6H2O(l)
SO2(g) + 2NaOH (aq) à Na2SO3(aq) + H2O(l)
SO3(l) + 2NaOH (aq) à Na2SO4(aq) + H2O(l)
Cl2O(g) + 2NaOH(aq) à 2NaClO(aq) + H2O
Cl2O7(g) + 2NaOH(aq) à 2NaClO4(aq) + H2O(l)
describe and explain the reactions of the chlorides with water
Na: NaCl dissolves in water, forming aqueous ions ( hydration), neutral pH=7
Mg: the high charge density in Mg2+ polarizes water molecules, giving H+ (hydration and hydrolysis) slightly acidic, pH = 6.5. MgCl2 undergoes hydration to form Mg2+ and Cl- ions. Mg2+ has a high charge density as it has a high charge and small ionic radius. (Charge density = charge / size of atom or ion). Hence, the Mg2+ forms a attraction force with the negative end of water molecules (polarises the water molecules). This weakens the O-H bond of water molecules and releases H+.
Al: Due to the high charge density of Al3+ ,it tends to form strong bond with oxygen and weakens the O-H bond of water molecules in the [Al(H2O)6]3+ ion. AlCl3 reacts with water to first form [Al(H2O)6]3+ ions . This Al-containing ion has a high charge density. (Charge density = charge / size of atom or ion). Hence, this ion forms a strong attraction force with the negative end of water molecules. (polarises the water molecules). This weakens the O-H bond of water molecules and releases H+.
Si and P: Si & P has vacant d orbitals available for dative bonding with water. Hence SiCl4 PCl3 and PCl5 can be hydrolysed by water.
interpret the variations and trends in (b), (c), (d), and (e) in terms of bonding and electronegativity
Oxides of elements (Bonding)
• Na2O, MgO, Al2O3: Exist as ionic structure. A large amount of energy is required to break the strong electrostatic attraction forces between the ions in the lattice
• SiO2: Exist as giant covalent structure where Si and O atoms are held by covalent bonds. A lot of energy is required to break the strong covalent bonds.
• P4O6 , P4O10, SO2, SO3: Exist as simple discrete molecules with Van der Waals’ forces and/or permanent dipole- permanent dipole between the molecules. Lower energy required to overcome these intermolecular forces.
Oxides of elements (Eletronegativity)
• Na2O, MgO, Al2O3: In molten state, the ions are mobile and therefore can carry charges around freely. Hence these oxides are able to conduct electricity.
• SiO2, P4O6 , P4O10, SO2, SO3: In giant covalent structure or simple discrete molecules, there are no mobile ions or electrons in all states. Hence they do not conduct electricity.
Reaction of elements with oxygen
4Na(s) + O2(g) -> 2Na2O(s)
On heating, burns in oxygen with a yellow flame to give a mixture of white sodium oxide and yellow sodium peroxide[2Na(s) + O2(g) ->Na2O2(s)]
2Mg(s) + O2(g) -> 2MgO(s)
On heating, burns in oxygen with brilliant white flame and a white magnesium oxide is produced.
4Al(s) + 3O2(g) -> 2Al2O3(s)
On heating, burns with white flame to give white aluminium oxide.
An oxide layer is formed to protect the metal against further oxidation under room temperature.
Si(s) + O2(g) -> SiO2(s)
On heating, burns to give white oxide. Does not undergo oxidation at room temperature.
Limited O2: P4(s) + 3O2(g)->P4O6(s)
Excess O2: P4(s) + 5O2(g)-> P4O10(s)
Red phosphorus: burns on heating. While white phosphorus burst into flame in oxygen to give a white oxide.
S(s) + O2(g) -> SO2(g)
On heating, burns with bright blue flame to give colourless gaseous oxide.
With Pt: 2SO2(g) + O2(g) -> 2SO3(g)
In the presence of a Pt catalyst, this oxide further reacts to give SO3.
Reaction of element with chlorine
2Na(s) +Cl2(g) ->NaCl(s)
On heating, reacts quite vigorously.
2Mg(s) +Cl2(g) -> MgCl2(s)
On heating, burns brightly.
3Al(s) +Cl2(g) -> Al2Cl6(s)
On heating under anhydrous conditions, forms solid chloride.
Si(s) +2Cl2(g) -> SiCl4(l)
On heating under anhydrous conditions, forms colourless covalent liquid.
P(s) +6Cl2(g) -> 4PCl3(l)
PCl3(l) + Cl2 (g)-> PCl5(s)
On heating, white phosphorus burns with pale green flame to give PCl3 and further reacts to give PCl5
Oxidation number in element of oxide and chloride
Na2O : +1 NaCl : +1 The variation between
MgO :+2 MgCl2 : +2 oxidation in element of oxide and
Al2O3 :+3 AlCl3 : +3 chloride in their oxidation number is in
SiO2 : +4 SiCl4 : +4accordance to their valance electron.As element Na Mg Al Si P AND S
P4O6 : +3 PCl3 : +3 are across the same period hence an electron is added to the previous
P4O10 : +5 PCl5 : +5 element electron number to be the next element electron number
SO2 : +4 (starting from Na electron number:11)
SO3 : +6
Reaction of element of oxide with water
Na2O + H2O -> 2NaOH
Basic oxides react with water to form alkaline solutions.
MgO + H2O -> Mg(OH)2
Basic oxides react with water to form alkaline solutions.
Al2O3 does not react with water due to high lattice energy.
SiO2 is insoluble in water due to its giant molecular structure.
P4O6 + 6H2O -> 4H3PO3
P4O10 + 6H2O -> 4H3PO4
Acidic oxides react with water to form acidic solutions.
SO2 + 2H2O -> H2SO3 + H2O
SO3 + 2H2O -> H2SO4 + H2O
Acidic oxides react with water to form acidic solutions.
Suggest types of chemical bonding present in chlorides and oxides from observations of their chemical and physical properties.
Elements- Na Mg Al Si P S Cl
O2 Na2O MgO Al2O3 SiO2 P4O10 SO2 /With ppt SO3
Cl2 NaCl MgCl2 Al2Cl6 SiCl4 PCL3/PCL5
Elements- Na2O MgO Al2O3 SiO2 P4O6 P4O10 SO2 SO3
Chem Bond ----------Giant Ionic------------- ---Giant Covalent----- ---------Simple Covalent----------------------
Elements- NaCl MgCl2 Al2Cl6 SiCl4 PCl3 PCL5
Chem Bond -----Ionic Bond----- -------------------Simple Covalent----------------------
Predict characteristic properties of an element in a given group by using knowledge of chemical periodicity.
Group 1 2 3 4 5 6 7 8
Elements- Na Mg Al Si P S Cl Ar
Structure -------Giant Metallic------ Giant Covalent --------------Simple Molecular------------
Bonding ------Metallic Bond------- ------Covalent---- ---------Van Der Waal's Forces----------
Mpt Pt. -----------HIGH-------------- ----- HIGH------- --------------------LOW------------------------
Electrical -----------GOOD------------ ------LOW-------- ---------NON-CONDUCTOR---------------
Electronegativity ------------Increases across a period & decreases down a group---------------------------
Deduce the nature, possible position in the Periodic Table and identity of unknown elements from given information of physical chemical properties.
Group 1 2 3 4 5 6 7 8
Elements- Na Mg Al Si P S Cl Ar
Physical ------------Metal------------ --Metalloid--- --------------Non-Metals--------------
Mpt Pt ---------HIGH--------------- -----HIGH----- ------------------LOW-----------------
Bonding ----Metallic Bond--------- ---Covalent-- ---------Van Der Waal's Forces---
Structure -----Giant Metallic-------- Giant Covalent -------Simple Molecular------------
Electrical ---------GOOD------------ -----LOW------- ----------NON-Conductor-----------
Chemical Energetics
Gaseous State
(a) state the basic assumptions of the kinetic theory as applied to an ideal gas
- gas consists of particles or molecules of negligible size (or volume)
- gas particles have negligible intermolecular forces of attraction
- gas particles are in continuous random motion
- collisions between molecules are perfectly elastic, ie the gas molecules bounce apart on collision with no loss in kinetic energy; the molecules do not stick together
(bi) explain qualitatively in terms of intermolecular forces and molecular size: the conditions necessary for a gas to approach ideal behaviour
1. at low pressures
- gas molecules are widely spaced, and therefore have negligible size (vol occupied by the gas is very large compared to the gas molecule)
- forces of attraction between gas molecules are zero
2. at high temperatures
- there are negligible intermolecular attractions since the gas particles have sufficient kinetic energy to overcome it
(bii) explain qualitatively in terms of intermolecular forces and molecular size: the limitations of ideality at very high pressures and very low temperatures
1. at high pressures
- gas molecules are packed close together, and the size of a gas molecule cannot be assumed to be negligible
- shown by increase in pV/RT value
2. at low temperatures
- force of attraction between gas molecules are significant
- as shown by the fall in pV/RT value
Formula:
pV=nRT
where p = pressure in Pa
V = volume in m3
n = no. of moles in gas
R = gas constant (8.31 JK-1 mol -1)
T = temperature in K
The enthalpy change of a reaction, ▲H, is defined as the heat change (heat energy absorbed or evolved) when the reaction takes place between the masses of the reagents indicated by the stoichiometric equation for the reaction..
An exothermic reaction gives out heat to the surrounding, ie heat energy is evolved. Hence, the surrounding temperature rises as the heat content of the system falls.
▲H is negative (< 0) since heat content of products < heat content of reactants
An endothermic reaction absorbs heat from the surrounding. Hence, the surrounding temperature falls as the heat content of the system rises.
▲H is positive (> 0) since heat content of products > heat content of reactants
Types of Enthalpy change
1. The standard enthalpy change of formation, ΔHof of a compound is defined as the enthalpy change when one mole of the compound is formed from its elements under standard conditions (at 25°C, 1 atm).
e.g. H2(g) + ½ O2(g) à H2O (l) ΔHof (H2O (l)) = -286 kJmol-1
? ΔHof are usually calculated indirectly from other enthalpy changes of reaction.
? ΔHof elements in its standard state is zero. E.g. ΔHof ((N2(g))) = 0.
? ΔHof is often used to predict the stability of a compound relative to its constituent elements.
If ΔHof < 0, compound is energetically more stable than its constituent elements.
If ΔHof > 0, compound is energetically more stable than its constituent elements.
2. The standard enthalpy change of combustion, ΔHoc , is defined as the enthalpy change when one mole of a substance is completely burnt in oxygen under standard conditions (at 25°C, 1 atm).
e.g. S(s) + O2(g) à SO2(g) ΔHoc (S(s)) = -297 kJmol-1
? ΔHoc is always negative, as heat is always evolved in the combustion.
? ΔHoc can be used to give the energy values of fuels and foods.
3. The standard enthalpy change of hydration, DHөhyd, of an ion is defined as the enthalpy change when one mole of the gaseous ions is dissolved in a large amount of water under standard conditions (at 25°C, 1 atm).
e.g. Na+(g) + aq à Na+(aq) DHөhyd = -406 kJmol-1
? DHөhyd is always negative, as heat is produced when bonds are formed between the ions and the dipoles on the water molecules.
? The hydration energies of the ions depend on the charge and size of the ions
The higher the charge and the smaller the size of the ions, the greater (i.e. more exothermic) is the hydration energy.
4. The standard enthalpy change of solution, DHөsoln , is defined as the enthalpy change when one mole of a substance dissolves in such a large volume of solvent that addition of more solvent produces no further heat change under standard conditions (at 25°C, 1 atm).
e.g. NH3(g) + aq à NH3 (aq) DHөsoln = -35.2kJmol-1
? DHөsoln can be positive or negative.
If DHөsoln is very positive, compound is insoluble in water.
If DHөsoln is negative, compound is soluble in water.
5. The standard enthalpy change of atomization, DHөatom, is defined as the enthalpy change when an element or a compound is converted into one mole of free gaseous atoms under standard conditions (at 25°C, 1 atm).
e.g. Na(s) à Na(g) DHөatom = +109 kJmol-1
? DHөatom is always positive, because energy must be absorbed to pull the atoms far apart and to break all the bonds between them/
? The enthalpy change of atomization is not the same as the enthalpy change of vapourisation.
6. The standard enthalpy change of neutralization, DHөneut , is defined as the enthalpy change when one mole of water is formed in the neutralization between an acid and an alkali, the reaction being carried out in aqueous solution under standard conditions (at 25°C, 1 atm).
e.g. HCl(aq) + NaOH(aq) à NaCl(s) + H2O(l) DHөneut = -57.1 kJmol-1
? DHөneut is always negative.
? For the neutralisation of strong acids with a strong base, the standard enthalpy of neutralisation is almost constant. (DHөneut = -57.3 kJmol-1 )
Ø Strong acids and strong bases are completely ionised in dilute solutions
HX(aq) à H+(aq) + X-(aq)
MOH(aq) à M+(aq) + OH-(aq)
Ø Thus, the reaction between any strong acid and any strong base involves simply the formation of water from H+ and OH- ions.
H+(aq) + OH-(aq) à H2O(l) (DHөneut = -57.3 kJmol-1 )
Ø Since the process is exactly the same for all, the enthalpy change of neutralisation must, therefore, be the same (and the value is -57.3 kJmol-1 )
? If a weak acid or weak base is used, or if both acid and base are weak, then the standard enthalpy of neutralisation differs significantly from -57.3 kJmol-1 .
CH3CO2H(aq) + NaOH (aq) à Ch3CO2Na(aq) + H2O(l) (DHөneut = -55.2 kJmol-1 )
Ø Weak acids and weak bases are only slightly ionised in aqueous solution.
Ø So, with weak acids or bases, neutralisation involves the enthalpy change due to the reaction
H+(aq) + OH-(aq) à H2O(l) (DH1 = -57.3 kJmol-1 )
(DH2 = ±)and an enthalpy change due to the unionised acid or base that has to be converted into ions (DH here may be positive or negative).
HA(aq) à H+(aq) + A-(aq)
BOH(aq) à B+(aq) +OH-(aq)
Ø Thus, the enthalpy change of neutralisation involving either weak acids or weak bases (DHөneut = DH1 + DH2)
? DHөneut can be determined by mixing solutions of acids and alkalis in a calorimeter, and measuring the rise in temperature.
Heat evolved
No. of moles of water formedDHөneut =
where m = mass of solution
c = specific heat capacity of solution
DT = change in temperature
Heat evolved = mc DT
and
7. Bond Dissociation Energy, B.E, is the energy required to break one mole of covalent bonds between 2 atoms in the gaseous state.
e.g. O=O(g) à 2O(g) B.E. = +497 kJmol-1
? Bond dissociation energy can be used to compare the strength of covalent bonds.
? The greater the bond energy, the stronger the bond.
8. Ionisation Energy, I.E. is the energy required to remove one mole of electron from one mole of atoms or ions in the gaseous state.
? First I.E. of an element is the energy required to remove one mole of electrons from one mole of atom of the element in gaseous state.
M(g) à M+(g) + e-
? Second I.E. of an element is the energy required to remove one mole of electron from one mole of singly-charged positive ion of the element in the gaseous state.
M+(g) à M2+ (g) + e-
9. Electron Affinity, E.A. is the enthalpy change when one mole of atom or negatively charged ion in gaseous state gains one mole of electron.
? First E.A. of an element is the energy required for one mole of gaseous atom acquires one mole of electron to form one mole of singly-charged negative ion.
A(g) + e à A-(g)
10. Lattice Energy is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions of 1atm and 298K.
q+ q-
r+ + r-M+(g) + A-(g) à MA (s)
? Magnitude of lattice energy
Where q+ : charge of cation & q- : charge of anion
r+ : radius of cation & r- : radius of anion
? Lattice energy cannot be measured directly. However, it can be determined using Hess’ Law
11. Hess’ Law states that the enthalpy change for a chemical reaction is the same, regardless of the route taken, provided that the initial states of the reactants and the final states of the products are the same.
Hess’s Law is used to determine enthalpy changes that cannot be found by direct experiments, through the construction of an energy cycle or an energy level diagram
(g)Entropy: ▲ (S) The measure of the disorder/ randomness in a system.
Any system in random motion tends to become more “mixed up” or disorderly as time passes. In an isolated environment, nature tends towards maximum entropy.
(h)Change in Temperature: Entropy increases as temperature increases.
E.g: H2O (l) at 25˚c à H2O (l) at 35˚c
Temperature increases from 25 - 35˚c.
K.E of particles increases resulting in higher movement rate of particles.
System becomes more disordered. ▲ (S) increases
As temperature increases, the molecules or ions undergo greater vibration in solid) and more rapid motion (in liquids and gases) and this reduces their orderliness.
(h)Change in Phase: Gas >> liquid >> solid
E.g: H2O (s) at 0˚c à H2O (l) at 0˚c
Change in phase
Crystalline structure of solid is broken, thus H20 molecules now more around freely in liquid state.
System becomes more disordered. ▲ (S) increases
A solid has low entropy due to its crystalline structure which is highly ordered and regular.
The entropy of a liquid is greater than that of a solid as molecules or ions in liquid state display less order than in solid state. Because particles/ions in liquid state move around more freely.
Entropy of gas is much greater than liquid because gases has a higher disorder due to the particles of gas having free movement and are not constrained to be adjacent to each other. Entropy of gas is far greater than that of solid.
(h)Change in number of particles: Entropy increases as the no. of particles increases.
E.g: CL2 (g)à 2CL(g)
Increase in number of particles
1mol of CL2 changes to 2mols of CL which contains more particles.
More ways to arrange CL particles
System becomes more disordered. ▲ (S) Increases.
Entropy increases as the no. of particles increases and the system becomes less orderly.
(h)Mixing of particles: Mixing process leads to disorder and so entropy increases.
E.g: NaCl (s) à Na (aq) + 2CL (aq)
Mixing of particles
Change in phase solid to liquid
Change in no. of particles
Entropy increases when two pure gases are mixed and allowed to diffuse into each other. Their orderliness is reduced as the molecules become randomly mixed
Entropy increases when a solid dissolves into a liquid. The resulting solution has less order than the original crystal because the particles are scattered throughout the solution and mix homogeneously with the solvent.
(i)The change in entropy (S) is given by the expression: ▲S = S(final) – S(initial)
▲S is positive when the system becomes less orderly S(final) > S(initial). This occurs when there is a change of state, or when a gas is produced by the decomposition of a liquid or solid.
▲S is negative when the system becomes more orderly S(final) < S(initial).
E.g: Ar at 2 atm à Ar at 1 atm .
▲S is positive because when pressure is decreased, Ar atoms are free to move in a larger volume and the system becomes less orderly.
(J)Every chemical reaction is accompanied by a heat (or energy) change ▲H, and a redistribution of matter, ▲S. The combined effects of these two factors are expressed by the quantity free energy, G.
Standard Gibbs free energy, ▲G is a state function of a system and is defined by means of the equation ▲G = ▲H - ▲S where T = temperature in K.
The sign of ▲G may be used to deduce whether a reaction or process will be spontaneous.
(K)
· If ▲G is negative (▲G < 0) the reaction is feasible and could take place spontaneously. The reaction is said to be exergonic or energy-giving.
· If ▲G is positive (▲G >0) the reaction is said to be not feasible and cannot take place spontaneously. The reaction is said to be endergonic or energy-giving.
· If ▲G is zero (▲G =0) the reaction is at equilibrium.
(L) When ▲G is positive, the reaction is non-spontaneous at r.t.p.
When ▲G is negative, the reaction is spontaneous at r.t.p.
(M)From the equation ▲G = H - T▲S the value of ▲G is dependent on temperature.
Hence ▲G may be negative when:
▲H - , ▲S+
Exothermic reaction, increase in entropy. Spontaneous at all temperatures.
▲H-, ▲S- :
Exothermic reaction, decrease in entropy as long as ▲H > T▲S. spontaneous at low temperatures.
▲H +, ▲S+:
Endothermic reaction, large increase in entropy where T▲S > ▲H. spontaneous only at high temperature.
▲H=0, ▲S+:
Reduction in Gibbs free energy arises completely from the increase in entropy.
▲G is positive when ▲H >0 (endothermic) and ▲S < 0 (decrease in entropy), the reaction is non-spontaneous at all temperature and has to be driven.
(a) state the basic assumptions of the kinetic theory as applied to an ideal gas
- gas consists of particles or molecules of negligible size (or volume)
- gas particles have negligible intermolecular forces of attraction
- gas particles are in continuous random motion
- collisions between molecules are perfectly elastic, ie the gas molecules bounce apart on collision with no loss in kinetic energy; the molecules do not stick together
(bi) explain qualitatively in terms of intermolecular forces and molecular size: the conditions necessary for a gas to approach ideal behaviour
1. at low pressures
- gas molecules are widely spaced, and therefore have negligible size (vol occupied by the gas is very large compared to the gas molecule)
- forces of attraction between gas molecules are zero
2. at high temperatures
- there are negligible intermolecular attractions since the gas particles have sufficient kinetic energy to overcome it
(bii) explain qualitatively in terms of intermolecular forces and molecular size: the limitations of ideality at very high pressures and very low temperatures
1. at high pressures
- gas molecules are packed close together, and the size of a gas molecule cannot be assumed to be negligible
- shown by increase in pV/RT value
2. at low temperatures
- force of attraction between gas molecules are significant
- as shown by the fall in pV/RT value
Formula:
pV=nRT
where p = pressure in Pa
V = volume in m3
n = no. of moles in gas
R = gas constant (8.31 JK-1 mol -1)
T = temperature in K
The enthalpy change of a reaction, ▲H, is defined as the heat change (heat energy absorbed or evolved) when the reaction takes place between the masses of the reagents indicated by the stoichiometric equation for the reaction..
An exothermic reaction gives out heat to the surrounding, ie heat energy is evolved. Hence, the surrounding temperature rises as the heat content of the system falls.
▲H is negative (< 0) since heat content of products < heat content of reactants
An endothermic reaction absorbs heat from the surrounding. Hence, the surrounding temperature falls as the heat content of the system rises.
▲H is positive (> 0) since heat content of products > heat content of reactants
Types of Enthalpy change
1. The standard enthalpy change of formation, ΔHof of a compound is defined as the enthalpy change when one mole of the compound is formed from its elements under standard conditions (at 25°C, 1 atm).
e.g. H2(g) + ½ O2(g) à H2O (l) ΔHof (H2O (l)) = -286 kJmol-1
? ΔHof are usually calculated indirectly from other enthalpy changes of reaction.
? ΔHof elements in its standard state is zero. E.g. ΔHof ((N2(g))) = 0.
? ΔHof is often used to predict the stability of a compound relative to its constituent elements.
If ΔHof < 0, compound is energetically more stable than its constituent elements.
If ΔHof > 0, compound is energetically more stable than its constituent elements.
2. The standard enthalpy change of combustion, ΔHoc , is defined as the enthalpy change when one mole of a substance is completely burnt in oxygen under standard conditions (at 25°C, 1 atm).
e.g. S(s) + O2(g) à SO2(g) ΔHoc (S(s)) = -297 kJmol-1
? ΔHoc is always negative, as heat is always evolved in the combustion.
? ΔHoc can be used to give the energy values of fuels and foods.
3. The standard enthalpy change of hydration, DHөhyd, of an ion is defined as the enthalpy change when one mole of the gaseous ions is dissolved in a large amount of water under standard conditions (at 25°C, 1 atm).
e.g. Na+(g) + aq à Na+(aq) DHөhyd = -406 kJmol-1
? DHөhyd is always negative, as heat is produced when bonds are formed between the ions and the dipoles on the water molecules.
? The hydration energies of the ions depend on the charge and size of the ions
The higher the charge and the smaller the size of the ions, the greater (i.e. more exothermic) is the hydration energy.
4. The standard enthalpy change of solution, DHөsoln , is defined as the enthalpy change when one mole of a substance dissolves in such a large volume of solvent that addition of more solvent produces no further heat change under standard conditions (at 25°C, 1 atm).
e.g. NH3(g) + aq à NH3 (aq) DHөsoln = -35.2kJmol-1
? DHөsoln can be positive or negative.
If DHөsoln is very positive, compound is insoluble in water.
If DHөsoln is negative, compound is soluble in water.
5. The standard enthalpy change of atomization, DHөatom, is defined as the enthalpy change when an element or a compound is converted into one mole of free gaseous atoms under standard conditions (at 25°C, 1 atm).
e.g. Na(s) à Na(g) DHөatom = +109 kJmol-1
? DHөatom is always positive, because energy must be absorbed to pull the atoms far apart and to break all the bonds between them/
? The enthalpy change of atomization is not the same as the enthalpy change of vapourisation.
6. The standard enthalpy change of neutralization, DHөneut , is defined as the enthalpy change when one mole of water is formed in the neutralization between an acid and an alkali, the reaction being carried out in aqueous solution under standard conditions (at 25°C, 1 atm).
e.g. HCl(aq) + NaOH(aq) à NaCl(s) + H2O(l) DHөneut = -57.1 kJmol-1
? DHөneut is always negative.
? For the neutralisation of strong acids with a strong base, the standard enthalpy of neutralisation is almost constant. (DHөneut = -57.3 kJmol-1 )
Ø Strong acids and strong bases are completely ionised in dilute solutions
HX(aq) à H+(aq) + X-(aq)
MOH(aq) à M+(aq) + OH-(aq)
Ø Thus, the reaction between any strong acid and any strong base involves simply the formation of water from H+ and OH- ions.
H+(aq) + OH-(aq) à H2O(l) (DHөneut = -57.3 kJmol-1 )
Ø Since the process is exactly the same for all, the enthalpy change of neutralisation must, therefore, be the same (and the value is -57.3 kJmol-1 )
? If a weak acid or weak base is used, or if both acid and base are weak, then the standard enthalpy of neutralisation differs significantly from -57.3 kJmol-1 .
CH3CO2H(aq) + NaOH (aq) à Ch3CO2Na(aq) + H2O(l) (DHөneut = -55.2 kJmol-1 )
Ø Weak acids and weak bases are only slightly ionised in aqueous solution.
Ø So, with weak acids or bases, neutralisation involves the enthalpy change due to the reaction
H+(aq) + OH-(aq) à H2O(l) (DH1 = -57.3 kJmol-1 )
(DH2 = ±)and an enthalpy change due to the unionised acid or base that has to be converted into ions (DH here may be positive or negative).
HA(aq) à H+(aq) + A-(aq)
BOH(aq) à B+(aq) +OH-(aq)
Ø Thus, the enthalpy change of neutralisation involving either weak acids or weak bases (DHөneut = DH1 + DH2)
? DHөneut can be determined by mixing solutions of acids and alkalis in a calorimeter, and measuring the rise in temperature.
Heat evolved
No. of moles of water formedDHөneut =
where m = mass of solution
c = specific heat capacity of solution
DT = change in temperature
Heat evolved = mc DT
and
7. Bond Dissociation Energy, B.E, is the energy required to break one mole of covalent bonds between 2 atoms in the gaseous state.
e.g. O=O(g) à 2O(g) B.E. = +497 kJmol-1
? Bond dissociation energy can be used to compare the strength of covalent bonds.
? The greater the bond energy, the stronger the bond.
8. Ionisation Energy, I.E. is the energy required to remove one mole of electron from one mole of atoms or ions in the gaseous state.
? First I.E. of an element is the energy required to remove one mole of electrons from one mole of atom of the element in gaseous state.
M(g) à M+(g) + e-
? Second I.E. of an element is the energy required to remove one mole of electron from one mole of singly-charged positive ion of the element in the gaseous state.
M+(g) à M2+ (g) + e-
9. Electron Affinity, E.A. is the enthalpy change when one mole of atom or negatively charged ion in gaseous state gains one mole of electron.
? First E.A. of an element is the energy required for one mole of gaseous atom acquires one mole of electron to form one mole of singly-charged negative ion.
A(g) + e à A-(g)
10. Lattice Energy is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions of 1atm and 298K.
q+ q-
r+ + r-M+(g) + A-(g) à MA (s)
? Magnitude of lattice energy
Where q+ : charge of cation & q- : charge of anion
r+ : radius of cation & r- : radius of anion
? Lattice energy cannot be measured directly. However, it can be determined using Hess’ Law
11. Hess’ Law states that the enthalpy change for a chemical reaction is the same, regardless of the route taken, provided that the initial states of the reactants and the final states of the products are the same.
Hess’s Law is used to determine enthalpy changes that cannot be found by direct experiments, through the construction of an energy cycle or an energy level diagram
(g)Entropy: ▲ (S) The measure of the disorder/ randomness in a system.
Any system in random motion tends to become more “mixed up” or disorderly as time passes. In an isolated environment, nature tends towards maximum entropy.
(h)Change in Temperature: Entropy increases as temperature increases.
E.g: H2O (l) at 25˚c à H2O (l) at 35˚c
Temperature increases from 25 - 35˚c.
K.E of particles increases resulting in higher movement rate of particles.
System becomes more disordered. ▲ (S) increases
As temperature increases, the molecules or ions undergo greater vibration in solid) and more rapid motion (in liquids and gases) and this reduces their orderliness.
(h)Change in Phase: Gas >> liquid >> solid
E.g: H2O (s) at 0˚c à H2O (l) at 0˚c
Change in phase
Crystalline structure of solid is broken, thus H20 molecules now more around freely in liquid state.
System becomes more disordered. ▲ (S) increases
A solid has low entropy due to its crystalline structure which is highly ordered and regular.
The entropy of a liquid is greater than that of a solid as molecules or ions in liquid state display less order than in solid state. Because particles/ions in liquid state move around more freely.
Entropy of gas is much greater than liquid because gases has a higher disorder due to the particles of gas having free movement and are not constrained to be adjacent to each other. Entropy of gas is far greater than that of solid.
(h)Change in number of particles: Entropy increases as the no. of particles increases.
E.g: CL2 (g)à 2CL(g)
Increase in number of particles
1mol of CL2 changes to 2mols of CL which contains more particles.
More ways to arrange CL particles
System becomes more disordered. ▲ (S) Increases.
Entropy increases as the no. of particles increases and the system becomes less orderly.
(h)Mixing of particles: Mixing process leads to disorder and so entropy increases.
E.g: NaCl (s) à Na (aq) + 2CL (aq)
Mixing of particles
Change in phase solid to liquid
Change in no. of particles
Entropy increases when two pure gases are mixed and allowed to diffuse into each other. Their orderliness is reduced as the molecules become randomly mixed
Entropy increases when a solid dissolves into a liquid. The resulting solution has less order than the original crystal because the particles are scattered throughout the solution and mix homogeneously with the solvent.
(i)The change in entropy (S) is given by the expression: ▲S = S(final) – S(initial)
▲S is positive when the system becomes less orderly S(final) > S(initial). This occurs when there is a change of state, or when a gas is produced by the decomposition of a liquid or solid.
▲S is negative when the system becomes more orderly S(final) < S(initial).
E.g: Ar at 2 atm à Ar at 1 atm .
▲S is positive because when pressure is decreased, Ar atoms are free to move in a larger volume and the system becomes less orderly.
(J)Every chemical reaction is accompanied by a heat (or energy) change ▲H, and a redistribution of matter, ▲S. The combined effects of these two factors are expressed by the quantity free energy, G.
Standard Gibbs free energy, ▲G is a state function of a system and is defined by means of the equation ▲G = ▲H - ▲S where T = temperature in K.
The sign of ▲G may be used to deduce whether a reaction or process will be spontaneous.
(K)
· If ▲G is negative (▲G < 0) the reaction is feasible and could take place spontaneously. The reaction is said to be exergonic or energy-giving.
· If ▲G is positive (▲G >0) the reaction is said to be not feasible and cannot take place spontaneously. The reaction is said to be endergonic or energy-giving.
· If ▲G is zero (▲G =0) the reaction is at equilibrium.
(L) When ▲G is positive, the reaction is non-spontaneous at r.t.p.
When ▲G is negative, the reaction is spontaneous at r.t.p.
(M)From the equation ▲G = H - T▲S the value of ▲G is dependent on temperature.
Hence ▲G may be negative when:
▲H - , ▲S+
Exothermic reaction, increase in entropy. Spontaneous at all temperatures.
▲H-, ▲S- :
Exothermic reaction, decrease in entropy as long as ▲H > T▲S. spontaneous at low temperatures.
▲H +, ▲S+:
Endothermic reaction, large increase in entropy where T▲S > ▲H. spontaneous only at high temperature.
▲H=0, ▲S+:
Reduction in Gibbs free energy arises completely from the increase in entropy.
▲G is positive when ▲H >0 (endothermic) and ▲S < 0 (decrease in entropy), the reaction is non-spontaneous at all temperature and has to be driven.
ATOMS, MOLECULES AND STOICHIOMETRY
ATOMS, MOLECULES AND STOICHIOMETRY
Relative Atomic Mass is defined as the mass of one atom of an element compared with 1/12 the mass of one atom of carbon-12.
Relative Isotopic Mass is defined as the mass of one atom of an isotope compared with 1/12 the mass of one atom of carbon- 12.
Relative Molecular Mass of a compound is the sum of the relative atomic masses of all the atoms in one molecule of a covalent compound.
Relative Formula Mass is the sum of all relative atomic masses from the formula of any given substance (atoms, molecules, ions, ionic compounds, etc.)
Avogadro Constant is a large mathematical constant defined as the number of carbon-12 atoms in 12g of carbon-12 (one mole).
Empirical Formula shows the simplest whole-number ratio for the atoms of different elements in a compound.
Molecular Formula shows the actual number of the different elements in one molecules of a compound.
ATOMIC STRUCTURE
Atoms are made up of protons, neutrons, and electrons.
Protons are positively charged. Each proton has a relative charge of +1 and a relative mass of 1 atomic mass unit.
Neutrons are uncharged. Each neutron has a relative charge of 0 and a relative mass of 1 atomic mass unit.
Electrons are negatively charged. Each electron has a relative charge of -1 and a relative mass of 1/1840 atomic mass unit.
Proton
Neutron
Electron
Relative charges
+1
0
-1
Relative masses
1
1
1/1840
A nuclide is a specific combination of protons and neutrons in a nucleus.
Atomic/Proton number---Number of protons
Mass/Nucleon number---Number of neutrons and protons
Shell
No. of orbitals (n2)
No. of each type of orbital
Maximum no. of electrons in shell (2n2)
s
p
d
f
n = 1
1
1
2
n = 2
4
1
3
8
n = 3
9
1
3
5
18
n = 4
16
1
3
5
7
32
? A sub-shell is a group of orbitals with the same energy level, but differ in their orientation in space, e.g. the second shell (n = 2) contains two sub-shells:
Ø A sub-shell containing one s orbital (2s), and
Ø A sub-shell containing three p orbitals (2pxy, 2py,2pz).
?
In an atom, the orbitals have definite
amounts of energy called energy levels.
These energy levels have a convergent pattern.
[NB. 4s orbital has slightly lower energy
than the 3d orbitals.]
Shapes of orbitals
s orbitals
? Each s sub-shell ha 1 orbital only
? Spherical shape
? Size increases as the principal quantum number increases.
p orbitals
? Each p sub-shell has 3 orbitals
? Dumb-bell shape
? Different axes of symmetry for each p orbital (each orbital is mutually perpendicular to the other)
? Orbitals with the same quantum number have the same energy i.e. they are degenerate.
? Size increases as the principal quantum number increases.
Relative Atomic Mass is defined as the mass of one atom of an element compared with 1/12 the mass of one atom of carbon-12.
Relative Isotopic Mass is defined as the mass of one atom of an isotope compared with 1/12 the mass of one atom of carbon- 12.
Relative Molecular Mass of a compound is the sum of the relative atomic masses of all the atoms in one molecule of a covalent compound.
Relative Formula Mass is the sum of all relative atomic masses from the formula of any given substance (atoms, molecules, ions, ionic compounds, etc.)
Avogadro Constant is a large mathematical constant defined as the number of carbon-12 atoms in 12g of carbon-12 (one mole).
Empirical Formula shows the simplest whole-number ratio for the atoms of different elements in a compound.
Molecular Formula shows the actual number of the different elements in one molecules of a compound.
ATOMIC STRUCTURE
Atoms are made up of protons, neutrons, and electrons.
Protons are positively charged. Each proton has a relative charge of +1 and a relative mass of 1 atomic mass unit.
Neutrons are uncharged. Each neutron has a relative charge of 0 and a relative mass of 1 atomic mass unit.
Electrons are negatively charged. Each electron has a relative charge of -1 and a relative mass of 1/1840 atomic mass unit.
Proton
Neutron
Electron
Relative charges
+1
0
-1
Relative masses
1
1
1/1840
A nuclide is a specific combination of protons and neutrons in a nucleus.
Atomic/Proton number---Number of protons
Mass/Nucleon number---Number of neutrons and protons
Shell
No. of orbitals (n2)
No. of each type of orbital
Maximum no. of electrons in shell (2n2)
s
p
d
f
n = 1
1
1
2
n = 2
4
1
3
8
n = 3
9
1
3
5
18
n = 4
16
1
3
5
7
32
? A sub-shell is a group of orbitals with the same energy level, but differ in their orientation in space, e.g. the second shell (n = 2) contains two sub-shells:
Ø A sub-shell containing one s orbital (2s), and
Ø A sub-shell containing three p orbitals (2pxy, 2py,2pz).
?
In an atom, the orbitals have definite
amounts of energy called energy levels.
These energy levels have a convergent pattern.
[NB. 4s orbital has slightly lower energy
than the 3d orbitals.]
Shapes of orbitals
s orbitals
? Each s sub-shell ha 1 orbital only
? Spherical shape
? Size increases as the principal quantum number increases.
p orbitals
? Each p sub-shell has 3 orbitals
? Dumb-bell shape
? Different axes of symmetry for each p orbital (each orbital is mutually perpendicular to the other)
? Orbitals with the same quantum number have the same energy i.e. they are degenerate.
? Size increases as the principal quantum number increases.
Saturday, April 11, 2009
Summary On Equilibrium Constants 2
Kc
Definition: Equilibrium constant is a measure of the ratio of concentrations of products and reactants at equilibrium.
What the value of K measures: equilibrium concentration
Type of system K applies to: molarity, system at equilibrium
Example of an equation (include state symbols): CO(g)+Br2(g) COBr2(g
)Expression for K: Kc=[CoBr2]/[CO][Br2]
Unit:mol dmˉ³
Kp
Definition: For reactions involving gases, the equilibrium constant of a gaseous reaction can also be expressed in terms of partial pressure.
What the value of K measures: equilibrium pressure
Type of system K applies to: reaction involve gases, system at equilibrium
Example of an equation (include state symbols): equilibrium
Expression for K:Kp=(PNH3)^2/(PN2)(PH2)^3
Unit Pa
Kw
Definition: The overall constant. Kw, is called teh ionic product for water.
What the value of K measures: dissociation of overall constant
Type of system K applies to: equilibrium
Example of an equation (include state symbols): H2O(l) H+(aq)+OH-(aq)
Expression for K: Kw=[H+][OH-]
Unit: mol² dm^6
Ka
Definition: Ka provides an accurate measure of the extent to which an acid is dissociated.
What the value of K measures: acid dissociation constant
Type of system K applies to: equilibrium
Example of an equation (include state symbols): HA(aq)+H2O(l) H3O+(aq)+A-(aq)
Expression for K: Ka=[H3O+][A-]/[HA]
Unit: mol dmˉ³
Kb
Definition: Kb provides an accurate measure of the extent to which an alkali is dissociated.
What the value of K measures: base dissociation constant
Type of system K applies to: equilibrium
Example of an equation (include state symbols): B(aq)+H2O(l) BH+(aq)+OH-(aq)
Expression for K: Kb=[BH+][OH-]/[B]
Unit: mol dmˉ³
Ksp
Definition: The solubility of a substance is the quantity that dissolves to form a saturated solution.
What the value of K measures: solubility product
Type of system K applies to: equilibrium
Example of an equation (include state symbols): PbBr2(s) Pb2+(aq)+2Br-(aq)
Expression for K:Ksp=[Pb2+][Br-]2
Unit: g dmˉ³;mol dmˉ³
Definition: Equilibrium constant is a measure of the ratio of concentrations of products and reactants at equilibrium.
What the value of K measures: equilibrium concentration
Type of system K applies to: molarity, system at equilibrium
Example of an equation (include state symbols): CO(g)+Br2(g) COBr2(g
)Expression for K: Kc=[CoBr2]/[CO][Br2]
Unit:mol dmˉ³
Kp
Definition: For reactions involving gases, the equilibrium constant of a gaseous reaction can also be expressed in terms of partial pressure.
What the value of K measures: equilibrium pressure
Type of system K applies to: reaction involve gases, system at equilibrium
Example of an equation (include state symbols): equilibrium
Expression for K:Kp=(PNH3)^2/(PN2)(PH2)^3
Unit Pa
Kw
Definition: The overall constant. Kw, is called teh ionic product for water.
What the value of K measures: dissociation of overall constant
Type of system K applies to: equilibrium
Example of an equation (include state symbols): H2O(l) H+(aq)+OH-(aq)
Expression for K: Kw=[H+][OH-]
Unit: mol² dm^6
Ka
Definition: Ka provides an accurate measure of the extent to which an acid is dissociated.
What the value of K measures: acid dissociation constant
Type of system K applies to: equilibrium
Example of an equation (include state symbols): HA(aq)+H2O(l) H3O+(aq)+A-(aq)
Expression for K: Ka=[H3O+][A-]/[HA]
Unit: mol dmˉ³
Kb
Definition: Kb provides an accurate measure of the extent to which an alkali is dissociated.
What the value of K measures: base dissociation constant
Type of system K applies to: equilibrium
Example of an equation (include state symbols): B(aq)+H2O(l) BH+(aq)+OH-(aq)
Expression for K: Kb=[BH+][OH-]/[B]
Unit: mol dmˉ³
Ksp
Definition: The solubility of a substance is the quantity that dissolves to form a saturated solution.
What the value of K measures: solubility product
Type of system K applies to: equilibrium
Example of an equation (include state symbols): PbBr2(s) Pb2+(aq)+2Br-(aq)
Expression for K:Ksp=[Pb2+][Br-]2
Unit: g dmˉ³;mol dmˉ³
Summary On Equilibrium Constants 1
Summary on Equilibrium Constants 1
Similarities between the equilibrium constants (Kc, Kp, Kw, Ka, Kb, Ksp)
1. System at equilibrium
2. Le Chatelier’s principle can be applied
3. Equation uses reversible signs
4. Temperature dependent
5. Involves concentration
Similarities between the equilibrium constants (Kc, Kp, Kw, Ka, Kb, Ksp)
1. System at equilibrium
2. Le Chatelier’s principle can be applied
3. Equation uses reversible signs
4. Temperature dependent
5. Involves concentration
Definition for Chemical Equilibrium
a.) A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as
aA + bB cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.Dynamic Equilibrium refers to a reversible reaction in which the rates of forward and reverse reactions become equal and there is no change in the concentration of the reactants and products.
b.) Le Chatelier's Principle can be used to predict the effect of a change in conditions on a chemical equilibrium.
c.) If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.
Concentration
Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. The chemical system will attempt to partially oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent and yield of products will be altered corresponding to the impact on the system.
This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.
CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced. This observation is supported by the "collision theory". As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.
Temperature
Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favour the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.
Changes in pressure owing to changes in volume
The equilibrium concentrations of the products and reactants do not directly depend on the pressure subjected to the system. However, a change in pressure due to a change in volume of the system will shift the equilibrium.
Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. When the volume of the system is changed, the partial pressures of the gases change. Because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.
Thus, an increase in pressure due to decreasing volume causes the reaction to shift to the side with the fewer moles of gas.[2] A decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical system (or equation)
g.) The Haber synthesis was developed into an industrial process.Conditions:medium temperature (~500oC) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).Osmium is a much better catalyst for the reaction but is very expensive.
This process produces an ammonia, NH3(g), yield of approximately 10-20%.
The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
N2(g)nitrogen+3H2(g)hydrogen heat, pressure, catalyst 2NH3(g) ammonia
H = -92.4 kJ mol-1
By Le Chetalier's Principle:
Increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation. Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules.
Decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as wellHowever, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia.
The equilibrium expression for this reaction is:
Keq =[NH3]2[N2][H2]3
h.) Chain Disproportionation occurs when two radicals meet, instead of coupling, they exchange a proton. That gives two terminated chains, one saturated and the other with a terminal double bond.
2R-(CH2CH2)n-CH2CH2• → R-(CH2CH2)n-CH=CH2 + R-(CH2CH2)n-CH2CH3
http://www.scribd.com/doc/6618373/InDepth-Overview-and-Comparison-of-2006-2007-AL-Chem-Syllabus
aA + bB cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.Dynamic Equilibrium refers to a reversible reaction in which the rates of forward and reverse reactions become equal and there is no change in the concentration of the reactants and products.
b.) Le Chatelier's Principle can be used to predict the effect of a change in conditions on a chemical equilibrium.
c.) If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.
Concentration
Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. The chemical system will attempt to partially oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent and yield of products will be altered corresponding to the impact on the system.
This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.
CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced. This observation is supported by the "collision theory". As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.
Temperature
Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favour the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.
Changes in pressure owing to changes in volume
The equilibrium concentrations of the products and reactants do not directly depend on the pressure subjected to the system. However, a change in pressure due to a change in volume of the system will shift the equilibrium.
Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. When the volume of the system is changed, the partial pressures of the gases change. Because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.
Thus, an increase in pressure due to decreasing volume causes the reaction to shift to the side with the fewer moles of gas.[2] A decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical system (or equation)
g.) The Haber synthesis was developed into an industrial process.Conditions:medium temperature (~500oC) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).Osmium is a much better catalyst for the reaction but is very expensive.
This process produces an ammonia, NH3(g), yield of approximately 10-20%.
The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
N2(g)nitrogen+3H2(g)hydrogen heat, pressure, catalyst 2NH3(g) ammonia
H = -92.4 kJ mol-1
By Le Chetalier's Principle:
Increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation. Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules.
Decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as wellHowever, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia.
The equilibrium expression for this reaction is:
Keq =[NH3]2[N2][H2]3
h.) Chain Disproportionation occurs when two radicals meet, instead of coupling, they exchange a proton. That gives two terminated chains, one saturated and the other with a terminal double bond.
2R-(CH2CH2)n-CH2CH2• → R-(CH2CH2)n-CH=CH2 + R-(CH2CH2)n-CH2CH3
http://www.scribd.com/doc/6618373/InDepth-Overview-and-Comparison-of-2006-2007-AL-Chem-Syllabus
Saturday, January 17, 2009
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