Introduction:
In AS we covered the characteristics of an equilibrium:
In AS we discovered that Le Chatelier’s Principle was given as a way of determining how changes in the conditions can affect the extent to which a reaction will go.
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Kc |
= |
[PRODUCTS]p | |
[REACTANTS]r |
aA | + | bB | D | cC | + | dD |
Kc |
= |
[C]c [D]d |
[A]a [B]b |
Approaching equilibrium
N2O4(g) | D | 2NO2(g) |
Initially: |
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As the reaction proceeds: |
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At equilibrium: |
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Writing expressions for Kc:
N2O4(g) | D | 2NO2(g) |
Kc |
= |
[NO2]2 |
[N2O4] |
2SO2(g) | + | O2(g) | D | 2SO3(g) |
Kc |
= |
[SO3]2 |
[SO2]2 [O2] |
H2(g) | + | I2(g) | D | 2HI(g) |
Kc |
= |
[HI]2 |
[H2] [I2] |
Units of Kc
Kc | = | [NO2]2 | Kc | = | [SO3]2 | Kc | = | [HI]2 | ||||||
[N2O4]1 | [SO2]2 [O2] | [H2] [I2] | ||||||||||||
Kc | = | mol dm-3 x mol dm-3 | Kc | = | (mol dm-3)2 | Kc | = | mol dm-3 mol dm-3 | ||||||
mol dm-3 | (mol dm-3)2 | mol dm-3 | mol dm-3 | mol dm-3 | ||||||||||
Kc | = |
mol dm-3 x |
Kc | = |
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Kc | = |
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mol dm-3 |
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Kc | = | mol dm-3 | Kc | = | 1 | Kc | = | No units | ||||||
mol dm-3 | ||||||||||||||
Kc | = | dm3 mol-1 |
Qu 1 P127
Determining Kc from equilibrium concentrations:
1) Hydrogen, Iodine, Hydrogen iodide equilibrium:
Reaction: | H2(g) | + | I2(g) | D | 2HI(g) |
Equilibrium concentrations: | 0.140 | 0.040 | 0.320 |
Kc | = | [HI]2 | Kc | = | [HI]2 | |||||
[H2] [I2] | [H2] [I2] | |||||||||
Kc | = | (0.320)2 | Kc | = | mol dm-3 mol dm-3 | |||||
0.140 x 0.040 | mol dm-3 | mol dm-3 | ||||||||
Kc | = | 18.3 | Kc | = |
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Kc | = | No units |
2) N2O4 / NO2 equilibrium: Given the number of moles at equilibrium in a volume of 2dm3.
Reaction: | N2O4(g) | D | 2NO2(g) |
Mole quantities at equilibrium: | 0.400 | 3.20 | |
Equilibrium concentrations: | 0.400 / 2 | 3.20 / 2 | |
Equilibrium concentrations: | 0.20 | 1.60 |
Kc | = | [NO2]2 | Kc | = | [NO2]2 | |||||
[N2O4] | [N2O4] | |||||||||
Kc | = | (1.60)2 | Kc | = | mol dm-3 mol dm-3 | |||||
0.20 | mol dm-3 | |||||||||
Kc | = | 12.8 | mol dm-3 | Kc | = |
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Kc | = | mol dm-3 |
Calculating the quantities and concentrations present at equilibrium:
3) Hydrogen, Iodine, Hydrogen iodide equilibrium:
Reaction: | H2(g) | + | I2(g) | D | 2HI(g) |
At start: | 0.60 | 0.40 | 0.0 | ||
At equilibrium: | 0.28 | 0.08 | 0.64 | ||
Reacted: | 0.32 | 0.32 | 0.64 | ||
Equilibrium concentrations: | 0.28 | 0.08 | 0.64 |
Kc | = | [HI]2 | Kc | = | [HI]2 | |||||
[H2] [I2] | [H2] [I2] | |||||||||
Kc | = | (0.64)2 | Kc | = | mol dm-3 mol dm-3 | |||||
0.28 x 0.08 | mol dm-3 | mol dm-3 | ||||||||
Kc | = | 18.3 | Kc | = |
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Kc | = | No units |
Qu 1 - 2 P129
The equilibrium position and Kc
What is the significance of a Kc value?
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Products favoured: Kc > 1
Reactants favoured: Kc < 1
How do changes in temperature affect Kc?
1) Endothermic reactions:
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Kc increases and vice versa |
2) Exothermic reactions:
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Kc decreases and vice versa |
Qu 1 P131
The equilibrium constant, Kc, and the rate constant, k
How does a change in concentration and pressure affect Kc?
Kc is unaltered by changes in concentration and pressure
1) Changes in concentration:
A change in Concentration has no effect on the equilibrium constant.
Remember if a reactant / product is added the equilibrium shifts to the opposite direction to keep the 'proportions' the same - Kc is unchanged:
Reaction: | N2O4(g) | D | 2NO2(g) |
[NO2] = 1.60 Mol dm-3 | [NO2] = 1.60 Mol dm-3 | ||||||||
[N2O4] = 0.20 Mol dm-3 | Double [ ] : | [N2O4] = 0.40 Mol dm-3 | |||||||
Kc | = | [NO2]2 | Kc | = | [NO2]2 | ||||
[N2O4] | [N2O4] | ||||||||
Kc | = | (1.60)2 | Kc | = | (3.20)2 | ||||
0.20 | 0.40 | ||||||||
Kc | = | 12.8 |
Mol dm-3 |
Kc | = | 6.4 | Mol dm-3 |
If the concentrations are doubled, the system is no longer at equilibrium, Kc = 12.8
To bring Kc back from 6.4 to 12.8:
The system must increase [NO2] and decrease [N2O4]
Remember if a reactant / product is added the equilibrium shifts to the opposite direction to keep the 'proportions' the same - Kc is unchanged:
2) Changes in Pressure:
A change in Pressure has no effect on the equilibrium constant.
If pressure is doubled - the volume is halved - meaning that the concentrations will have doubled
Reaction: | N2O4(g) | D | 2NO2(g) |
[NO2] = 1.60 Mol dm-3 | Double [ ] : | [NO2] = 3.20 Mol dm-3 | |||||||
[N2O4] = 0.20 Mol dm-3 | [N2O4] = 0.40 Mol dm-3 | ||||||||
Kc | = | [NO2]2 | Kc | = | [NO2]2 | ||||
[N2O4] | [N2O4] | ||||||||
Kc | = | (1.60)2 | Kc | = | (3.20)2 | ||||
0.20 | 0.40 | ||||||||
Kc | = | 12.8 |
Mol dm-3 |
Kc | = | 25.6 | Mol dm-3 |
If the pressure is doubled, the system is no longer at equilibrium, Kc = 12.8
To bring Kc back from 25.6 to 12.8:
The system must increase [NO2] and decrease [N2O4]
Remember if a reactant / product is added the equilibrium shifts to the opposite direction to keep the 'proportions' the same - Kc is unchanged:
How does the presence of a catalyst affect Kc?
A catalyst has no effect on the equilibrium constant.
A catalyst speeds up both the forward and reverse reaction.
Equilibrium is achieved more quickly.
The equilibrium constant Kc, and the rate constant, k
These 2 constants tell us the most important things in the chemical industry:
a) | Equilibrium | How far |
b) | Rates | How fast |
a) The equilibrium constant, Kc
Kc indicates the position of the equilibrium:
Large Kc | Products predominate | |
Small Kc | Reactants predominate |
Remember LCP:
Endothermic | Kc increases with an increase in temperature (increases products) | |
Exothermic | Kc decreases with an increase in temperature (decreases products) |
Kc can be written from the balanced chemical equation
b) The rate constant, k
k is a measure of the rate of a reaction:
Large k | Fast rate | |
Small k | Slow rate |
Remember:
k increases with an increase in temperature - Rate increases with an increase in temperature | |
k decreases with a decrease in temperature - Rate decreases with a decrease in temperature |
k can only be determined experimentally from the rate equation
The importance of compromise:
The 2 desirable outcomes are a) increasing rate and b) increasing the amount:
a) Increasing the rate:
Increasing temperature: Increases the rate of production of product - desirable |
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Increasing temperature: Decreases the amount of product made - undesirable |
b) Increasing the amount:
Decreasing the temperature: Increases the amount of product made - desirable |
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Decreasing temperature: Decreases the rate of production of product - undesirable |
The compromise is:
Moderate temperature: | k increases moderately - Rate increases by a moderate amount |
Moderate temperature: | Kc decreases by a moderate amount - allowing a moderate yield |
Qu 1 P133 / Qu 4 - 5 P159 / Qu 3 - 4 P161