1. The Concept of Dynamic Equilibrium
A reversible reaction in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.
Equilibrium is dynamic because reactions continue occurring at the microscopic level in both directions simultaneously at identical rates.
2. Le Chatelier's Principle
If a chemical system at equilibrium is subjected to a change in conditions, the system adjusts to oppose that change and restore equilibrium.
| Condition Changed | Equilibrium Shift | Opposing Response of the System |
|---|---|---|
| Increase Temperature | Shifts in the endothermic direction (+ΔH) | Absorbs thermal energy to lower system temperature. |
| Decrease Temperature | Shifts in the exothermic direction (−ΔH) | Releases thermal energy to raise system temperature. |
| Increase Pressure | Shifts to side with fewer gas moles | Reduces total gas molecules to lower pressure. |
| Decrease Pressure | Shifts to side with more gas moles | Increases total gas molecules to raise pressure. |
| Increase Concentration of a Reactant | Shifts to the right (forward direction) | Uses up some of the added reactant, to remove it. |
| Decrease Concentration of a Product | Shifts to the right (forward direction) | Makes more product, to replace what was removed. |
| Add Catalyst | No shift in position | Increases rates of forward and reverse reactions equally. |
- The position of equilibrium shifts to the right, to remove some of the added ethanol.
- So the amount of ethyl ethanoate increases.
- The temperature has not changed, so \(K_c\) does not change.
3. Compromise Conditions in Industry
Industry needs a good yield and a fast rate, at a sensible cost. These aims often conflict, so the conditions chosen are a compromise.
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \qquad \Delta H = -92 \text{ kJ mol}^{-1} \]
The Haber process makes ammonia. The forward reaction is exothermic, and 4 mol of gas form 2 mol of gas.
| Condition | Effect on yield | Effect on rate and cost | Condition used |
|---|---|---|---|
| Temperature | A low temperature gives a higher yield (the forward reaction is exothermic). | A low temperature gives a slow rate. | About 450 °C |
| Pressure | A high pressure gives a higher yield (fewer moles of gas on the right). | A high pressure gives a faster rate, but it is expensive and needs strong equipment. | About 200 atm |
| Catalyst | No change to the yield. | An iron catalyst increases the rate. | Iron |
4. The Equilibrium Constant (\(K_c\))
For a general homogeneous reversible reaction:
\[ a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \]
The equilibrium constant expression is:
\[ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
Where square brackets [X] denote equilibrium concentrations in mol dm⁻³.
Determining Units of \(K_c\)
Substitute mol dm⁻³ for each concentration term and cancel algebraically:
- If numerator and denominator have equal powers, \(K_c\) has no units.
- For \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\): \[ \text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \frac{1}{(\text{mol dm}^{-3})^2} = \mathbf{\text{mol}^{-2}\text{dm}^6} \]
5. ICE Calculation Tables (Initial, Change, Equilibrium)
Step 1: Construct the ICE Table:
| Row | CH₃COOH | C₂H₅OH | CH₃COOC₂H₅ | H₂O |
|---|---|---|---|---|
| Initial (mol) | 1.00 | 1.00 | 0.00 | 0.00 |
| Change (mol) | -0.670 | -0.670 | +0.670 | +0.670 |
| Equilibrium (mol) | 0.330 | 0.330 | 0.670 | 0.670 |
| Concentration (mol/V) | 0.330 / V | 0.330 / V | 0.670 / V | 0.670 / V |
Step 2: Substitute into \(K_c\) expression:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} = \frac{(0.670 / V)(0.670 / V)}{(0.330 / V)(0.330 / V)} \]
The volume terms \(V\) cancel out completely:
\[ K_c = \frac{0.670 \times 0.670}{0.330 \times 0.330} = \frac{0.4489}{0.1089} = \mathbf{4.12} \text{ (no units)} \]
Step 1: Moles at equilibrium. 0.200 mol of PCl₅ reacted, so there is 0.200 mol each of PCl₃ and Cl₂.
| Row | PCl₅ | PCl₃ | Cl₂ |
|---|---|---|---|
| Initial (mol) | 0.500 | 0 | 0 |
| Change (mol) | -0.200 | +0.200 | +0.200 |
| Equilibrium (mol) | 0.300 | 0.200 | 0.200 |
| Concentration (mol dm⁻³) | 0.150 | 0.100 | 0.100 |
Step 2: Divide by the volume (2.00 dm³). The moles on each side are not equal, so the volume does not cancel.
Step 3: Substitute.
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.100 \times 0.100}{0.150} = \mathbf{0.0667 \text{ mol dm}^{-3}} \]
Units: \(\frac{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}{\text{mol dm}^{-3}} = \text{mol dm}^{-3}\).
Step 1: Rearrange the expression.
\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \quad \Rightarrow \quad [\text{HI}]^2 = K_c \times [\text{H}_2] \times [\text{I}_2] \]
Step 2: Substitute.
\[ [\text{HI}]^2 = 50.0 \times 0.0200 \times 0.0200 = 0.0200 \]
Step 3: Take the square root.
\[ [\text{HI}] = \sqrt{0.0200} = \mathbf{0.141 \text{ mol dm}^{-3}} \]
6. The Effect of Conditions on the Value of \(K_c\)
| Forward reaction | Temperature increased | Temperature decreased |
|---|---|---|
| Exothermic (ΔH negative) | \(K_c\) decreases | \(K_c\) increases |
| Endothermic (ΔH positive) | \(K_c\) increases | \(K_c\) decreases |
\(K_c\) is not changed by a change in concentration or by a catalyst. Only temperature changes \(K_c\).
- Temperature: If temperature change shifts equilibrium in the forward direction, \(K_c\) increases. If it shifts equilibrium in the reverse direction, \(K_c\) decreases.
- Concentration / Pressure: Changing concentrations or total pressure shifts the position of equilibrium, but does not change the numerical value of \(K_c\).
- Catalysts: Catalysts speed up forward and reverse rates equally; they alter neither the position of equilibrium nor the value of \(K_c\).
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
7. Practice Questions
Show Answer & Mark Scheme
Answer:
- The value of \(K_c\) decreases. (1 mark)
- Explanation: Because the forward reaction is exothermic, an increase in temperature shifts the equilibrium in the endothermic reverse direction to absorb heat. This reduces the concentration of products and increases reactants, decreasing the value of the fraction. (1 mark)
Show Answer & Mark Scheme
Answer:
- 0.100 mol of N₂O₄ reacted, so there is 2 × 0.100 = 0.200 mol of NO₂. [N₂O₄] = 0.300 and [NO₂] = 0.200 mol dm⁻³. (1 mark)
- \(K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{0.200^2}{0.300} = 0.133\). (1 mark)
- Units: mol dm⁻³. (1 mark)
Show Answer & Mark Scheme
Answer:
- \([\text{I}_2] = \frac{[\text{HI}]^2}{K_c \times [\text{H}_2]} = \frac{0.500^2}{50.0 \times 0.0400}\). (1 mark)
- \([\text{I}_2] = \frac{0.250}{2.00} = 0.125 \text{ mol dm}^{-3}\). (1 mark)
Show Answer & Mark Scheme
Answer:
- The forward reaction is exothermic, so a lower temperature would give a higher yield. (1 mark)
- But at a lower temperature the rate would be too slow. (1 mark)
- 450 °C is a compromise: an acceptable yield at an acceptable rate. (1 mark)