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 (\(+\Delta H\)) | Absorbs thermal energy to lower system temperature. |
| Decrease Temperature | Shifts in the exothermic direction (\(-\Delta H\)) | 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. |
| Add Catalyst | No shift in position | Increases rates of forward and reverse reactions equally. |
3. 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 \([\text{X}]\) denote equilibrium concentrations in \(\text{mol dm}^{-3}\).
Determining Units of \(K_c\)
Substitute \(\text{mol dm}^{-3}\) for each concentration term and cancel algebraically:
- If numerator and denominator have equal powers, \(K_c\) has no units.
- For \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\): \[ \text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \frac{1}{(\text{mol dm}^{-3})^2} = \mathbf{\text{mol}^{-2}\text{dm}^6} \]
4. 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:
\[ K_c = \frac{0.670 \times 0.670}{0.330 \times 0.330} = \frac{0.4489}{0.1089} = \mathbf{4.12} \text{ (no units)} \]
5. The Effect of Conditions on the Value of \(K_c\)
- 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 \(K_c\).
6. Practice Questions
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)