For any reversible reaction in a closed system, the position of equilibrium can be expressed quantitatively by the equilibrium constant, \( K_c \). Deducing the expression for \( K_c \), determining its units, and calculating its value are core quantitative skills for AQA A-Level Chemistry.
🔑 Key Principle
The value of \( K_c \) is a constant for a specific reaction at a constant temperature. Changing the concentration of reactants or products, changing the pressure of gaseous mixtures, or introducing a catalyst does not alter the value of \( K_c \). Only a change in temperature will change the value of \( K_c \).
Constructing Kc Expressions
For a general reversible reaction in a homogeneous system:
The equilibrium constant expression is:
Where:
- Square brackets, e.g. \([\text{A}]\), represent the equilibrium concentration of substance A in \(\text{mol dm}^{-3}\).
- The numerator contains the concentrations of the products multiplied together, each raised to the power of their stoichiometric balancing number.
- The denominator contains the concentrations of the reactants multiplied together, each raised to the power of their stoichiometric balancing number.
A ratio of the concentrations of products to reactants at equilibrium for a homogeneous system, with each concentration raised to the power of its stoichiometric balancing coefficient in the balanced equation.
A reaction mixture in which all reactants and products are in the same physical state or phase (for example, all are gases, or all are dissolved in aqueous solution).
Determining the Units of Kc
The units of \( K_c \) are not fixed: they depend on the powers of the concentration terms in the expression. To determine the units, substitute the concentration unit \(\text{mol dm}^{-3}\) into the \( K_c \) expression and cancel terms:
Step 1: Write the \( K_c \) expression:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
Step 2: Substitute units (\(\text{mol dm}^{-3}\)) into the expression:
\[ \text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4} \]
Step 3: Cancel down the terms:
\[ \text{Units} = \frac{1}{(\text{mol dm}^{-3})^2} = (\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\text{ dm}^6 \]
Always write units starting with the positive index. For instance, write \(\text{dm}^6\text{ mol}^{-2}\) instead of \(\text{mol}^{-2}\text{ dm}^6\), although both are mathematically correct. If the concentration terms on the top and bottom cancel out completely, you must write no units. Never leave the units blank without clarifying that there are no units.
Calculations Using ICE Tables
Often, you are given the initial amounts of reactants and only one equilibrium value. You must determine the equilibrium concentrations of all species to calculate \( K_c \). The standard method to solve these problems is the ICE table:
- Initial moles
- Change in moles
- Equilibrium moles
- Equilibrium concentration (moles divided by total volume, \(V\))
Step 1: Write the \( K_c \) expression:
\[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} \]
Step 2: Construct the ICE table for moles:
| Species | \(2\text{SO}_2\) | \(\text{O}_2\) | \(2\text{SO}_3\) |
|---|---|---|---|
| Initial Moles | 0.200 | 0.200 | 0.000 |
| Change in Moles | -0.160 | -0.080 | +0.160 |
| Equilibrium Moles | 0.040 | 0.120 | 0.160 |
| Equilibrium Concentration (\(\text{mol dm}^{-3}\)) | \(0.040 / 2 = 0.020\) | \(0.120 / 2 = 0.060\) | \(0.160 / 2 = 0.080\) |
How the changes were calculated:
- The equilibrium moles of \(\text{SO}_3\) is given as 0.160. Since initial moles were 0, the change for \(\text{SO}_3\) is \(+0.160\).
- According to the 2:1 stoichiometry, the moles of \(\text{O}_2\) reacting must be half of the moles of \(\text{SO}_3\) formed: \(\frac{0.160}{2} = 0.080\). The change is \(-0.080\).
- According to the 2:2 stoichiometry, the moles of \(\text{SO}_2\) reacting must equal the moles of \(\text{SO}_3\) formed: 0.160. The change is \(-0.160\).
- Equilibrium moles = Initial moles + Change.
- Equilibrium concentration = Equilibrium moles / Volume (2.00 \(\text{dm}^3\)).
Step 3: Calculate the value of \( K_c \):
\[ K_c = \frac{(0.080)^2}{(0.020)^2 \times (0.060)} = \frac{0.0064}{0.0004 \times 0.060} = \frac{0.0064}{0.000024} = 266.7 \]
Step 4: Determine the units of \( K_c \):
\[ \text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \frac{1}{\text{mol dm}^{-3}} = \text{mol}^{-1}\text{ dm}^3 \]
Therefore, \( K_c = 267\text{ dm}^3\text{ mol}^{-1}\) (to 3 significant figures).
A very common student error is failing to divide the equilibrium moles by the volume to find concentrations. If you put moles directly into the \( K_c \) expression when the volume does not cancel, you will get an incorrect numerical value and lose multiple marks. Always write down the volume explicitly, and divide by it before plugging numbers into the \( K_c \) equation.
Factors Affecting the Value of Kc
Understanding which parameters change the actual value of \( K_c \) is a common topic in AQA conceptual multiple-choice and short-answer questions. The effects are summarised in the table below:
| Variable Change | Effect on the Position of Equilibrium | Effect on the Value of \( K_c \) |
|---|---|---|
| Temperature | Shifts to oppose temperature change (Le Chatelier's Principle) | Changes |
| Concentration | Shifts to keep the concentration ratio constant | No Effect |
| Pressure | Shifts to side with fewer/more gas moles | No Effect |
| Catalyst | No shift (rates of forward and reverse reactions increase equally) | No Effect |
Why Concentration and Pressure Do Not Change Kc
If you increase the concentration of a reactant, Le Chatelier's principle states that the system will shift to the right to produce more products. This shift changes the equilibrium concentrations of both reactants and products. However, they change in such a way that when they are substituted back into the \( K_c \) expression, the overall ratio remains exactly the same. Thus, the position of equilibrium changes, but the value of \( K_c \) remains constant.
Why a Catalyst Has No Effect on Kc
A catalyst provides an alternative reaction pathway with lower activation energy. It increases the rate of both the forward and reverse reactions by the exact same factor. As a result, the system reaches equilibrium faster, but the concentrations of reactants and products at equilibrium are identical to the uncatalysed reaction. Therefore, a catalyst has no effect on either the position of equilibrium or the value of \( K_c \).
How Temperature Changes Kc
To determine the effect of temperature, you must know the sign of the enthalpy change (\( \Delta H \)) of the forward reaction:
Exothermic Forward Reaction
Increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb heat. This increases reactant concentrations and decreases product concentrations. The numerator decreases and the denominator increases, so \( K_c \) decreases.
Endothermic Forward Reaction
Increasing the temperature shifts the equilibrium in the endothermic direction (to the right) to absorb heat. This increases product concentrations and decreases reactant concentrations. The numerator increases and the denominator decreases, so \( K_c \) increases.
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