Exploring the Equilibrium Constant $K$
Equilibrium constant
The equilibrium constant $K$ is a ratio that provides a snapshot of the relative concentrations of products and reactants at equilibrium for a reversible chemical reaction.
- It is derived from the stoichiometry of the reaction and remains constant as long as the temperature is unchanged.
- Consider the general reaction: $$
aA + bB \rightleftharpoons cC + dD
$$ - At equilibrium, the concentrations of $ A $, $ B $, $ C $, and $ D $ remain constant.
- The equilibrium constant $ K $ is expressed as: $$
K = \frac{[C]^c[D]^d}{[A]^a[B]^b}
$$ where:- $[X]$ represents the equilibrium concentration of species $ X $ in $ \text{mol dm}^{-3} $.
- $ a, b, c, d $ are the stoichiometric coefficients from the balanced chemical equation.
- Before writing the $ K $ expression, ensure the chemical equation is balanced.
- The stoichiometric coefficients directly determine the exponents in the $ K $ expression.
Deriving the Expression for $ K $: Step-by-Step
- Let’s illustrate this with an example: the synthesis of ammonia. $$
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
$$ - Identify the Reactants and Products:
- Reactants: $ N_2 $ and $ H_2 $
- Product: $ NH_3 $
- Write the General Form of $ K $:
- According to the equilibrium law, the concentrations of the products are placed in the numerator, while the concentrations of the reactants are placed in the denominator.
- Each species is raised to the power of its stoichiometric coefficient: $$
K = \frac{[NH_3]^2}{[N_2][H_2]^3}
$$
- Interpret the Expression:
- If $ K $ is large ($ K > 1 $), the numerator dominates, indicating that the equilibrium mixture contains more products than reactants.
- If $ K $ is small ($ K < 1 $), the denominator dominates, meaning the equilibrium mixture contains more reactants than products.
Write the $ K $ expression for the reaction:$$
2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)
$$
Solution
The equilibrium constant expression is:$$
K = \frac{[SO_3]^2}{[SO_2]^2[O_2]}
$$
Properties of the Equilibrium Constant
- Temperature Dependence:
- $ K $ remains constant only at a fixed temperature.
- Altering the temperature shifts the equilibrium position and changes the value of $ K $.
- No Units:
- $ K $ is often treated as a dimensionless quantity because the units of concentration cancel out when the expression is written correctly.
- Homogeneous vs. Heterogeneous Equilibria:
- For homogeneous equilibria (all species in the same phase), all reactants and products appear in the $ K $ expression.
- For heterogeneous equilibria (species in different phases), the concentrations of pure solids and liquids are omitted because they are constant.
In aqueous reactions, the concentration of water is often omitted from the $ K $ expression if it is the solvent and present in large excess.
Interpreting $ K $: What Does It Tell Us?
The value of $ K $ provides insight into the extent of the reaction at equilibrium:
- $ K > 1 $:
- Products are favored.
- The reaction proceeds significantly in the forward direction.
- $ K < 1 $:
- Reactants are favored.
- The reaction does not proceed significantly in the forward direction.
- $ K = 1 $:
- Neither reactants nor products are favored.
- The concentrations of reactants and products are comparable.
Calculating $ K $
The equilibrium concentrations for the reaction $ N_2(g) + 3H_2(g) \leftrightarrow 2NH_3(g) $ at 475 K are:
- $[N_2] = 0.50 \, \text{mol dm}^{-3}$,
- $[H_2] = 1.50 \, \text{mol dm}^{-3}$,
- $[NH_3] = 1.00 \, \text{mol dm}^{-3}$.
Calculate $ K $ for the reaction.
Solution
- Write the $ K $ expression:
$$
K = \frac{[NH_3]^2}{[N_2][H_2]^3}
$$ - Substitute the equilibrium concentrations:
$$
K = \frac{(1.00)^2}{(0.50)(1.50)^3}
$$ - Perform the calculation:
$$
K = \frac{1.00}{0.50 \times 3.375} = \frac{1.00}{1.6875} \approx 0.59
$$ - Answer: $ K = 0.59 $