Equilibrium Calculations in Chemistry
- As discussed previously, the equilibrium constant ($ K $) provides a numerical measure of the balance between reactants and products at equilibrium.
- For a general reaction: $$
aA + bB \rightleftharpoons cC + dD
$$ - The equilibrium constant expression is: $$
K = \frac{[C]^c [D]^d}{[A]^a [B]^b}
$$ Here:- $ [... ] $ represents molar concentrations.
- The exponents correspond to the stoichiometric coefficients in the balanced equation.
For the reaction $ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) $, the equilibrium constant expression is:$$
K = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
$$
Step-by-Step: Calculating Equilibrium Concentrations
To calculate equilibrium concentrations, follow these steps:
- Write the $ K $ Expression:
- Start by writing the equilibrium constant expression for the reaction.
- Define Initial and Equilibrium Concentrations:
- Use an ICE (Initial, Change, Equilibrium) table to organize the data.
- Define the initial concentrations, the changes during the reaction, and the equilibrium concentrations.
- Substitute into the $ K $ Expression:
- Substitute the equilibrium concentrations into the $ K $ expression, often in terms of $ x $, the change in concentration.
- Solve for $ x $:
- Solve the resulting equation to find $ x $, representing the change in concentration.
- Calculate Equilibrium Concentrations:
- Use $ x $ to calculate the equilibrium concentrations of all species.
Calculate the equilibrium concentrations for the reaction $ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) $, where $ K = 50.0 $ and the initial concentrations are:
- $[H_2] = 0.100 \, \text{mol dm}^{-3}$,
- $[I_2] = 0.100 \, \text{mol dm}^{-3}$,
- $[HI] = 0 \, \text{mol dm}^{-3}$.
Solution
- Write the $ K $ expression:$$
K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}
$$ - Define the changes using an ICE table:
| Species | Initial ($\text{mol dm}^{-3}$) | Change ($\text{mol dm}^{-3}$) | Equilibrium ($\text{mol dm}^{-3}$) |
|---|---|---|---|
| $H_2$ | 0.100 | $-x$ | $0.100-x$ |
| $I_2$ | 0.100 | $-x$ | $0.100-x$ |
| $HI$ | 0 | $+2x$ | $2x$ |
- Substitute into the $ K $ expression:$$
50.0 = \frac{(2x)^2}{(0.100 - x)(0.100 - x)}
$$ - To solve for $ x $, expand and simplify:$$
50.0 = \frac{4x^2}{(0.100 - x)^2}
$$ $$
x = 0.078 \, \text{mol dm}^{-3}
$$ - Calculate equilibrium concentrations:
$$[H_2] = [I_2] = 0.100 - 0.078 = 0.022 \, \text{mol dm}^{-3}$$ $$[HI] = 2x = 2(0.078) = 0.156 \, \text{mol dm}^{-3}$$
Always double-check your calculations and ensure units are consistent throughout the problem.
Approximation for Small $ K $
- When $ K $ is very small (e.g., $ K< 10^{-3} $), the reaction heavily favors the reactants.
- In such cases, the change in reactant concentration ($ x $) is often negligible compared to the initial concentration.
- This allows for simplifications in calculations.
Steps for Approximation
- Set Up the $ K $ Expression: Write the equilibrium constant expression as usual.
- Assume Negligible Change: If $ K $ is small, assume $ [\text{Reactant}] - x \approx [\text{Reactant}] $.
- Solve for $ x $: Substitute the approximation into the $ K $ expression and solve for $ x $.
- Verify the Approximation:
- Check whether $ x $ is less than 5% of the initial concentration.
- If true, the approximation is valid.
For the reaction $ 2NO_2(g) \rightleftharpoons N_2O_4(g) $, $ K = 0.001 $ and the initial concentration of $ NO_2 $ is $ 0.100 \, \text{mol dm}^{-3} $. Calculate the equilibrium concentrations.
Solution
- Write the $ K $ expression: $$
K = \frac{[N_2O_4]}{[NO_2]^2}
$$ - Define equilibrium concentrations: $$[NO_2] = 0.100 - 2x$$ $$[N_2O_4] = x$$
- Substitute into the $ K $ expression:$$
0.001 = \frac{x}{(0.100)^2}
$$ - Solve for $ x $:$$
x = 0.001 \times 0.100^2 = 0.00001 \, \text{mol dm}^{-3}
$$ - Verify the approximation:
- $ 2x = 0.00002 \, \text{mol dm}^{-3} $,
- $ 0.00002 \ll 0.100 $, so the approximation is valid.
- Calculate equilibrium concentrations: $$[NO_2] \approx 0.100 \, \text{mol dm}^{-3}$$ $$[N_2O_4] = x = 0.00001 \, \text{mol dm}^{-3}$$


