R2.3.6 Quantifying equilibrium concentrations (Higher Level Only)

Equilibrium Calculations in Chemistry

Using $K$ to Calculate Equilibrium Concentrations

As discussed previously, the equilibrium constant ( $K$ ) provides a numerical measure of the balance between reactants and products at equilibrium.
For a general reaction:

$a A + b B ⇌ c C + d D$

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.

Example

For the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , the equilibrium constant expression is: $K = \frac{[{NH}_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}$

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.

Example question

Calculate the equilibrium concentrations for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ , where $K = 50.0$ and the initial concentrations are:

$[H_{2}] = 0.100 {mol dm}^{- 3}$ ,
$[I_{2}] = 0.100 {mol dm}^{- 3}$ ,
$[H I] = 0 {mol dm}^{- 3}$ .

Solution

Write the $K$ expression: $K = \frac{[HI]^{2}}{[H_{2}] [I_{2}]}$
Define the changes using an ICE table:

Species	Initial ( ${mol dm}^{- 3}$ )	Change ( ${mol dm}^{- 3}$ )	Equilibrium ( ${mol dm}^{- 3}$ )
$H_{2}$	0.100	$- x$	$0.100 - x$
$I_{2}$	0.100	$- x$	$0.100 - x$
$H I$	0	$+ 2 x$	$2 x$

Substitute into the $K$ expression: $50.0 = \frac{(2 x)^{2}}{(0.100 - x) (0.100 - x)}$
Solve for $x$ :
Expand and simplify: $50.0 = \frac{4 x^{2}}{(0.100 - x)^{2}}$ $x = 0.078 {mol dm}^{- 3}$
Calculate equilibrium concentrations:
$[H_{2}] = [I_{2}] = 0.100 - 0.078 = 0.022 {mol dm}^{- 3}$ $[H I] = 2 x = 2 (0.078) = 0.156 {mol dm}^{- 3}$

Tip

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 $[Reactant] - x \approx [Reactant]$ .
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.

Example question

For the reaction $2 N O_{2} (g) ⇌ N_{2} O_{4} (g)$ , $K = 0.001$ and the initial concentration of $N O_{2}$ is $0.100 {mol dm}^{- 3}$ . Calculate the equilibrium concentrations.

Solution

Write the $K$ expression: $K = \frac{[N_{2} O_{4}]}{[N O_{2}]^{2}}$
Define equilibrium concentrations:
- $[N O_{2}] = 0.100 - 2 x$ ,
- $[N_{2} O_{4}] = x$ .
Substitute into the $K$ expression: $0.001 = \frac{x}{(0.100)^{2}}$
Solve for $x$ : $x = 0.001 \times {0.100}^{2} = 0.00001 {mol dm}^{- 3}$
Verify the approximation:
- $2 x = 0.00002 {mol dm}^{- 3}$ ,
- $0.00002 ≪ 0.100$ , so the approximation is valid.
Calculate equilibrium concentrations:
- $[N O_{2}] \approx 0.100 {mol dm}^{- 3}$ ,
- $[N_{2} O_{4}] = x = 0.00001 {mol dm}^{- 3}$ .

Common Mistake

Many students forget to verify the approximation by checking whether $x$ is less than 5% of the initial concentration.

Reflection

Theory of Knowledge

How does the use of approximations in equilibrium problems reflect the broader scientific practice of simplifying complex systems for analysis?

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