Equilibrium (OCR A-Level Chemistry A): Revision Notes

Worked Example 1: Hydrogen Iodide Formation

Consider the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$

The equilibrium constant expression is:

$K_c = \frac{[\text{HI}(g)]^2}{[\text{H}_2(g)][\text{I}_2(g)]}$

Substituting units:

$K_c = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}$

When we work this out, the numerator gives (mol dm⁻³)² and the denominator also gives (mol dm⁻³)², so all units cancel completely.

Result: $K_c$ has no units for this reaction.

Worked Example 2: Dinitrogen Tetroxide Decomposition

Consider the reaction: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$

The equilibrium constant expression is:

$K_c = \frac{[\text{NO}_2(g)]^2}{[\text{N}_2\text{O}_4(g)]}$

Substituting units:

$K_c = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})}$

This simplifies to (mol dm⁻³)¹, since the power 2 minus power 1 leaves one mol dm⁻³ term.

Result: Units = mol dm⁻³

Worked Example 3: Sulfur Trioxide Formation

Consider the reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$

The equilibrium constant expression is:

$K_c = \frac{[\text{SO}_3(g)]^2}{[\text{SO}_2(g)]^2[\text{O}_2(g)]}$

Substituting units:

$K_c = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})}$

The numerator has (mol dm⁻³)² and the denominator has (mol dm⁻³)³, so one (mol dm⁻³) term remains in the denominator, which we write as (mol dm⁻³)⁻¹ or dm³ mol⁻¹.

Result: Units = dm³ mol⁻¹

Worked Example: Nitrogen Dioxide Formation

Nitrogen monoxide (NO) and oxygen (O₂) react together in a reversible reaction:

$2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g)$

Initially, 1.60 mol of NO(g) and 1.40 mol of O₂(g) are mixed in a sealed container with volume 4.0 dm³. When equilibrium is established, 1.20 mol of NO₂(g) has formed. Calculate $K_c$ for this reaction.

Step 1: Work out equilibrium amounts of all species

We'll use a table to organize our information systematically:

How to fill in the table:

• The question tells us 1.20 mol of NO₂ has formed, so the change for NO₂ is +1.20 mol
• Using the stoichiometry (the balancing numbers), if 2 mol of NO₂ forms, then 2 mol of NO reacts and 1 mol of O₂ reacts
• Therefore, the change for NO is −1.20 mol and for O₂ is −0.60 mol (half of 1.20)
• Work out equilibrium amounts by adding initial amounts and changes

Step 2: Convert amounts to concentrations

The total volume is 4.0 dm³, so we divide each equilibrium amount by 4.0 to get concentrations in mol dm⁻³:

$[\text{NO}(g)] = \frac{0.40}{4} = 0.10 \text{ mol dm}^{-3}$

$[\text{O}_2(g)] = \frac{0.80}{4} = 0.20 \text{ mol dm}^{-3}$

$[\text{NO}_2(g)] = \frac{1.20}{4} = 0.30 \text{ mol dm}^{-3}$

Step 3: Write the Kc expression and calculate the value

$K_c = \frac{[\text{NO}_2(g)]^2}{[\text{NO}(g)]^2[\text{O}_2(g)]}$

Substituting the equilibrium concentrations:

$K_c = \frac{(0.30)^2}{(0.10)^2 \times 0.20} = \frac{0.09}{0.01 \times 0.20} = \frac{0.09}{0.002} = 45 \text{ dm}^3 \text{ mol}^{-1}$

Worked Example: Calculating Kc from Experimental Data

Step 1: Determine the equilibrium amount of CH₃COOH

The titration of the equilibrium mixture measures all the acid present - both the HCl catalyst and the unreacted CH₃COOH:

Equilibrium amount of CH₃COOH = 0.115 − 0.0500 = 0.065 mol

Step 2: Use stoichiometry to find equilibrium amounts of all components

Let's set up a table:

Working out the changes:

• We found that 0.065 mol of CH₃COOH remains at equilibrium
• Starting with 0.100 mol, the change must be 0.100 − 0.065 = −0.035 mol
• According to the 1:1:1:1 stoichiometry, the same changes apply to all species
• Therefore, 0.035 mol of ester has formed

Step 3: Convert to concentrations

The volume is 20 cm³ = 0.0200 dm³ (remember to convert cm³ to dm³ by dividing by 1000)

$[\text{CH}_3\text{COOH}] = \frac{0.065}{0.0200} = 3.25 \text{ mol dm}^{-3}$

$[\text{C}_2\text{H}_5\text{OH}] = \frac{0.065}{0.0200} = 3.25 \text{ mol dm}^{-3}$

$[\text{CH}_3\text{COOC}_2\text{H}_5] = \frac{0.035}{0.0200} = 1.75 \text{ mol dm}^{-3}$

$[\text{H}_2\text{O}] = \frac{0.535}{0.020} = 26.75 \text{ mol dm}^{-3}$

Step 4: Calculate Kc

$K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$

$K_c = \frac{1.75 \times 26.75}{3.25 \times 3.25} = \frac{46.8125}{10.5625} = 4.43$

Note that for this particular reaction, all the units cancel out, so $K_c$ has no units.