Worked Solutions
Module 5: Equilibrium and Acid Reactions — Worked Solutions (HSC Chemistry)
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Worked examples for HSC Chemistry Module 5: Equilibrium and Acid Reactions. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Le Chatelier's principle
Question
The following gaseous equilibrium is exothermic in the forward direction:
$$\text{N}_2\text{O}_4{}_{(g)} \rightleftharpoons 2\text{NO}_2{}_{(g)} \quad \Delta H = +57\ \text{kJ mol}^{-1}$$
$\text{N}_2\text{O}_4$ is colourless and $\text{NO}_2$ is brown. Predict and explain the effect on the position of equilibrium and the colour of the mixture when (i) the temperature is increased and (ii) the total pressure is increased at constant temperature.
Solution
Le Chatelier: the system shifts to oppose any imposed change.
(i) The forward reaction is endothermic ($\Delta H = +57\ \text{kJ mol}^{-1}$). Adding heat is opposed by absorbing heat, so equilibrium shifts right. More $\text{NO}_2{}_{(g)}$ forms, so the mixture goes darker brown.
(ii) Raising the pressure is opposed by shifting to the side with fewer gas moles. The left has 1 mol, the right has 2 mol, so equilibrium shifts left. $\text{N}_2\text{O}_4{}_{(g)}$ increases, so the mixture goes paler.
Always quote the direction and the reason (heat term or mole count). A bare "shifts right" loses the explain mark.
Le Chatelier's principle says that when we disturb a system at equilibrium, it responds in the direction that partly undoes the disturbance — so let's apply that idea twice.
(i) The forward reaction is endothermic, meaning it absorbs heat ($\Delta H = +57\ \text{kJ mol}^{-1}$). When we add heat by raising the temperature, the system "uses up" that extra heat by favouring the forward reaction, so equilibrium shifts right. Since $\text{NO}_2{}_{(g)}$ is the brown gas, making more of it turns the mixture a darker brown.
(ii) Increasing the pressure squeezes the gas, so the system relieves that by moving to whichever side has fewer gas molecules. Counting moles: 1 mol of $\text{N}_2\text{O}_4{}_{(g)}$ on the left versus 2 mol of $\text{NO}_2{}_{(g)}$ on the right, so it shifts left to reduce the number of particles. More colourless $\text{N}_2\text{O}_4{}_{(g)}$ forms, so the colour fades to a paler shade.
Disturbance → shift opposing it.
(i) Increase $T$:
- Forward is endothermic ($\Delta H > 0$)
- Shift consumes added heat → right
- More $\text{NO}_2{}_{(g)}$ → darker brown
(ii) Increase $P$:
- Shift to fewer gas mol: left = 1, right = 2
- Shift → left
- More $\text{N}_2\text{O}_4{}_{(g)}$ → paler
Where the marks go
- 1 mark: States equilibrium shifts right on heating with a colour change to darker brown
- 1 mark: Justifies (i) using the endothermic forward reaction / heat term
- 1 mark: States equilibrium shifts left on increased pressure with a paler colour
- 1 mark: Justifies (ii) by comparing the moles of gas on each side (1 vs 2)
Key idea
Le Chatelier: a system shifts to oppose change — toward the endothermic side when heated, and toward fewer gas moles when pressure rises.
Example 2 — Calculating the equilibrium constant
Question
At a fixed temperature the following reaction reaches equilibrium in a $2.0\ \text{L}$ container:
$$\text{H}_2{}_{(g)} + \text{I}_2{}_{(g)} \rightleftharpoons 2\text{HI}_{(g)}$$
At equilibrium the container holds $0.20\ \text{mol}$ of $\text{H}_2{}_{(g)}$, $0.20\ \text{mol}$ of $\text{I}_2{}_{(g)}$ and $1.6\ \text{mol}$ of $\text{HI}_{(g)}$. Calculate the value of the equilibrium constant $K_{eq}$ at this temperature.
Solution
$K_{eq}$ uses equilibrium concentrations, so convert moles to $\text{mol L}^{-1}$ first by dividing by $2.0\ \text{L}$.
$[\text{H}_2] = 0.10$, $[\text{I}_2] = 0.10$, $[\text{HI}] = 0.80\ \text{mol L}^{-1}$.
$$K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.80)^2}{(0.10)(0.10)} = \frac{0.64}{0.010} = 64$$
$K_{eq} = 64$ (no units — the powers cancel here). Don't forget to square $[\text{HI}]$; that coefficient of 2 is where marks go missing.
The equilibrium constant compares products to reactants, but it's built from concentrations, not moles — so our first job is to convert. Dividing each amount by the $2.0\ \text{L}$ volume:
$[\text{H}_2] = 0.20 / 2.0 = 0.10$, $[\text{I}_2] = 0.10$, and $[\text{HI}] = 1.6 / 2.0 = 0.80\ \text{mol L}^{-1}$.
Now we write the expression. Each species is raised to the power of its coefficient in the balanced equation, and $\text{HI}$ has a coefficient of 2, so it gets squared:
$$K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.80)^2}{(0.10)(0.10)} = \frac{0.64}{0.010} = 64$$
So $K_{eq} = 64$. Here the units cancel because there are two moles of gas on each side, which is why we leave it as a plain number.
Convert mol → $\text{mol L}^{-1}$ (÷ 2.0 L):
- $[\text{H}_2] = 0.10$
- $[\text{I}_2] = 0.10$
- $[\text{HI}] = 0.80$
$$K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.80)^2}{(0.10)(0.10)} = 64$$
$K_{eq} = 64$ (dimensionless).
Where the marks go
- 1 mark: Converts moles to concentrations using the 2.0 L volume
- 1 mark: Writes the correct $K_{eq}$ expression with $[\text{HI}]$ squared
- 1 mark: Correct value $K_{eq} = 64$
Key idea
$K_{eq}$ uses equilibrium concentrations (not moles), with each species raised to its balancing coefficient.