Module 4: Drivers of Reactions — Worked Solutions (Preliminary Chemistry)
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Worked examples for Preliminary Chemistry Module 4: Drivers of 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.
In short: full step-by-step worked solutions for Preliminary Chemistry — Module 4: Drivers of Reactions. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
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 — Enthalpy of combustion by calorimetry
Question
In a calorimetry experiment, burning 1.15 g of ethanol ($\text{C}_2\text{H}_5\text{OH}$, $M = 46.08\ \text{g mol}^{-1}$) raised the temperature of 250.0 g of water by 32.0 °C. Using $q = mc\Delta T$ with $c = 4.18\ \text{J g}^{-1}\,°\text{C}^{-1}$, calculate the molar heat of combustion of ethanol in $\text{kJ mol}^{-1}$.
Solution
First the heat absorbed by the water:
$q = mc\Delta T = 250.0 \times 4.18 \times 32.0 = 33\,440\ \text{J} = 33.44\ \text{kJ}$.
Then the moles of ethanol burned:
$n = m/M = 1.15 / 46.08 = 0.02496\ \text{mol}$.
Molar heat of combustion is heat released per mole:
$\Delta H_c = q / n = 33.44 / 0.02496 = 1340\ \text{kJ mol}^{-1}$ (3 s.f.).
Combustion is exothermic, so report it as $\Delta H_c = -1340\ \text{kJ mol}^{-1}$. Don't forget the negative sign — energy is released to the surroundings.
The plan is: find how much heat the water gained, find how many moles of fuel produced it, then divide to get the heat per mole.
Start with the water, using $q = mc\Delta T$. The water gained:
$q = 250.0 \times 4.18 \times 32.0 = 33\,440\ \text{J}$, which is $33.44\ \text{kJ}$.
This heat came from burning the ethanol, so now find the moles of ethanol with $n = m/M$:
$n = 1.15 / 46.08 = 0.02496\ \text{mol}$.
The molar heat of combustion is simply the energy divided by the amount that produced it:
$\Delta H_c = q / n = 33.44 / 0.02496 = 1340\ \text{kJ mol}^{-1}$.
Because combustion gives out heat, it's exothermic, so we write it as a negative: $\Delta H_c = -1340\ \text{kJ mol}^{-1}$. The sign is the chemistry telling us energy flowed out of the reaction into the water.
Heat to water:
- $q = mc\Delta T = 250.0 \times 4.18 \times 32.0 = 33\,440\ \text{J} = 33.44\ \text{kJ}$
Moles of ethanol:
- $n = 1.15 / 46.08 = 0.02496\ \text{mol}$
Molar heat of combustion:
- $\Delta H_c = q/n = 33.44 / 0.02496 = 1340\ \text{kJ mol}^{-1}$
- Exothermic → $\Delta H_c = -1340\ \text{kJ mol}^{-1}$
Where the marks go
- 1 mark: Correct heat absorbed by water using $q = mc\Delta T$ ($33.44\ \text{kJ}$)
- 1 mark: Correct moles of ethanol ($0.02496\ \text{mol}$)
- 1 mark: Correct molar heat of combustion magnitude ($1340\ \text{kJ mol}^{-1}$, 3 s.f.)
- 1 mark: States the value as negative (exothermic): $-1340\ \text{kJ mol}^{-1}$
Key idea
Molar heat of combustion $= q/n$, where $q = mc\Delta T$ is the heat gained by the water and $n$ is the moles of fuel burned; combustion is exothermic so $\Delta H$ is negative.
Example 2 — Gibbs free energy and spontaneity
Question
For the decomposition $\text{CaCO}_3{}_{(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_2{}_{(g)}$, $\Delta H = +178\ \text{kJ mol}^{-1}$ and $\Delta S = +161\ \text{J K}^{-1}\,\text{mol}^{-1}$. Using $\Delta G = \Delta H - T\Delta S$, determine whether the reaction is spontaneous at 298 K, and find the temperature above which it becomes spontaneous.
Solution
Use $\Delta G = \Delta H - T\Delta S$, but first make the units match — convert $\Delta S$ to $\text{kJ}$: $161\ \text{J K}^{-1} = 0.161\ \text{kJ K}^{-1}\,\text{mol}^{-1}$.
At 298 K:
$\Delta G = 178 - (298)(0.161) = 178 - 47.98 = +130\ \text{kJ mol}^{-1}$.
$\Delta G > 0$, so the reaction is non-spontaneous at 298 K.
It becomes spontaneous when $\Delta G < 0$, i.e. at the crossover $\Delta G = 0$:
$T = \dfrac{\Delta H}{\Delta S} = \dfrac{178}{0.161} = 1106\ \text{K}$.
So it's spontaneous above about 1106 K (≈ 833 °C). Watch the unit mismatch — $\Delta S$ in joules, $\Delta H$ in kilojoules — it catches people out every year.
Spontaneity is decided by the sign of $\Delta G$ in $\Delta G = \Delta H - T\Delta S$: negative means spontaneous, positive means not.
The first thing to fix is units — $\Delta H$ is in kJ but $\Delta S$ is in J, so convert $\Delta S$ to kJ: $161\ \text{J K}^{-1}\,\text{mol}^{-1} = 0.161\ \text{kJ K}^{-1}\,\text{mol}^{-1}$.
Now at 298 K:
$\Delta G = 178 - (298)(0.161) = 178 - 47.98 = +130\ \text{kJ mol}^{-1}$.
Since $\Delta G$ is positive, the reaction is not spontaneous at room temperature.
But notice both $\Delta H$ and $\Delta S$ are positive — the entropy term $T\Delta S$ grows as we heat the system, and eventually it overtakes $\Delta H$. The turning point is exactly where $\Delta G = 0$:
$T = \dfrac{\Delta H}{\Delta S} = \dfrac{178}{0.161} = 1106\ \text{K}$.
Above this temperature $T\Delta S$ wins, $\Delta G$ becomes negative, and decomposition becomes spontaneous — which is why limestone is "cooked" at high temperatures industrially.
Convert: $\Delta S = 161\ \text{J K}^{-1} = 0.161\ \text{kJ K}^{-1}\,\text{mol}^{-1}$.
At 298 K:
- $\Delta G = 178 - (298)(0.161) = +130\ \text{kJ mol}^{-1}$
- $\Delta G > 0$ → non-spontaneous
Crossover ($\Delta G = 0$):
- $T = \Delta H / \Delta S = 178 / 0.161 = 1106\ \text{K}$
- Spontaneous above ≈ 1106 K
Where the marks go
- 1 mark: Converts $\Delta S$ to consistent units ($0.161\ \text{kJ K}^{-1}\,\text{mol}^{-1}$)
- 1 mark: Calculates $\Delta G = +130\ \text{kJ mol}^{-1}$ at 298 K and concludes non-spontaneous
- 1 mark: Finds the crossover temperature $T = \Delta H/\Delta S \approx 1106\ \text{K}$
Key idea
A reaction is spontaneous when $\Delta G = \Delta H - T\Delta S < 0$; when $\Delta H$ and $\Delta S$ are both positive, raising temperature eventually makes $\Delta G$ negative at $T = \Delta H/\Delta S$.
Frequently asked questions
Step-by-step solutions to Drivers of Reactions questions in Preliminary Chemistry, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Chemistry, using the methods and notation expected in exams and assessments.