Worked Solutions
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.
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 $T = \Delta H/\Delta S$.