2023 H2 Chem P1 Q9 — Thermochemistry Solution

The answer is A (\(-56.4\text{ kJ mol}^{-1}\)).
\(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)
Amount of \(\text{H}_2\text{SO}_4 = \frac{50.0}{1000} \times 2.00 = 0.100\) mol
Amount of NaOH \(= \frac{100}{1000} \times 1.00 = 0.100\) mol
0.100 mol of \(\text{H}_2\text{SO}_4\) would need 0.200 mol of NaOH, so NaOH is the limiting reagent and 0.100 mol of water is formed.
Heat released, \(q = mc\Delta T = (100 + 50.0)(4.18)(29.0 - 20.0) = 5643\) J
The enthalpy change of neutralisation is per mole of water formed, and the reaction is exothermic (the temperature rose), so
\(\Delta H = -\frac{5643}{0.100} = -56430\text{ J mol}^{-1} = -56.4\text{ kJ mol}^{-1}\)
• B is wrong: \(-28.2\text{ kJ mol}^{-1}\) comes from dividing 5643 J by 0.200 mol, as if all the \(\text{H}_2\text{SO}_4\) reacted to form 0.200 mol of water. Only 0.100 mol of water forms because NaOH runs out.
• C is wrong: the temperature rose, so heat was released and \(\Delta H\) must be negative; the value 28.2 also uses the wrong amount of water (0.200 mol).
• D is wrong: the size (56.4) is right, but the sign is not; an exothermic neutralisation has a negative \(\Delta H\).