2021 H2 Chem P1 Q11 — Thermochemistry Solution

The answer is C (\(-41.8\) kJ mol\(^{-1}\)).

\(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)
\(n_{\text{NaOH}} = \frac{20.0}{1000} \times 5.00 = 0.100\) mol and \(n_{\text{H}_2\text{SO}_4} = \frac{20.0}{1000} \times 5.00 = 0.100\) mol.
0.100 mol of NaOH needs only 0.0500 mol of \(\text{H}_2\text{SO}_4\), so \(\text{H}_2\text{SO}_4\) is in excess and NaOH is limiting.
\(n_{\text{H}_2\text{O}}\) formed = \(n_{\text{NaOH}}\) = 0.100 mol

Mass of solution = 20.0 + 20.0 = 40.0 g (density 1 g cm\(^{-3}\))
\(q = -mc\Delta T = -40.0 \times 4.18 \times (50.0 - 25.0) = -4180\) J
\(\Delta H_{\text{neutralisation}} = \frac{q}{n_{\text{H}_2\text{O}}} = \frac{-4180}{0.100} = -41\,800\) J mol\(^{-1}\) = \(-41.8\) kJ mol\(^{-1}\)

• A is wrong: \(-4.18\) kJ is the heat change \(q\) for the whole experiment. It has not been divided by the 0.100 mol of water formed, so it is not a molar enthalpy change.
• B is wrong: \(-20.9\) kJ mol\(^{-1}\) = \(\frac{-4180}{0.200}\) J mol\(^{-1}\), which divides by 0.200 mol of water, the amount 0.100 mol of \(\text{H}_2\text{SO}_4\) would give if it all reacted. NaOH is the limiting reagent, so only 0.100 mol of water forms.
• D is wrong: \(-83.6\) kJ mol\(^{-1}\) = \(\frac{-4180}{0.0500}\) J mol\(^{-1}\), which is per mole of \(\text{H}_2\text{SO}_4\) reacted. The enthalpy change of neutralisation is defined per mole of water formed.