IChO 2003 · Theory problem 11

Heat of neutralisation in a calorimeter

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10.0 cm\(^3\) of 0.50 M \(\ce{HCl}\) and 10.0 cm\(^3\) of 0.50 M \(\ce{NaOH}\) solutions, both at the same temperature, are mixed in a calorimeter. A temperature increase of \(\Delta T\) is recorded. Estimate the temperature increase if 5.0 cm\(^3\) of 0.50 M \(\ce{NaOH}\) were used instead of 10.0 cm\(^3\). Thermal losses are negligible and the specific heats of both solutions are taken as equal.

(a) \((1/2)\,\Delta T\) ( )

(b) \((2/3)\,\Delta T\) ( )

(c) \((3/4)\,\Delta T\) ( )

(d) \(\Delta T\) ( )

Official English text of the problem.

Answers

  • 11: (b) \(\boxed{\Delta T_{\mathrm{new}}=\frac{2}{3}\Delta T}\). Mark the parentheses beside (b).

Solution

11. The net neutralisation reaction is

\[ \ce{H+(aq) + OH-(aq) -> H2O(l)}. \]

The species are hydrogen ion [H+], hydroxide ion [OH-], and water O.

In the original experiment, the amounts of acid and base are each

\[ n_1=(0.50\ \mathrm{mol\,L^{-1}})(10.0\times10^{-3}\ \mathrm{L}) =5.0\times10^{-3}\ \mathrm{mol}. \]

They react completely in a \(1:1\) ratio. With only 5.0 cm\(^3\) of base, hydroxide becomes the limiting reactant:

\[ n_2=(0.50\ \mathrm{mol\,L^{-1}})(5.0\times10^{-3}\ \mathrm{L}) =2.5\times10^{-3}\ \mathrm{mol}. \]

The amount neutralised is halved, so the heat released is halved:

\[ \frac{q_2}{q_1}=\frac{n_2}{n_1}=\frac12. \]

For the estimate, take the dilute solutions to have equal densities. The ratio of their total masses is then

\[ \frac{m_2}{m_1} =\frac{10.0+5.0}{10.0+10.0} =\frac34. \]

Neglecting the calorimeter’s heat capacity, \(q=mc\Delta T\). Since the specific heat capacity \(c\) is the same in both experiments,

\[ \frac{\Delta T_{\mathrm{new}}}{\Delta T} =\frac{q_2}{q_1}\frac{m_1}{m_2} =\frac12\times\frac43 =\frac23. \]

Thus the correct choice is (b).

The answers were checked against the official answers published by the organisers.