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Example 5.8: Measurement of an Enthalpy Change

When 0.0500 mol of \(\ce{HCl(aq)}\) reacts with 0.0500 mol of \(\ce{NaOH(aq)}\) to form 0.0500 mol of \(\ce{NaCl(aq)}\), 2.9 kJ of heat are produced. What is the molar enthalpy change \(ΔH\) (i.e. enthalpy per mole of reaction) for the acid-base reaction run under the conditions described in Example 5.4?

Solution

\(ΔH_{\mathrm{rxn}}\) = ?


\(n_{\mathrm{\ce{HCl},rxn}}\) \(= 0.0500\ \mathrm{mol}\)


\(n_{\mathrm{\ce{NaOH},rxn}}\) \(= 0.0500\ \mathrm{mol}\)


\(n_{\mathrm{\ce{NaCl},rxn}}\) \(= 0.0500\ \mathrm{mol}\)


\(q_{\mathrm{rxn}}\) \(= -2.9\ \mathrm{kJ}\)


The heat generated depends on the amount of substance that reacts (double the reaction, double the heat), so it is an extensive quantity. On the other hand, the molar enthalpy change \(ΔH\) is per amount of reaction \(n_{\mathrm{\ce{->}}}\), so it is an intensive quantity:


Plan: 1) \(n_{\mathrm{\ce{->}}}\) 2) \(ΔH_{\mathrm{rxn}}\)
In this case, calculating \(n_{\mathrm{\ce{->}}}\) is easy because it says how much of each substance is made or used up (and their ratios correspond to the 1:1:1 ratio given by the chemical equation).

\(ν_{\mathrm{\ce{HCl}}}\) \(= 1\)


\(n_{\mathrm{\ce{->},rxn}}\) \(= \dfrac{n_{\mathrm{\ce{HCl},rxn}}}{ν_{\mathrm{\ce{HCl}}}}\)

\(\ \ \ =\dfrac{0.0500\ \mathrm{mol}}{1}\)

\(\ \ \ =0.0500\ \mathrm{mol}\)


\(ΔH\) \(= \dfrac{q_{\mathrm{rxn}}}{n_{\mathrm{\ce{->},rxn}}}\)

\(\ \ \ =\dfrac{-2.9\ \mathrm{kJ}}{0.0500\ \mathrm{mol}}\)

\(\ \ \ =-58.\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)


We could have used \(n_{\mathrm{\ce{->}}} = \dfrac{n_{\mathrm{\ce{NaOH},rxn}}}{ν_{\mathrm{\ce{NaOH}}}}\) or \(n_{\mathrm{\ce{->}}} = \dfrac{n_{\mathrm{\ce{NaCl},rxn}}}{ν_{\mathrm{\ce{NaCl}}}}\) instead...