Question 14.3: Calculate the value of ΔU°rxn at 25°C for the reaction descr......

Calculate the value of ΔU_{rxn}^{\circ} at 25°C for the reaction described by

2\ H_{2}(g) + O_{2}(g) → 2\ H_{2}O(l) \qquad ΔH_{rxn}^{\circ} = –571.6\ kJ·mol^{–1}
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In this case, water is a liquid and so

\Delta v_{gas} = 0  –  (v_{H_{2}} + v_{O_{2}}) = 0  –  (2 + 1) = – 3

Thus at 25°C,

P\Delta V_{rxn}^{\circ} = \Delta v_{gas}RT = – 3RT = (–3)(8.3145\ J·mol^{–1}·K^{–1})(298\ K)= –7.43\ kJ·mol^{–1}

and, therefore, ΔH_{rxn}^{\circ} and ΔU_{rxn}^{\circ} differ by about 7\ kJ·mol^{–1}. We now use Equation 14.10 to calculate the value of ΔU_{rxn}^{\circ}:

ΔH_{rxn} = ΔU_{rxn} + P\Delta V_{rxn} \qquad (constant\ pressure)                    (14.10)

ΔU_{rxn}^{\circ} = ΔH_{rxn}^{\circ}  –  P\Delta V_{rxn}^{\circ} = –571.6\ kJ·mol^{–1}  –  (–7.43\ kJ·mol^{–1}) = –564.2\ kJ·mol^{–1}

Hence, in this case, the values of ΔU_{rxn}^{\circ} and ΔH_{rxn}^{\circ} differ by only about 1%; this will usually be the case.

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