Question
ΔfH° for CO2(g) and H2O(l) are −94.0 and −68.0 kcal/mol. ΔfH° for the propanoic acid is one-third of the enthalpy of combustion of the acid and has the same sign as the latter. The value of ΔfH° of propanoic acid is
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Solution
$CH_{3}CH_{2}COOH(l) + \frac{7}{2}O_{2}(g) \rightarrow 3CO_{2}(g) + 3H_{2}O(l)$
$\Delta_{C}H_{CH_{3}CH_{2}COOH(l)} = \left\lbrack 3 \times \Delta_{f}H_{CO_{2}(g)} + 3 \times \Delta_{f}H_{H_{2}O(l)} \right\rbrack - \left\lbrack \Delta_{f}H_{CH_{3}CH_{2}COOH(l)} + \frac{7}{2} \times \Delta_{f}H_{O_{2}(g)} \right\rbrack$
$\text{Or, }3 \times \Delta_{f}H_{CH_{3}CH_{2}COOH(l)} = \left\lbrack 3( - 94) + 3( - 68) \right\rbrack - \left\lbrack \Delta_{f}H_{CH_{3}CH_{2}COOH(l)} + O \right\rbrack $$$\therefore\Delta_{f}H_{CH_{3}CH_{2}COOH(l)} = - 121.5\text{ kcal/mol}$$
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