Heat of formation of water is -68.3K.cal, heat of combustion of acetylene and ethylene are -310.6, -337.2 K.cal respectively, calculate the heat of reaction for the hydrogenation of acetylene at constant volume, at 25°C in K.cal

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-41.1

-32.2

-56.7

-22.4

answer is A.

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Detailed Solution

$\begin{array}{l} H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H_{1} = - 68.3 K . cal \\ C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 {CO}_{2} + H_{2} O; Δ H_{2} = - 310.6 K . cal \\ C_{2} H_{4} + 3 O_{2} \to 2 {CO}_{2} + 2 H_{2} O; Δ H_{3} = - 337.2 K . cal \\ C_{2} H_{2} + H_{2} \to C_{2} H_{4}; Δ H = ? \\ Δ H = Δ H_{1} + Δ H_{2} - Δ H_{3} \\ ∴ C_{2} H_{2} (g) + H_{2} (g) \to C_{2} H_{4} (g); Δ H = - 41.7 K . Cal \\ Δ H = Δ U + Δ n R T \\ Δ n = 1 - 2 = - 1; R = 2 cal, T = 298 K \\ Δ U = - 41.7 + 1 \times \frac{2}{1000} \times 298 K \\ = - 41.1 K . Cal \end{array}$