The molar enthalpies of combustion of $C_{2} H_{2} (g)$ C(graphite) and $H_{2} (g)$ and $- 1300, - 394$ and $- 286 kJ {mol}^{- 1}$ respectively. The standard enthalpy of formation of $C_{2} H_{2} (g)$ is

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$- 226 kJ {mol}^{- 1}$

$620 kJ {mol}^{- 1}$

$226 kJ {mol}^{- 1}$

$- 620 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

It is given that

$\begin{array}{ll} C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l) Δ H_{1} = - 1300 kJ {mol}^{- 1} \\ C (graphite) + O_{2} (g) \to {CO}_{2} (g) Δ H_{2} = - 394 kJ {mol}^{- 1} \\ H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H_{3} = - 286 kJ {mol}^{- 1} \end{array}$

Hence, for the equation

$\begin{array}{l} 2C(graphite) + H_{2} (g) \to C_{2} H_{2} (g) \\ Δ H = 2 Δ H_{2} + Δ H_{3} - Δ H_{1} = [- 2 \times 394 - 286 + (1300)] kJ {mol}^{- 1} = 226 kJ {mol}^{- 1} \end{array}$