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The enthalpy of combustion of C (graphite), $H_{2} (g)$ and ${CH}_{4} (g)$ -393,-242 and -803 kJ $m o l^{- 1}$ respectively. The enthalpy of formation of ${CH}_{4} (g)$ is

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-49.0 KJ $m o l^{- 1}$

-66.0 kJ $m o l^{- 1}$

-67.0 kJ $m o l^{- 1}$

-74.0 kJ $m o l^{- 1}$

answer is D.

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Detailed Solution

$C (graphite) + O_{2} (g) \to {CO}_{2} (g) + 393 kJ$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g) + 242 kJ$

${CH}_{4} (g) + O_{2} (g) \to {CO}_{2} + 2 H_{2} O (g) + 803 kJ$

$C (graphite) + 2 H_{2} (g) \to {CH}_{4} (g) + x kJ$

a + 2b – c = d, x = 393 +(242 x 2) – 803 = 74.0

$∴ △ H_{1} = - 74 .0 kJ {mol}^{- 1}$

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