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$∆ H_{1}^{0} for {CO}_{2} (g), CO (g) and H_{2} O (g) are - 393.5, - 110.5 and - 241.8 kJ {mol}^{- 1}$ respectively. Standard enthalpy change for the reaction
${CO}_{2} (g) + H_{2} (g) \to CO (g) + H_{2} O (g) is$

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$524.1 kJ {mol}^{- 1}$

$41.2 kJ {mol}^{- 1}$

$- 262.5 kJ {mol}^{- 1}$

$- 41.2 kJ {mol}^{- 1}$

answer is B.

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Detailed Solution

$∆ H_{reaction} = [∆ H_{f}^{0} (CO) + ∆ H_{f}^{0} (H_{2} O)] - [∆ H_{f}^{0} ({CO}_{2}) + ∆ H_{f}^{0} (H_{2})] = [- 110.5 + (- 241.8)] - [393.5 + 0] = 41.2 kJ$

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