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Q.

The $Δ H_{f}^{\circ}$ for ${CO}_{2} (g)$ , $CO (g)$ and $H_{2} O (g)$ are $- 393.5$ $- 110.5$ and $- 241.8 kJ mol$ respectively. The standard enthalpy change (in $kJ$ ) for the reaction ${CO}_{2} (g) + H_{2} (g) ⟶ CO (g) + H_{2} O (g)$ is :

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a

524.1

b

41.2

c

- 262.5

d

-41.2

answer is B.

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Detailed Solution

The general formula is

$ΔH = H_{f}^{\circ} (products) - H_{f}^{\circ} (reactants)$

Reactants are : ${CO}_{2} H_{2} $ and product are $CO$ and $H_{2} O$

$ΔH = - 100.5 + (- 241.8) - (- 393.5 + 0) = 41.2 kJ$

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The ΔHf∘for CO2 (g), CO (g) and H2O (g) are -393.5 -110.5 and -241.8 kJ mol respectively. The standard enthalpy change (in kJ) for the reaction CO2(g)+H2(g)⟶CO(g)+H2O(g) is :