CO(g) + H₂O(g) → CO₂(g) + H₂(g). Calculate the standard heat change if the standard enthalpies of formation of CO_2g, H₂O_(g) and CO_(g) in KJ mol^-1 are −394, −242 and −110 kJ mol⁻¹ respectively.

Correct answer: −42 kJ mol⁻¹ (Option D)

Explanation: Use standard enthalpies of formation and apply
\(\Delta H^\circ_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})\).

• Products: CO₂(g) = −394; H₂(g) = 0 → sum = −394 kJ mol⁻¹.
• Reactants: CO(g) = −110; H₂O(g) = −242 → sum = −352 kJ mol⁻¹.

\(\Delta H^\circ_{\text{rxn}} = (-394) - (-352) = -394 + 352 = \mathbf{-42\ kJ\ mol^{-1}}\).

Negative sign means the reaction is exothermic.