The delta_f H° for (g), (g) and (g) are -393.5, -110.5 and -241.8 kJ mol^-1, respectively. The standard enthalpy change (in kJ) for the reaction: (g) + (g) - (g) + (g) is

Question

The delta_f H° for $CO_2$(g), $CO$(g) and $H_2O$(g) are -393.5, -110.5 and -241.8 kJ mol^-1, respectively. The standard enthalpy change (in kJ) for the reaction: $CO_2$(g) + $H_2$(g) -> $CO$(g) + $H_2O$(g) is

Prashant Kotian · Accepted Answer

Correct answer: B. The standard enthalpy of reaction is calculated as: δ H° = sum(delta_f H°_products) - sum(delta_f H°_reactants). δ H° = [delta_f H°(, g) + delta_f H°(, g)] - [delta_f H°(, g) + delta_f H°(, g)]. Using the given values (and delta_f H°(, g) = 0): δ H° = [-110.5 + (-241.8)] - [-393.5 + 0] = -352.3 + 393.5 = +41.2 kJ.

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