Consider the stepwise reduction: (1) kJ. (2) kJ. What is the enthalpy change for ? — Thermodynamics and Thermochemistry Chemistry Question
Question
Consider the stepwise reduction: (1) $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g), \Delta H_1 = -26.8$ kJ. (2) $FeO(s) + CO(g) \rightarrow Fe(s) + CO_2(g), \Delta H_2 = -16.5$ kJ. What is the enthalpy change for $Fe_2O_3(s) + CO(g) \rightarrow 2FeO(s) + CO_2(g)$?
Answer: A
💡 Solution & Explanation
Apply Hess's Law. Reverse equation 2 and multiply by 2: $2Fe + 2CO_2 \rightarrow 2FeO + 2CO$ ($\Delta H = +33.0$ kJ). Add this to equation 1: $Fe_2O_3 + 3CO + 2Fe + 2CO_2 \rightarrow 2Fe + 3CO_2 + 2FeO + 2CO$. Net: $Fe_2O_3 + CO \rightarrow 2FeO + CO_2$. $\Delta H = -26.8 + 33.0 = +6.2$ kJ.
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