Q.Calculate the enthalpy change for the reaction $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$ from the following data. $2Fe + \tfrac{3}{2}O_2 \rightarrow Fe_2O_3$; $\Delta H = -741$ kJ; $C + \tfrac{1}{2}O_2 \rightarrow CO$; $\Delta H = -137$ kJ; $C + O_2 \rightarrow CO_2$; $\Delta H = -394.5$ kJ

Step 1. Target: $Fe_2O_3+3CO\rightarrow2Fe+3CO_2$. Reverse reaction 1 ($2Fe+\tfrac{3}{2}O_2\rightarrow Fe_2O_3$, $\Delta H=-741$ kJ) to get $Fe_2O_3\rightarrow2Fe+\tfrac{3}{2}O_2$, $\Delta H=+741$ kJ.

Step 2. Build $3CO+\tfrac{3}{2}O_2\rightarrow3CO_2$ from reactions 2 and 3: per mole, $CO+\tfrac{1}{2}O_2\rightarrow CO_2 = (C+O_2\rightarrow CO_2) - (C+\tfrac{1}{2}O_2\rightarrow CO)$, giving $\Delta H=-394.5-(-137)=-257.5$ kJ per mole CO; for 3 moles, $3\times(-257.5)=-772.5$ kJ. …

Reverse reaction 1 (Fe$_2$O$_3$ is a reactant in the target but a product in the given equation), build the CO$\rightarrow$CO$_2$ leg from reactions 2 …

• Forgetting to multiply the CO$\rightarrow$CO$_2$ per-mole enthalpy by 3 (since the target needs 3 moles of …