Question

The standard enthalpy of formation of H₂O(l) and Fe₂O₃(s) are respectively −286 kJ mol⁻¹ and −824 kJ mol⁻¹. What is the standard enthalpy change for the following reaction?

\[ \text{Fe}_2\text{O}_3(s) + 3\text{H}_2(g) \rightarrow 3\text{H}_2\text{O}(l) + 2\text{Fe}(s) \]

• A. $-538$ kJ mol$^{-1}$
• B. $+538$ kJ mol$^{-1}$
• C. $-102$ kJ mol$^{-1}$
• D. $+34$ kJ mol$^{-1}$
• E. $-34$ kJ mol$^{-1}$

Hint:

Elements in standard state have zero enthalpy of formation.

Answer 1

Answer: (E)

Concept: Enthalpy of reaction using formation enthalpies
\[ \Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \] ---

Step 1: Write values
\[ \Delta H_f^\circ (\text{H}_2\text{O}) = -286 \] \[ \Delta H_f^\circ (\text{Fe}_2\text{O}_3) = -824 \] \[ \Delta H_f^\circ (\text{H}_2) = 0, \quad \Delta H_f^\circ (\text{Fe}) = 0 \] ---

Step 2: Products side
\[ 3 \times (-286) + 2 \times 0 = -858 \] ---

Step 3: Reactants side
\[ -824 + 3 \times 0 = -824 \] ---

Step 4: Calculate enthalpy change
\[ \Delta H = (-858) - (-824) \] \[ = -858 + 824 = -34 \text{ kJ mol}^{-1} \] --- Final Answer: \[ \boxed{-34 \text{ kJ mol}^{-1}} \]