Question

\( \mathrm{Fe_2O_{3(s)} + 3H_{2(g)} \rightarrow 2Fe_{(s)} + 3H_2O_{(l)}} \) \(\text{(Given: } \Delta H^\circ(\mathrm{H_2O_{(l)}}) = -285.83\,\text{kJ mol}^{-1},\ \Delta H^\circ(\mathrm{Fe_2O_{3(s)}}) = -824.2\,\text{kJ mol}^{-1}\text{)}\)

• A. -33.29
• B. +33.29
• C. +3.33
• D. -3.33
• E. -30.29

Hint:

Always remember: elements in their standard states have \(\Delta_f H^\circ = 0\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Enthalpy of reaction \(\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})\).

Step 2: Detailed Explanation:
\(\Delta_f H^\circ\) for Fe(s) and H2(g) is 0 by definition.
\(\Delta_r H^\circ = [2(0) + 3(-285.83)] - [1(-824.2) + 3(0)]\)
\(= [-857.49] - [-824.2] = -857.49 + 824.2 = -33.29 \text{ kJ/mol}\).

Step 3: Final Answer:
The enthalpy of reaction is -33.29 kJ.