The standard enthalpy of formation of `H_(2)O_((l)) and Fe_(2)O_(3(s))` are respectively `- 286kJ mol^(-1) and - 824 kJ mol^(-1)`. What is the standard enthalpy change for the following reaction `Fe_(2)O_(3(s)) + 3H_(2(g)) rarr 3H_(2)O_((l)) + 2Fe_((s))`

Options

A. `-538 kJ mol^(-1)`
B. `+538 kJ mol^(-1)`
C. `-102 kJ mol^(-1)`
D. `-34kJ mol^(-1)`

Explanation

To calculate the standard enthalpy change for the reaction:

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

we will use the standard enthalpy of formation values provided for the reactants and products.

### Step-by-step solution:

1. **Identify the standard enthalpy of formation values:**
- For water, \( \Delta H_f^\circ (\text{H}_2\text{O}(l)) = -286 \, \text{kJ/mol} \)
- For iron(III) oxide, \( \Delta H_f^\circ (\text{Fe}_2\text{O}_3(s)) = -824 \, \text{kJ/mol} \)
- The standard enthalpy of formation for elemental iron \( \Delta H_f^\circ (\text{Fe}(s)) = 0 \, \text{kJ/mol} \) (since it is in its standard state).
- The standard enthalpy of formation for hydrogen gas \( \Delta H_f^\circ (\text{H}_2(g)) = 0 \, \text{kJ/mol} \) (since it is in its standard state).

2. **Write the formula for the standard enthalpy change of the reaction:**
\[ \Delta H_{reaction} = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

3. **Calculate the enthalpy of formation for the products:**
- For \( 3 \text{H}_2\text{O}(l) \):
\[ 3 \times (-286 \, \text{kJ/mol}) = -858 \, \text{kJ} \]
- For \( 2 \text{Fe}(s) \):
\[ 2 \times 0 \, \text{kJ/mol} = 0 \, \text{kJ} \]
- Total for products:
\[ -858 \, \text{kJ} + 0 \, \text{kJ} = -858 \, \text{kJ} \]

4. **Calculate the enthalpy of formation for the reactants:**
- For \( \text{Fe}_2\text{O}_3(s) \):
\[ -824 \, \text{kJ/mol} \]
- For \( 3 \text{H}_2(g) \):
\[ 3 \times 0 \, \text{kJ/mol} = 0 \, \text{kJ} \]
- Total for reactants:
\[ -824 \, \text{kJ} + 0 \, \text{kJ} = -824 \, \text{kJ} \]

5. **Substitute into the enthalpy change formula:**
\[ \Delta H_{reaction} = (-858 \, \text{kJ}) - (-824 \, \text{kJ}) \]
\[ \Delta H_{reaction} = -858 \, \text{kJ} + 824 \, \text{kJ} = -34 \, \text{kJ} \]

### Final Answer:
The standard enthalpy change for the reaction is \( \Delta H_{reaction} = -34 \, \text{kJ} \).