What will be the heat of reaction for the following reaction? Will the reaction be exothermic or endothermic?
`Fe_(2)O_(3(s))+3H_(2(g))rarr 2Fe_((s))+3H_(2)O_((l))`
`Delta_(f)H^(@)(H_(2)O,l)=-"285.83 kJ mol"^(-1)`
`Delta_(f)H^(@)(Fe_(2)O_(3),s)=-"824.2 kJ mol"^(-1)`

Options

A. `-"824. 2 kJ mol"^(-1),` exothermic
B. `+"33.3 kJ mol"^(-1),` endothermic
C. `-"33.3 kJ mol"^(-1)`, exothermic
D. `+"824.2 kJ mol"^(-1),` endothermic

Explanation

To find the heat of reaction for the given chemical equation:

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

we will use the enthalpy of formation values provided:

- \(\Delta_f H^\circ(H_2O, l) = -285.83 \, \text{kJ/mol}\)
- \(\Delta_f H^\circ(Fe_2O_3, s) = -824.2 \, \text{kJ/mol}\)

### Step 1: Write the formula for the heat of reaction.

The heat of reaction (\(\Delta H_{reaction}\)) can be calculated using the formula:

\[ \Delta H_{reaction} = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

### Step 2: Identify the products and reactants.

**Products:**
- 2 moles of Fe (solid)
- 3 moles of H2O (liquid)

**Reactants:**
- 1 mole of Fe2O3 (solid)
- 3 moles of H2 (gas)

### Step 3: Substitute the values into the formula.

Since the enthalpy of formation for elements in their standard state is zero, we have:

- \(\Delta H_f(Fe) = 0 \, \text{kJ/mol}\)
- \(\Delta H_f(H_2) = 0 \, \text{kJ/mol}\)

Now, substituting the values:

\[ \Delta H_{reaction} = [2 \times \Delta H_f(Fe) + 3 \times \Delta H_f(H_2O)] - [\Delta H_f(Fe_2O_3) + 3 \times \Delta H_f(H_2)] \]

Substituting the known values:

\[ \Delta H_{reaction} = [2 \times 0 + 3 \times (-285.83)] - [-824.2 + 3 \times 0] \]

\[ \Delta H_{reaction} = [0 - 857.49] - [-824.2] \]

\[ \Delta H_{reaction} = -857.49 + 824.2 \]

\[ \Delta H_{reaction} = -33.29 \, \text{kJ/mol} \]

### Step 4: Determine if the reaction is exothermic or endothermic.

Since the heat of reaction is negative (\(-33.29 \, \text{kJ/mol}\)), this indicates that the reaction releases heat. Therefore, the reaction is **exothermic**.

### Final Answer:
The heat of reaction is \(-33.29 \, \text{kJ/mol}\) and the reaction is **exothermic**.

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