Given the standard enthalpies at 298 k in kj/mol for the following two reactions
`Fe_2O_(3(s))+3/2C_(s) hArr 3/2CO_(2(g))+2Fe DeltaH=+234` …………(1)
`C_(s)+O_(2(g)) hArr CO_(2(g)) DeltaH=-393`
the value of `DeltaH` for `4Fe_((s))+3O_(2(g))hArr 2Fe_2O_3` is

Options

A. `3/2 (-393)+2xx234`
B. `3/2 (-393)+234`
C. `3(-393+2xx234`)
D. `3(-393)-2(234)`

Answer

To find the value of \( \Delta H \) for the reaction \( 4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \), we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step 1: Write down the given reactions and their enthalpy changes.

1. **Reaction 1:**
\[ Fe_2O_{3(s)} + \frac{3}{2}C_{(s)} \rightleftharpoons \frac{3}{2}CO_{2(g)} + 2Fe_{(s)} \quad \Delta H_1 = +234 \, \text{kJ/mol} \]

2. **Reaction 2:**
\[ C_{(s)} + O_{2(g)} \rightleftharpoons CO_{2(g)} \quad \Delta H_2 = -393 \, \text{kJ/mol} \]

### Step 2: Manipulate the reactions to obtain the desired reaction.

To derive the reaction \( 4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \), we need to reverse and adjust the coefficients of the given reactions.

- **Reverse Reaction 1** and multiply it by 2:
\[ 2Fe_{(s)} + \frac{3}{2}CO_{2(g)} \rightarrow Fe_2O_{3(s)} + \frac{3}{2}C_{(s)} \quad \Delta H = -2 \times 234 = -468 \, \text{kJ/mol} \]

- **Multiply Reaction 2** by 3:
\[ 3C_{(s)} + 3O_{2(g)} \rightarrow 3CO_{2(g)} \quad \Delta H = 3 \times (-393) = -1179 \, \text{kJ/mol} \]

### Step 3: Add the manipulated reactions.

Now we add the two manipulated reactions:

1. From the reversed Reaction 1:
\[ 2Fe_{(s)} + \frac{3}{2}CO_{2(g)} \rightarrow Fe_2O_{3(s)} + \frac{3}{2}C_{(s)} \quad \Delta H = -468 \, \text{kJ/mol} \]

2. From the multiplied Reaction 2:
\[ 3C_{(s)} + 3O_{2(g)} \rightarrow 3CO_{2(g)} \quad \Delta H = -1179 \, \text{kJ/mol} \]

When we add these two reactions, we get:
\[ 4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \]

### Step 4: Calculate the total enthalpy change.

Now we sum the enthalpy changes:
\[ \Delta H = -468 + (-1179) = -1647 \, \text{kJ/mol} \]

### Final Answer:
The value of \( \Delta H \) for the reaction \( 4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \) is \( \Delta H = -1647 \, \text{kJ/mol} \).

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