Given the standard enthalpies at 298K in kJ `//` mol for the following two reactions
`Fe_(2) O_(3) ( s) + ( 3)/( 2) C ( s) rarr ( 3)/( 2) CO_(2) ( g) + 2Fe ` `Delta H = +234` .....(1)
`C ( s) + O_(2)(g) rarr CO_(2)(g)` `Delta H = - 393 ` ....(2)
The value of `Delta H ` for `4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3) ( s)` is

Options

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

Answer

To find the standard enthalpy change (ΔH) for the reaction:

\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]

we will use the given reactions and their enthalpy changes.

### Step 1: Write down the given reactions and their ΔH values.

1. **Reaction 1**:
\[ \text{Fe}_2\text{O}_3(s) + \frac{3}{2} \text{C}(s) \rightarrow \frac{3}{2} \text{CO}_2(g) + 2 \text{Fe} \]
\[ \Delta H = +234 \text{ kJ} \]

2. **Reaction 2**:
\[ \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \]
\[ \Delta H = -393 \text{ kJ} \]

### Step 2: Manipulate Reaction 1 to match the target reaction.

We need to have 4 Fe on the reactant side, so we will multiply Reaction 1 by 2:

\[ 2 \text{Fe}_2\text{O}_3(s) + 3 \text{C}(s) \rightarrow 3 \text{CO}_2(g) + 4 \text{Fe} \]

Now, the ΔH for this modified reaction will be:

\[ \Delta H = 2 \times 234 = +468 \text{ kJ} \]

### Step 3: Reverse Reaction 1.

To find the enthalpy change for the formation of Fe2O3, we need to reverse the modified Reaction 1:

\[ 4 \text{Fe} + 3 \text{CO}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]

When we reverse a reaction, the sign of ΔH also changes:

\[ \Delta H = -468 \text{ kJ} \]

### Step 4: Use Reaction 2.

Now we need to use Reaction 2 to get the required O2. We need 3 O2, so we will multiply Reaction 2 by 3:

\[ 3 \text{C}(s) + 3 \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) \]

The ΔH for this reaction will be:

\[ \Delta H = 3 \times (-393) = -1179 \text{ kJ} \]

### Step 5: Combine the reactions.

Now we can add the two reactions:

1. \( 4 \text{Fe} + 3 \text{CO}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \) (ΔH = -468 kJ)
2. \( 3 \text{C}(s) + 3 \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) \) (ΔH = -1179 kJ)

When we add these two reactions, the \(3 \text{CO}_2(g)\) cancels out:

\[ 4 \text{Fe} + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]

### Step 6: Calculate the total ΔH.

Now, we can calculate the total ΔH:

\[ \Delta H = -468 \text{ kJ} + (-1179 \text{ kJ}) = -1647 \text{ kJ} \]

### Conclusion

The value of ΔH for the reaction \( 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \) is:

\[ \Delta H = -1647 \text{ kJ} \]

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