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 ΔH for the reaction \( 4Fe_{(s)} + 3O_{2(g)} \rightleftharpoons 2Fe_2O_{3(s)} \), we will use the given reactions and their enthalpy changes.

### Step-by-Step Solution:

**Step 1: Identify the reactions and their enthalpy changes.**

We have the following reactions:

1. \( Fe_2O_{3(s)} + \frac{3}{2}C_{(s)} \rightleftharpoons \frac{3}{2}CO_{2(g)} + 2Fe \)
\( \Delta H = +234 \, \text{kJ} \) (Reaction 1)

2. \( C_{(s)} + O_{2(g)} \rightleftharpoons CO_{2(g)} \)
\( \Delta H = -393 \, \text{kJ} \) (Reaction 2)

**Step 2: Manipulate the first reaction.**

We need to obtain \( 4Fe \) from the first reaction. Since the first reaction produces \( 2Fe \), we will multiply the entire reaction by 2:

\[ 2Fe_2O_{3(s)} + 3C_{(s)} \rightleftharpoons 3CO_{2(g)} + 4Fe \]

Now, we also need to adjust the enthalpy change:

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

**Step 3: Reverse the second reaction.**

Next, we need to consider the formation of \( O_2 \) from \( CO_2 \). We will reverse the second reaction to get \( O_2 \) on the left side:

\[ 3CO_{2(g)} \rightleftharpoons 3C_{(s)} + 3O_{2(g)} \]

Reversing the reaction changes the sign of ΔH:

\[ \Delta H = +3 \times 393 = +1179 \, \text{kJ} \]

**Step 4: Combine the reactions.**

Now we add the modified first reaction and the reversed second reaction:

1. \( 2Fe_2O_{3(s)} + 3C_{(s)} \rightleftharpoons 3CO_{2(g)} + 4Fe \) (ΔH = +468 kJ)
2. \( 3CO_{2(g)} \rightleftharpoons 3C_{(s)} + 3O_{2(g)} \) (ΔH = +1179 kJ)

When we add these reactions, \( 3CO_{2(g)} \) cancels out:

\[ 2Fe_2O_{3(s)} + 3O_{2(g)} \rightleftharpoons 4Fe + 3C_{(s)} \]

**Step 5: Calculate the total ΔH.**

Now we sum the ΔH values:

\[ \Delta H = +468 + 1179 = +1647 \, \text{kJ} \]

**Step 6: Write the final reaction.**

Now we can write the final reaction:

\[ 4Fe + 3O_{2(g)} \rightleftharpoons 2Fe_2O_{3(s)} \]

### Final Answer:

The value of \( \Delta H \) for the reaction \( 4Fe_{(s)} + 3O_{2(g)} \rightleftharpoons 2Fe_2O_{3(s)} \) is \( +1647 \, \text{kJ} \).

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