The following teo reaction are known :
`Fe_(2)O_(3)(s)+3CO(g)rarr2Fe(s)+3CO_(2)(g)`,
`Delta H= -26.8 kJ`
`FeO(s)+CO(g)rarr Fe(s)+CO_(2)(g)` ,
`Delta H = - 16.5 kJ`
Correct target equation is
`Fe_(2)O_(3)(s)+CO(g)rarr 2FeO(s)+CO_(2)(g), Delta H = ?`

Options

A. `+10.3 kJ`
B. `-43.3 kJ`
C. `-10.3 kJ`
D. `+6.2 kJ`

Answer

To solve the problem, we need to manipulate the given reactions and their enthalpy changes to find the enthalpy change for the target reaction. Here are the steps:

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

1. **Reaction 1:**
\[ \text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g), \quad \Delta H_1 = -26.8 \, \text{kJ} \]

2. **Reaction 2:**
\[ \text{FeO}(s) + \text{CO}(g) \rightarrow \text{Fe}(s) + \text{CO}_2(g), \quad \Delta H_2 = -16.5 \, \text{kJ} \]

### Step 2: Reverse Reaction 2 to match the target equation.

When we reverse Reaction 2, we get:
\[ \text{Fe}(s) + \text{CO}_2(g) \rightarrow \text{FeO}(s) + \text{CO}(g), \quad \Delta H = +16.5 \, \text{kJ} \]

### Step 3: Multiply the reversed Reaction 2 by 2.

To match the stoichiometry of the target reaction, we need to multiply the entire reversed Reaction 2 by 2:
\[ 2\text{Fe}(s) + 2\text{CO}_2(g) \rightarrow 2\text{FeO}(s) + 2\text{CO}(g), \quad \Delta H = 2 \times 16.5 = +33.0 \, \text{kJ} \]

### Step 4: Add the modified Reaction 2 to Reaction 1.

Now we add the modified Reaction 2 to Reaction 1:
\[ \text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g) \quad (\Delta H_1 = -26.8 \, \text{kJ}) \]
\[ 2\text{Fe}(s) + 2\text{CO}_2(g) \rightarrow 2\text{FeO}(s) + 2\text{CO}(g) \quad (\Delta H = +33.0 \, \text{kJ}) \]

When we add these reactions, the \(2\text{Fe}(s)\) and \(2\text{CO}_2(g)\) will cancel out:
\[ \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2\text{FeO}(s) + \text{CO}_2(g) \]

### Step 5: Calculate the overall enthalpy change.

Now we can calculate the overall enthalpy change (\(\Delta H\)):
\[ \Delta H = \Delta H_1 + \Delta H_2 = -26.8 \, \text{kJ} + 33.0 \, \text{kJ} = +6.2 \, \text{kJ} \]

### Final Answer:

The enthalpy change for the target reaction is:
\[ \Delta H = +6.2 \, \text{kJ} \]

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