The following two 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. (a) `-43.3 kJ`
B. (b) `-10.3 kJ`
C. (c) `+6.2 kJ`
D. (d) `+10.3 kJ`

Answer

To solve the problem, we need to manipulate the given reactions to derive the target equation and calculate the corresponding enthalpy change (ΔH). Here’s a step-by-step breakdown of the solution:

### 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
Since the target equation has FeO on the product side, we need to reverse Reaction 2:
\[ \text{Fe}(s) + \text{CO}_2(g) \rightarrow \text{FeO}(s) + \text{CO}(g) \]
When we reverse a reaction, the sign of ΔH also changes:
\[ \Delta H_3 = +16.5 \, \text{kJ} \]

### Step 3: Multiply the Reversed Reaction by 2
To get 2 moles of FeO as required in the target equation, we multiply the reversed Reaction 2 by 2:
\[ 2\text{Fe}(s) + 2\text{CO}_2(g) \rightarrow 2\text{FeO}(s) + 2\text{CO}(g) \]
The enthalpy change for this reaction will also be multiplied by 2:
\[ \Delta H_4 = 2 \times 16.5 \, \text{kJ} = 33.0 \, \text{kJ} \]

### Step 4: Add Reaction 1 and the Modified Reaction 4
Now we add Reaction 1 and Reaction 4:
1. **Reaction 1**:
\[ \text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g) \]

2. **Modified Reaction 4**:
\[ 2\text{Fe}(s) + 2\text{CO}_2(g) \rightarrow 2\text{FeO}(s) + 2\text{CO}(g) \]

When we add these two reactions, we get:
\[ \text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) + 2\text{Fe}(s) + 2\text{CO}_2(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g) + 2\text{FeO}(s) + 2\text{CO}(g) \]

### Step 5: Cancel Out Common Species
On both sides, we can cancel out the common species:
- 2Fe on both sides
- 2CO2 on both sides

This simplifies to:
\[ \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2\text{FeO}(s) + \text{CO}_2(g) \]

### Step 6: Calculate the Overall Enthalpy Change
Now we can calculate the overall ΔH for the target reaction:
\[ \Delta H = \Delta H_1 + \Delta H_4 = -26.8 \, \text{kJ} + 33.0 \, \text{kJ} = 6.2 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change for the target equation is:
\[ \Delta H = +6.2 \, \text{kJ} \]