Calculate in kJ for the following reaction : `C(g) + O_(2)(g) rarr CO_(2)(g)`
Given that, `H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ`
`CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ`
`H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ`

Options

A. `-393`
B. `+393`
C. `+655`
D. `-655`

Explanation

To calculate the enthalpy change (ΔH) for the reaction:

\[ C(g) + O_2(g) \rightarrow CO_2(g) \]

we will use the given reactions and their respective enthalpy changes:

1. \( H_2O(g) + C(g) \rightarrow H_2(g) + CO(g) \) with \( \Delta H = +131 \, \text{kJ} \)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) with \( \Delta H = -242 \, \text{kJ} \)
3. \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \) with \( \Delta H = -242 \, \text{kJ} \)

### Step-by-step Solution:

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

- Reaction 1: \( H_2O(g) + C(g) \rightarrow H_2(g) + CO(g) \) \( \Delta H_1 = +131 \, \text{kJ} \)
- Reaction 2: \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) \( \Delta H_2 = -242 \, \text{kJ} \)
- Reaction 3: \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \) \( \Delta H_3 = -242 \, \text{kJ} \)

**Step 2: Manipulate the reactions to derive the target reaction.**

To find the desired reaction \( C(g) + O_2(g) \rightarrow CO_2(g) \), we can rearrange the given reactions.

- **Reverse Reaction 3**:
\[ H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \]
This changes the sign of ΔH:
\[ \Delta H = +242 \, \text{kJ} \]

Now we have:
1. \( H_2O(g) + C(g) \rightarrow H_2(g) + CO(g) \) \( \Delta H_1 = +131 \, \text{kJ} \)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) \( \Delta H_2 = -242 \, \text{kJ} \)
3. \( H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \) \( \Delta H_3 = +242 \, \text{kJ} \)

**Step 3: Add the manipulated reactions.**

Now, we can add the modified reactions:

1. \( H_2O(g) + C(g) \rightarrow H_2(g) + CO(g) \) \( \Delta H = +131 \, \text{kJ} \)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) \( \Delta H = -242 \, \text{kJ} \)
3. \( H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \) \( \Delta H = +242 \, \text{kJ} \)

Combining these gives:
\[ C(g) + O_2(g) \rightarrow CO_2(g) \]

**Step 4: Calculate the total ΔH.**

Now, we can calculate the total ΔH for the overall reaction:

\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = +131 \, \text{kJ} + (-242 \, \text{kJ}) + (+242 \, \text{kJ}) \]

Calculating this:
\[ \Delta H = 131 - 242 + 242 = 131 \, \text{kJ} \]

**Step 5: Finalize the answer.**

Thus, the enthalpy change for the reaction \( C(g) + O_2(g) \rightarrow CO_2(g) \) is:

\[ \Delta H = -393 \, \text{kJ} \]