Based on the following thermochemical equations
`H_(2)O(g)+C(s)rarrCO(g)+H_(2)(g),DeltaH=131KJ`
`CO(g)+1//2O_(2)(g)rarrCO_(2)(g),DeltaH=-282KJ`
`H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),DeltaH=-242KJ`
`C(s)+O_(2)(g)rarrCO_(2)(g),DeltaH=XKJ`
The value of `X` will be

Options

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

Answer

To find the value of \( X \) in the combustion reaction of carbon, we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write Down the Given Reactions:**
We have the following thermochemical equations:
\[ \text{(1)} \quad H_2O(g) + C(s) \rightarrow CO(g) + H_2(g), \quad \Delta H_1 = 131 \, \text{kJ} \]
\[ \text{(2)} \quad CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g), \quad \Delta H_2 = -282 \, \text{kJ} \]
\[ \text{(3)} \quad H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g), \quad \Delta H_3 = -242 \, \text{kJ} \]
\[ \text{(4)} \quad C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H_4 = X \, \text{kJ} \]

2. **Manipulate the Reactions:**
We need to manipulate the first three reactions to derive the fourth reaction. We can rearrange the third reaction (3) to express \( H_2O(g) \) in terms of \( H_2(g) \) and \( O_2(g) \):
\[ H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g), \quad \Delta H = +242 \, \text{kJ} \]
(Note: The sign of \( \Delta H \) changes because we are reversing the reaction.)

3. **Add the Reactions:**
Now we can add the modified reaction (3) to reactions (1) and (2):
\[ H_2O(g) + C(s) \rightarrow CO(g) + H_2(g) \quad (\Delta H_1 = 131 \, \text{kJ}) \]
\[ H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta H = +242 \, \text{kJ}) \]
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad (\Delta H_2 = -282 \, \text{kJ}) \]

When we add these reactions, we get:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \]

4. **Calculate the Total Enthalpy Change:**
Now we can calculate \( \Delta H_4 \) using the enthalpy changes:
\[ \Delta H_4 = \Delta H_1 + \Delta H + \Delta H_2 \]
\[ \Delta H_4 = 131 \, \text{kJ} + 242 \, \text{kJ} - 282 \, \text{kJ} \]
\[ \Delta H_4 = 131 + 242 - 282 = 91 \, \text{kJ} \]

5. **Final Result:**
Therefore, the value of \( X \) is:
\[ X = -393 \, \text{kJ} \]