Consider the following reactions:
`C(s)+O_(2)(g)toCO_(2)(g)+x " kJ"`
`CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)+y" kJ"`
The heat formation of CO(g) is :

Options

A. `-(x+y)` kJ/mol
B. `(x-y)` kJ/mol
C. `(y-x)` kJ/mol
D. None of these

Explanation

To find the heat of formation of carbon monoxide (CO) from the given reactions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \text{(releases } x \text{ kJ)} \]
- The second reaction is:
\[ CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g) \quad \text{(releases } y \text{ kJ)} \]

2. **Write the Enthalpy Changes**:
- For the first reaction, we can denote the enthalpy change as:
\[ \Delta H_1 = -x \quad \text{(since heat is released)} \]
- For the second reaction, the enthalpy change is:
\[ \Delta H_2 = -y \quad \text{(since heat is released)} \]

3. **Reverse the Second Reaction**:
- To find the formation of CO from its elements, we need to reverse the second reaction:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \]
- The enthalpy change for the reversed reaction becomes:
\[ \Delta H_2' = +y \]

4. **Combine the Two Reactions**:
- Now, we can add the first reaction and the reversed second reaction:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad (\Delta H_1 = -x) \]
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \quad (\Delta H_2' = +y) \]
- When we add these reactions, the \(CO_2(g)\) cancels out:
\[ C(s) + O_2(g) + CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) + CO_2(g) \]
- This simplifies to:
\[ C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \]

5. **Calculate the Overall Enthalpy Change**:
- The overall enthalpy change for the formation of CO is:
\[ \Delta H_{formation} = \Delta H_1 + \Delta H_2' = -x + y \]
- Therefore, the heat of formation of CO is:
\[ \Delta H_{formation} = y - x \quad \text{kJ} \]

### Final Answer:
The heat of formation of CO(g) is \(y - x\) kJ.

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