The following thermochemical reactions are given :
`M+(1)/(2)O_(2)to MO+351.4 kJ " " X+(1)/(2)O_(2)to XO+90.8 kJ`
It follows that the heat of reaction for the following process `M+XO to MO +X` is given by

Options

A. 422.2 kJ
B. 268.7 kJ
C. `-442.2` kJ
D. `-260.6 kJ`

Answer

To find the heat of reaction for the process \( M + XO \to MO + X \), we can 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. **Identify the Given Reactions**:
- Reaction 1:
\[ M + \frac{1}{2}O_2 \to MO \quad \Delta H_1 = +351.4 \, \text{kJ} \]
- Reaction 2:
\[ X + \frac{1}{2}O_2 \to XO \quad \Delta H_2 = +90.8 \, \text{kJ} \]

2. **Write the Target Reaction**:
- We need to find the heat of reaction for:
\[ M + XO \to MO + X \]

3. **Manipulate the Given Reactions**:
- To obtain the target reaction, we can reverse Reaction 1 and Reaction 2:
- Reverse Reaction 1:
\[ MO \to M + \frac{1}{2}O_2 \quad \Delta H = -351.4 \, \text{kJ} \]
- Reverse Reaction 2:
\[ XO \to X + \frac{1}{2}O_2 \quad \Delta H = -90.8 \, \text{kJ} \]

4. **Add the Reversed Reactions**:
- Now, add the reversed reactions:
\[ MO \to M + \frac{1}{2}O_2 \quad (-351.4 \, \text{kJ}) \]
\[ XO \to X + \frac{1}{2}O_2 \quad (-90.8 \, \text{kJ}) \]
- When we add these two reactions, the \(\frac{1}{2}O_2\) cancels out:
\[ M + XO \to MO + X \]

5. **Calculate the Total Enthalpy Change**:
- The total enthalpy change for the target reaction is:
\[ \Delta H = -351.4 - 90.8 = -442.2 \, \text{kJ} \]

### Final Answer:
The heat of reaction for \( M + XO \to MO + X \) is:
\[ \Delta H = -442.2 \, \text{kJ} \]