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. **Reverse Reaction 2:**
To find the enthalpy change for the reaction \( M + XO \to MO + X \), we need to reverse Reaction 2:
\[ XO \to X + \frac{1}{2} O_2 \quad \Delta H = +90.8 \, \text{kJ} \]

3. **Combine the Reactions:**
Now we can add Reaction 1 and the reversed Reaction 2:
\[ (M + \frac{1}{2} O_2 \to MO) + (XO \to X + \frac{1}{2} O_2) \]
This simplifies to:
\[ M + XO \to MO + X \]

4. **Calculate the Total Enthalpy Change:**
The total enthalpy change for the combined reaction is:
\[ \Delta H = \Delta H_1 + \Delta H_2' = -351.4 \, \text{kJ} + 90.8 \, \text{kJ} \]
\[ \Delta H = -351.4 + 90.8 = -260.6 \, \text{kJ} \]

5. **Final Result:**
The heat of reaction for the process \( M + XO \to MO + X \) is:
\[ \Delta H = -260.6 \, \text{kJ} \]