Enthalpy of `CH_(4)+(1)/(2)O_(2)rarrCH_(3)OH` is
negative. If enthalpy of combustion of `CH_(4)` and `CH_(3)OH` are `x` and `y` respectively, then which relation is correct?

Options

A. `x gt y`
B. `x lt y`
C. `x=y`
D. `x ge y`

Explanation

To solve the problem, we need to analyze the enthalpy changes associated with the combustion of methane (CH₄) and methanol (CH₃OH) and relate them to the given reaction.

### Step-by-Step Solution:

1. **Identify the combustion reactions:**
- The combustion of methane (CH₄) can be represented as:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
The enthalpy change for this reaction is denoted as \( -X \).

- The combustion of methanol (CH₃OH) can be represented as:
\[ \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
The enthalpy change for this reaction is denoted as \( -Y \).

2. **Write the reaction of interest:**
- The reaction given in the question is:
\[ \text{CH}_4 + \frac{1}{2} \text{O}_2 \rightarrow \text{CH}_3\text{OH} \]
The enthalpy change for this reaction is given as negative, which we denote as \( \Delta H = -\Delta H_3 \).

3. **Apply Hess's Law:**
- According to Hess's Law, the enthalpy change of a reaction can be calculated by summing the enthalpy changes of the individual steps.
- We can express the enthalpy change for the reaction of interest using the two combustion reactions:
\[ \Delta H_3 = \Delta H_{\text{combustion of CH}_4} - \Delta H_{\text{combustion of CH}_3\text{OH}} \]
This translates to:
\[ -\Delta H_3 = -X + Y \]

4. **Set up the equation:**
- Since the enthalpy change for the reaction is negative, we have:
\[ -X + Y < 0 \]
Rearranging this gives:
\[ Y < X \]

5. **Conclusion:**
- The correct relationship between the enthalpies of combustion of methane and methanol is:
\[ X > Y \]

### Final Answer:
The correct relation is \( X > Y \).