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

Options

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

Answer

To solve the problem, we need to analyze the given reactions and their enthalpy changes. Let's break it down step by step.

### Step 1: Write the combustion reactions
1. **Combustion of Methane (CH₄)**:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
The enthalpy change for this reaction is denoted as \( \Delta H = -X \) (since heat is released).

2. **Combustion of Methanol (CH₃OH)**:
\[ \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
The enthalpy change for this reaction is denoted as \( \Delta H = -Y \) (since heat is also released).

### Step 2: Reverse the combustion of Methanol
When we reverse the combustion of methanol, the reaction becomes:
\[ \text{CO}_2 + 2 \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \]
The enthalpy change for this reversed reaction is \( \Delta H = +Y \).

### Step 3: Add the two reactions
Now, we add the combustion of methane and the reversed combustion of methanol:
1. From the combustion of methane:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \quad (\Delta H = -X) \]
2. From the reversed combustion of methanol:
\[ \text{CO}_2 + 2 \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \quad (\Delta H = +Y) \]

When we add these two equations, the \(\text{CO}_2\) and \(2 \text{H}_2\text{O}\) on the left side cancel out:
\[ \text{CH}_4 + \frac{1}{2} \text{O}_2 \rightarrow \text{CH}_3\text{OH} \]

### Step 4: Write the overall enthalpy change
The overall enthalpy change for the reaction is:
\[ \Delta H = -X + Y \]
According to the problem, this enthalpy change is negative:
\[ -X + Y < 0 \]
This implies:
\[ Y < X \]

### Conclusion
Thus, the correct relationship is:
\[ X > Y \]