If enthalpy of hydrogenation of cyclohexene is `[-x kJ mol^-1]` and resonance energy of benzene is `[-y kJ mol^-1]` then the enthalpy of hydrogenation of benzene will be

Options

A. `[(y - 3x) kJ mol^-1]`
B. `[-(x + y) kJ mol^-1]`
C. `[(y + 3x) kJ mol^-1]`
D. `[(x + y) kJ mol^-1]`

Answer

To find the enthalpy of hydrogenation of benzene based on the given information, we can follow these steps:

### Step 1: Understand the Given Information
We know:
- The enthalpy of hydrogenation of cyclohexene is \(-x \, \text{kJ mol}^{-1}\).
- The resonance energy of benzene is \(-y \, \text{kJ mol}^{-1}\).

### Step 2: Write the Reaction for Hydrogenation of Benzene
The hydrogenation of benzene can be represented as:
\[ \text{C}_6\text{H}_6 + 3 \text{H}_2 \rightarrow \text{C}_6\text{H}_{12} \]
This reaction shows that benzene (C6H6) reacts with 3 moles of hydrogen (H2) to form cyclohexane (C6H12).

### Step 3: Relate the Enthalpy of Hydrogenation
The enthalpy change for the hydrogenation of benzene can be expressed in terms of the enthalpy of hydrogenation of cyclohexene and the resonance energy of benzene:
\[ \Delta H_{\text{hydrogenation of benzene}} = 3 \times \Delta H_{\text{hydrogenation of cyclohexene}} - \text{Resonance Energy of Benzene} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ \Delta H_{\text{hydrogenation of benzene}} = 3 \times (-x) - (-y) \]
This simplifies to:
\[ \Delta H_{\text{hydrogenation of benzene}} = -3x + y \]
or rearranging gives:
\[ \Delta H_{\text{hydrogenation of benzene}} = y - 3x \]

### Step 5: Final Expression
Thus, the enthalpy of hydrogenation of benzene is:
\[ \Delta H_{\text{hydrogenation of benzene}} = - (3x - y) \, \text{kJ mol}^{-1} \]

### Conclusion
The enthalpy of hydrogenation of benzene can be expressed as:
\[ \Delta H_{\text{hydrogenation of benzene}} = -y - 3x \, \text{kJ mol}^{-1} \]