Q22 / 150 · MHT CET 2026 (13 April Shift 2) · Chemistry
The enthalpy of combustion of cyclohexane, cyclohexene and $\text{H}_2$ are respectively $-3920, -3800$ and $-241$ KJ mol$^{-1}$. The enthalpy of hydrogenation of cyclohexene is
- Correct answerA$-121$ KJ mol$^{-1}$
- B$+121$ KJ mol$^{-1}$
- C$-242$ KJ mol$^{-1}$
- D$+242$ KJ mol$^{-1}$
Solution
The hydrogenation reaction of cyclohexene is given by:
$\text{C}_6\text{H}_{10} + \text{H}_2 \rightarrow \text{C}_6\text{H}_{12}$
The enthalpy of hydrogenation can be calculated using the enthalpies of combustion of the reactants and products:
$\Delta H = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$
$\Delta H = [\Delta H_c(\text{C}_6\text{H}_{10}) + \Delta H_c(\text{H}_2)] - \Delta H_c(\text{C}_6\text{H}_{12})$
Substituting the given values:
$\Delta H = [-3800 + (-241)] - (-3920)$
$\Delta H = -4041 + 3920$
$\Delta H = -121 \text{ kJ mol}^{-1}$
Answer: $-121$ KJ mol$^{-1}$
$\text{C}_6\text{H}_{10} + \text{H}_2 \rightarrow \text{C}_6\text{H}_{12}$
The enthalpy of hydrogenation can be calculated using the enthalpies of combustion of the reactants and products:
$\Delta H = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$
$\Delta H = [\Delta H_c(\text{C}_6\text{H}_{10}) + \Delta H_c(\text{H}_2)] - \Delta H_c(\text{C}_6\text{H}_{12})$
Substituting the given values:
$\Delta H = [-3800 + (-241)] - (-3920)$
$\Delta H = -4041 + 3920$
$\Delta H = -121 \text{ kJ mol}^{-1}$
Answer: $-121$ KJ mol$^{-1}$
Hence, the correct answer is option: 'A'.