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NEET CHEMISTRYMedium

The enthalpy of combustion of H2\text{H}_2, cyclohexene (C6H10\text{C}_6\text{H}_{10}) and cyclohexane (C6H12\text{C}_6\text{H}_{12}) are 241-241, 3800-3800 and 3920 kJ per mol-3920 \text{ kJ per mol} respectively. Heat of hydrogenation of cyclohexene is:

A

121 kJ per mol-121 \text{ kJ per mol}

B

+121 kJ per mol+121 \text{ kJ per mol}

C

+242 kJ per mol+242 \text{ kJ per mol}

D

242 kJ per mol-242 \text{ kJ per mol}

Step-by-Step Solution

The reaction for the hydrogenation of cyclohexene is: C6H10+H2C6H12\text{C}_6\text{H}_{10} + \text{H}_2 \rightarrow \text{C}_6\text{H}_{12} According to Hess's Law, the enthalpy of reaction can be calculated from the enthalpies of combustion of reactants and products: ΔHhydrogenation=ΣΔHc(reactants)ΣΔHc(products)\Delta H_{\text{hydrogenation}} = \Sigma \Delta H_c (\text{reactants}) - \Sigma \Delta H_c (\text{products}) ΔHhydrogenation=[ΔHc(C6H10)+ΔHc(H2)][ΔHc(C6H12)]\Delta H_{\text{hydrogenation}} = [\Delta H_c(\text{C}_6\text{H}_{10}) + \Delta H_c(\text{H}_2)] - [\Delta H_c(\text{C}_6\text{H}_{12})] Given: ΔHc(C6H10)=3800 kJ mol1\Delta H_c(\text{C}_6\text{H}_{10}) = -3800 \text{ kJ mol}^{-1} ΔHc(H2)=241 kJ mol1\Delta H_c(\text{H}_2) = -241 \text{ kJ mol}^{-1} ΔHc(C6H12)=3920 kJ mol1\Delta H_c(\text{C}_6\text{H}_{12}) = -3920 \text{ kJ mol}^{-1} Substituting the values: ΔHhydrogenation=[3800+(241)][3920]\Delta H_{\text{hydrogenation}} = [-3800 + (-241)] - [-3920] ΔHhydrogenation=4041+3920=121 kJ mol1\Delta H_{\text{hydrogenation}} = -4041 + 3920 = -121 \text{ kJ mol}^{-1}

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