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NEET CHEMISTRYThermodynamicsMedium

Question

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}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsenthalpycombustioncyclohexenetextctexthcyclohexane

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