The standard enthalpy of combustion of liquid cyclopentane ( C ₅ H ₁₀ ) is -3290 kJ mol ⁻¹ . Given that the standard enthalpies of formation of CO ₂ (g) and H ₂ O(l) are -394 kJ mol ⁻¹ and -286 kJ mol ⁻¹ respectively, the standard enthalpy

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The standard enthalpy of combustion of liquid cyclopentane ( C ₅ H ₁₀ ) is -3290 kJ mol ⁻¹ . Given that the standard enthalpies of formation of CO ₂ (g) and H ₂ O(l) are -394 kJ mol ⁻¹ and -286 kJ mol ⁻¹ respectively, the standard enthalpy of formation of liquid cyclopentane is

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Correct answer: C. -110 kJ mol ⁻¹

First, write the balanced chemical equation for the complete combustion of liquid cyclopentane: C ₅ H ₁₀ (l) + 15 2 O ₂ (g) 5 CO ₂ (g) + 5 H ₂ O(l) The standard enthalpy of combustion is related to the standard enthalpies of formation by the equation: H_ c = H_ f ( products ) - H_ f ( reactants ) H_ c = [5 H_ f ( CO ₂) + 5 H_ f ( H ₂ O )] - [ H_ f ( C ₅ H ₁₀) + 15 2 H_ f ( O ₂)] Since the standard enthalpy of formation of an element in its standard state (like O ₂ ) is zero: -3290 = [5(-394) + 5(-286)] - H_ f ( C ₅

The standard enthalpy of combustion of liquid cyclopentane ( C ₅ H ₁₀ ) is -3290 kJ mol ⁻¹ . Given that the standard enthalpies of formation of CO ₂ (g) and H ₂ O(l) are -394 kJ mol ⁻¹ and -286 kJ mol ⁻¹ respectively, the standard enthalpy of formation of liquid cyclopentane is

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