The standard enthalpy of the reaction CO(g) + 2 H ₂ (g) CH ₃ OH(l) is -128 kJ mol ⁻¹ . If the standard enthalpies of combustion of H ₂ (g) and CH ₃ OH(l) are -286 kJ mol ⁻¹ and -726 kJ mol ⁻¹ respectively, the standard enthalpy of combustio

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The standard enthalpy of the reaction CO(g) + 2 H ₂ (g) CH ₃ OH(l) is -128 kJ mol ⁻¹ . If the standard enthalpies of combustion of H ₂ (g) and CH ₃ OH(l) are -286 kJ mol ⁻¹ and -726 kJ mol ⁻¹ respectively, the standard enthalpy of combustion of CO(g) is

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Correct answer: B. -282 kJ mol ⁻¹

The enthalpy of a reaction can be calculated from the standard enthalpies of combustion of the reactants and products using the relation: H_ reaction = H_ c ( reactants ) - H_ c ( products ) For the given reaction CO(g) + 2 H ₂ (g) CH ₃ OH(l) : H_ reaction = [ H_ c ( CO ) + 2 H_ c ( H ₂)] - [ H_ c ( CH ₃ OH )] Substituting the given values: -128 = [ H_ c ( CO ) + 2(-286)] - [-726] -128 = H_ c ( CO ) - 572 + 726 -128 = H_ c ( CO ) + 154 Solving for H_ c ( CO ) : H_ c ( CO ) = -128 - 154 = -282 kJ mol ⁻¹ Answer: -282

The standard enthalpy of the reaction CO(g) + 2 H ₂ (g) CH ₃ OH(l) is -128 kJ mol ⁻¹ . If the standard enthalpies of combustion of H ₂ (g) and CH ₃ OH(l) are -286 kJ mol ⁻¹ and -726 kJ mol ⁻¹ respectively, the standard enthalpy of combustion of CO(g) is

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