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JEE MainChemistryThermodynamics (C)

The standard enthalpies of combustion of C(graphite) and H ₂( g ) are -x kJ mol ⁻¹ and -y kJ mol ⁻¹ respectively. If the standard enthalpy of formation of CH ₄( g ) is -z kJ mol ⁻¹ , the standard enthalpy of combustion of CH ₄( g ) is:

Options

  1. Ax + 2y - z
  2. Bz - x - y
  3. Cz - x - 2y
  4. Dz + x + 2y

Correct answer

C. z - x - 2y

Step-by-step solution

The combustion reaction of methane is: CH ₄( g ) + 2 O ₂( g ) CO ₂( g ) + 2 H ₂ O ( l ) The enthalpy of combustion of CH ₄( g ) can be calculated using the standard enthalpies of formation: H_ c ( CH ₄) = H_ f ( CO ₂) + 2 H_ f ( H ₂ O ) - H_ f ( CH ₄) The standard enthalpy of formation of CO ₂( g ) is equal to the standard enthalpy of combustion of C(graphite) , so H_ f ( CO ₂) = -x . The standard enthalpy of formation of H ₂ O ( l ) is equal to the standard enthalpy of combustion of H ₂( g ) , so H_ f ( H ₂ O ) =

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