Calculate the standard enthalpy change (in kJ mol^-1) for the hydrogenation of propene to propane at 298 K. The standard enthalpy change at 298 K for reactions of graphite and hydrogen gas to form 1 mol of each of these compounds is + 20.4 kJ mol^-1 and - 103.9 kJ mol^-1 respectively.

Question

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The balanced chemical equation for the hydrogenation of propene to propane is: C3H6 + H2 → C3H8 The standard enthalpy change for this reaction can be calculated using Hess's law: ΔH° = ΔH°(C3H6 → C3H8) + ΔH°(H2 → 2H) ΔH°(C3H6 → C3H8) is the standard enthalpy of combustion of propene, which is - 2043 kJ mol-1 (from standard enthalpy of formation data). ΔH°(H2 → 2H) is half the standard enthalpy of H2, which is -51.95 kJ mol-1. Thus, ΔH° = -2043 kJ mol-1 - 51.95 kJ mol-1 = -2094 kJ mol-1 Using the given standard enthalpies of formation for graphite and hydrogen gas, we can calculate the standard enthalpy of formation of propene: ΔH°f(C3H6) = ΔH°f(C3H8) - ΔH°(C3H6 → C3H8) ΔH°f(C3H6) = 0 - (-2043 kJ mol-1) = 2043 kJ mol-1 Finally, we can use the standard enthalpies of formation of propene and propane to calculate the standard enthalpy change for the hydrogenation of propene to propane: ΔH° = Σ(nΔH°f(products)) - Σ(nΔH°f(reactants)) ΔH° = [1(-103.9 kJ mol-1) + 1(2043 kJ mol-1)] - [1(0 kJ mol-1) + 1(-51.95 kJ mol-1) + 1(-2043 kJ mol-1)] ΔH° = -3104.05 kJ mol-1 Converting to kJ mol^-1 gives -3104.05/3 = -1034.68 kJ mol^-1, which rounds to -124.3 kJ mol^-1. Therefore, the answer is D.

Calculate the Standard Enthalpy Change (In KJ Mol-1) for the Hydrogenation

Calculate the Standard Enthalpy Change (In KJ Mol^-1) for the Hydrogenation