Calculate the standard enthalpy change (in kJ mol^-1) for the hydrogenation of ethyne to ethene 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 + 226.7 kJ mol^-1 and + 52.5 kJ mol^-1 respectively.

Question

Quizplus · Accepted Answer

The hydrogenation of ethyne to ethene can be represented by the following equation:

C2H2(g) + H2(g) → C2H4(g)

The balanced chemical equation indicates that 1 mole of C2H2 reacts with 1 mole of H2 to form 1 mole of C2H4. Therefore, the standard enthalpy change for this reaction is equal in magnitude but opposite in sign to the difference between the standard enthalpies of formation of the products and reactants:

ΔH° = ΣΔH°f (products) - ΣΔH°f (reactants)

Using the given standard enthalpies of formation:

ΔH° = [ΔH°f (C2H4) - ΔH°f (C2H2)] + ½ [ΔH°f (H2)]
ΔH° = [(+52.5) - (+226.7)] + ½ [0]
ΔH° = -174.2 kJ mol^-1

Therefore, the correct answer is D (-174.2).

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

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