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For the reversible reaction A2 + B2 ⇌ 2AB, the activation energies are 180 kJ mol⁻¹ for the forward direction and 200 kJ mol⁻¹ for the reverse direction. If a catalyst reduces the activation energy of both directions by 100 kJ mol⁻¹, what will be the enthalpy change of the reaction A2 + B2 ⇌ 2AB in the catalyzed pathway (in kJ mol⁻¹)?

1. 20
2. 300
3. 120
4. 280

Correct answer: 20

Solution

deltaH = Ea(forward) - Ea(reverse) = 180 - 200 = -20 kJ/mol. A catalyst lowers both activation energies by the same amount (100), so the difference and hence deltaH is unchanged at -20 kJ/mol (magnitude 20).

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