For a reversible reaction A ⇌ B, the Δ H_(forward reaction) = 20 kJ mol⁻¹. The activation energy of the uncatalysed forward reaction is 300 kJ mol⁻¹. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at 27° C is found to be same as that of the uncatalysed reaction at 327° C.
The activation energy of the catalysed backward reaction is ___________ kJ mol⁻¹.

Solution

Chemistry ma’am · 3 steps

Equal rates mean equal Eₐ/T, so the catalysed barrier is 150. Then subtract Δ H to go backwards.

1. Equal rates. The reactant concentration is the same and a catalyst does not change A. So equal rates need equal k = A e^(-Eₐ/RT), which means the exponents must be equal: E_(a,cat)/T₁ = E_(a,uncat)/T₂.

   E_(a,cat)/(300 K) = (300 kJ mol⁻¹)/(600 K) ⇒ E_(a,cat) = 150 kJ mol⁻¹

2. Forward and backward. For any one path, Δ H = E_(a,f) - E_(a,b). The catalyst lowers both barriers by the same amount and leaves Δ H unchanged. So E_(a,b) = E_(a,f) - Δ H = 150 - 20.
3. So. The activation energy of the catalysed backward reaction is 130 kJ mol⁻¹.

The common slip: 170. Adding Δ H instead of subtracting it gives 150 + 20 = 170. The reaction is endothermic, so the backward barrier is the smaller one.

The idea to keep. Same rate with the same A means Eₐ/T is the same, and temperatures must be in kelvin. For an endothermic step the backward barrier is lower: E_(a,b) = E_(a,f) - Δ H.

Answer from the NTA final answer key. Solution by PrepWiser.