Answer: B
With equal pre-exponential ratios, ln(kE/kS)=−ΔEa/RT+const, so ΔEa=Rln(0.40/0.25)/(1/300−1/330). This is 8.314×0.4700/(3.030×10−4)=1.29×104 J mol−1=12.9 kJ mol−1. The ratio rises with temperature, so the sign is positive: elimination has the higher activation energy.