Consider the two reactions, whose pre-exponential factor is same. A 700 ~K products A [ catalyst ] 500 ~K products In the presence of catalyst, it was found that the activation energy Ea is decreased by 30 ~kJ / mol and the rate constant remains unchanged. The activation energy for the catalysed reaction is:

Step-by-step solution

The Arrhenius equation is given by k = A e^ -E_a / (RT) . For the uncatalyzed reaction at T₁ = 700 K with activation energy E_ a1 , the rate constant is k₁ = A e^ -E_ a1 / (R 700) . For the catalyzed reaction at T₂ = 500 K with activation energy E_ a2 , the rate constant is k₂ = A e^ -E_ a2 / (R 500) . Given that the rate constants are equal ( k₁ = k₂ ) and the pre-exponential factor A is the same, we have e^ -E_ a1 / (700R) = e^ -E_ a2 / (500R) . Equating the exponents, E_ a1 700 = E_ a2 500 , which simplifies to