The hydrogenation of Ethyne is carried out at 600 K . The same reaction when carried out in presence of a catalyst maintaining the same rate constant, the temperature required is only 400 K . If the catalyst lowers the Activation energy of the reaction by 20 ~kJ / mol , what is the value of E _ a ?

Step-by-step solution

The Arrhenius equation is given by k = A e^ -E_a / RT . For the reaction without a catalyst at T₁ = 600 K with activation energy E_ a1 = E_a , the rate constant is k₁ = A e^ -E_a / (R 600) . For the reaction with a catalyst at T₂ = 400 K with activation energy E_ a2 = E_a - 20 kJ/mol , the rate constant is k₂ = A e^ -(E_a - 20) / (R 400) . Since the rate constants are the same, k₁ = k₂ , which implies the exponents must be equal: E_a 600 R = E_a - 20 400 R Canceling R from both sides and simplifying the fraction: E