A reaction has an activation energy of 120 kJ mol ⁻¹ at 300 K . The addition of a catalyst lowers the activation energy by 30 kJ mol ⁻¹ . The temperature at which the catalysed reaction will have the same rate constant as the uncatalysed reaction at 300 K is _____ K . [Assume the pre-exponential factor remains the same]

Step-by-step solution

According to the Arrhenius equation, the rate constant is k = A e^ - E_a RT . For the uncatalysed reaction at T₁ = 300 K : k_ uncat = A e^ - E_ a,uncat RT₁ For the catalysed reaction at T₂ : k_ cat = A e^ - E_ a,cat RT₂ Since the rate constants are equal ( k_ uncat = k_ cat ): A e^ - E_ a,uncat RT₁ = A e^ - E_ a,cat RT₂ Taking the natural logarithm on both sides: E_ a,uncat T₁ = E_ a,cat T₂ Given E_ a,uncat = 120 kJ mol ⁻¹ and the catalyst lowers it by 30 kJ mol ⁻¹ : E_ a,cat = 120 - 30 = 90 kJ mol ⁻¹ Substituting