Two reactions of the same order have identical pre-exponential factors. The activation energy of the first reaction exceeds that of the second reaction by 16.6 kJ mol ⁻¹ . If the rate constant of the second reaction is e^5 times the rate constant of the first reaction at a certain temperature T , then the value of T is _____ K. (Given: R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

Given: A₁ = A₂ E_ a1 - E_ a2 = 16.6 kJ mol ⁻¹ = 16600 J mol ⁻¹ k₂ = k₁ e^5 k₂ k₁ = e^5 R = 8.3 J K ⁻¹ mol ⁻¹ From the Arrhenius equation, k = A e^ -E_a/RT . Taking the ratio of the rate constants for the two reactions: k₂ k₁ = A₂ e^ -E_ a2 /RT A₁ e^ -E_ a1 /RT Since A₁ = A₂ , this simplifies to: k₂ k₁ = e^ (E_ a1 - E_ a2 )/RT Taking the natural logarithm on both sides: ( k₂ k₁ ) = E_ a1 - E_ a2 RT Substitute the given values into the equation: (e^5) = 16600 8.3 T 5 = 16600 8.3 T Rearranging to solve for T : T = 166