Two different first-order reactions, I and II, have identical half-lives at 27^ C . The activation energies for reactions I and II are 60 kJ mol ⁻¹ and 40 kJ mol ⁻¹ , respectively. When the temperature is raised to 127^ C , the ratio of the half-life of reaction II to the half-life of reaction I becomes e^x . The value of x is ________. [Given: R = 25 3 J K ⁻¹ mol ⁻¹ ]

Step-by-step solution

For a first-order reaction, the half-life is inversely proportional to the rate constant: t_ 1/2 = 2 k . At T₁ = 27^ C = 300 K , the half-lives are equal, so k₁ = k₂ . Using the Arrhenius equation k = A e^ -E_a/RT : A₁ e^ -E_ a1 /(R 300) = A₂ e^ -E_ a2 /(R 300) Taking the natural logarithm: ( A₁ A₂ ) = E_ a1 - E_ a2 R 300 Given E_ a1 = 60000 J mol ⁻¹ and E_ a2 = 40000 J mol ⁻¹ : ( A₁ A₂ ) = 20000 ( 25 3 ) 300 = 20000 2500 = 8 At T₂ = 127^ C = 400 K , the ratio of the rate constants is: ( k₁' k₂' ) = ( A₁ A₂ ) - E_