The rate constant of a reaction at 600 K with an activation energy of 191.47 kJ mol ⁻¹ is 5.0 10⁻⁵ s ⁻¹ . What is the temperature at which the half-life of the reaction becomes 152 s? [Consider pre-exponential factor and activation energy to be independent of temperature. R = 8.314 J K ⁻¹ mol ⁻¹ ]

Step-by-step solution

From the unit of the rate constant (s ⁻¹ ), the reaction is of first order. The rate constant at temperature T₂ is: k₂ = 0.693 t_ 1/2 = 0.693 152 4.56 10⁻³ s ⁻¹ Using the Arrhenius equation: ( k₂ k₁ ) = E_a 2.303 R ( 1 T₁ - 1 T₂ ) Substituting the given values: E_a 2.303 R = 191.47 10^3 2.303 8.314 = 191470 19.147 = 10000 K ( 4.56 10⁻³ 5.0 10⁻⁵ ) = 10000 ( 1 600 - 1 T₂ ) (91.2) = 10000 ( 1 600 - 1 T₂ ) 1.96 = 10000 ( 1 600 - 1 T₂ ) 1 600 - 1 T₂ = 0.000196 1 T₂ = 1 600 - 0.000196 = 0.001667 - 0.000196 = 0.001471 K ⁻