The half-life of a first-order reaction is 500 minutes at 400 K and 50 minutes at 500 K . The activation energy of the reaction is ______ kJ mol ⁻¹ . (Given 2.303 R = 19 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

For a first-order reaction, the rate constant k is inversely proportional to the half-life t_ 1/2 : k = 2 t_ 1/2 The ratio of rate constants at the two temperatures is: k₂ k₁ = (t_ 1/2 )₁ (t_ 1/2 )₂ = 500 50 = 10 Using the Arrhenius equation in base-10 logarithm form: ( k₂ k₁ ) = E_a 2.303 R [ T₂ - T₁ T₁ T₂ ] Substitute the known values: (10) = 1 T₁ = 400 K T₂ = 500 K 2.303 R = 19 J K ⁻¹ mol ⁻¹ The equation becomes: 1 = E_a 19 [ 500 - 400 400 500 ] 1 = E_a 19 [ 100 200000 ] 1 = E_a 19 2000 E_a = 38000 J mol ⁻¹ Conv