CBSE Class 12-science Answered

The Arrhenius equation is

                 log₁₀ k₂/k₁ = E_a/(R×2.303) [(T₂-T₁)/(T₁ T₂ )]

                 Given:  k²/k¹ = 3; R=8.314 JK^-1 mol–1; T1 = 20 + 273 = 293 K

                 and   T₂ = 50 + 273 = 323 K

                 Subtracting the given values in the Arrhenius equation,

                 log₁₀ 3 = E_a/(8.314×2.303) [(323-293)/(323×293)]

                 E_a = (2.303 × 8.314×323 × 293×0.477)/30

                 = 28811.8 J mol^-1

                 = 28.8118 kJ mol^-1