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Asked by Sunny Acharjee | 13th Jan, 2013, 01:31: PM

Expert Answer:

The Arrhenius equation is

                 log10 k2/k1 = Ea/(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 = Ea/(8.314×2.303) [(323-293)/(323×293)]

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

                 = 28811.8 J mol-1

                 = 28.8118 kJ mol-1

Answered by  | 13th Jan, 2013, 05:46: PM

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