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The Arrhenius equation is log10k2k1=Ea(R×2.303)[(T2–T1)(T1T2)]
Given: k2k1 = 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
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