A first-order reaction has rate constants of 1.5 ✕ 10-2 s-1 and 7.8 ✕ 10-2 s-1 at 0°C and 20.°C, respectively. What is the value of the activation energy?

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Answer 1 · 2018-02-16T00:30:24

Consider the relationship between both first order reactions,

#ln(k_2/k_1) = E_a/R *(1/T_1 - 1/T_2)#

#=> (R*ln(k_2/k_1))/(1/T_1 - 1/T_2) = E_a#

Hence,

#E_a = ((8.314J)/(mol*K) * ln((7.8*10^-2s^-1)/(1.5*10^-2s^-1)))/(1/(273K) - 1/(293K)#

#therefore E_a approx 54.8kJ#

is the activation energy of the reaction under study given your data.