Question

某反应在500K时，速率常数为k1，600K时为k2=10k1，计算该反应的活化能。

Answer 1

【答案】：T₁=500K k₁ T₂=600K k₂=10k₁
lg(k₂/k₁)=E_a(T₂-T₁)/(2.303RT₁T₂)
E_a=2.303RT₁T₂lg(k₂/k₁)/(T₂-T₁)
=2.303×8.314J·K^-1·mol^-1×500K×600K×lg10/(600-500)K
=57.44kJ·mol^-1