Question #613c4

Question

1038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-24T17:57:20

The activation energy is #=9*10^5Jmol^-1#

We apply the following equation

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

#R=8.34Jmol^-1K^-1#

#T_1=350+273=623K#

#T_2=320+273=593K#

#k_1=4.6*10^-4s^-1#

#k_2=1.87*10^-4s^-1#

Therefore,

#ln(4.6/1.87)=E_a/8.34(1/593-1/623)#

#0.9=9.7*10^-6E_a#

#E_a=0.9/(9.7*10^-6)=0.09*10^7=9*10^5Jmol^-1#