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The activation energy of a reaction is 94.14 KJ/mol and the value of rate constant at 40^@ C is 1.8 xx 10^(-1) sec^(-1). Calculate the frequency factor A. |
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Answer» SOLUTION :Given, `E_a=94.14xx10^(-3) J MOL^(-1), T=40+273 =313` K , `K=1.8xx10^(-1) "SEC"^(-1)` By using , `K=Ae^(-E_a//RT) rArr ` In K = In A -`E_a/(RT)` Or log K= log A -`E_a/(2.303RT)` Or `log(1.8xx10^(-1))+(94.19xx10^3)/(2.303xx8.314xx313)`=log A Or A = ANTILOG (10.9635) `=9.194xx10^10 "sec"^(-1)` |
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