1.

The activation energy of a reaction is 225 k cal mol"^(-1)and the value of rate constant at 40^@C "is "1.8xx10^(-5)s^(-1).Calculate the frequency factor , a , .

Answer»

Solution :Here, we are given that `E_a = 22.5"k CAL mol"^(-1) = 22500 "cal mol"^(-1)`
`T = 40^@C = 40+273=313K`
`k=1.8xx10^(-5) "sec"^(-1)`
Substituting the value in the equation
`LOGA = log k + ((E_a)/(2.303RT))`
`log A = log (1.8xx10^(-5))+((22500)/(2.303xx1.987xx313))`
`logA =log (1.8)-5+(15.7089)`
A = antilog (10.9642)
`A = 9.208 xx10^(10) "collisions s"^(-1)`


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