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The rate constant for the first order de...

The rate constant for the first order decompoistion of a certain reaction is described by the equation
`log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)`
(a) What is the energy of activation for the reaction?
(b) At what temperature will its half-life periof be `256 min`?

Text Solution

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According to Arrhenius equation, `k="A e"^(-E_(a)//"RT")`
or `lnk=lnA-(E_(a))/(RT)" or "logk=logA-(E_(a))/(2.303"RT")`
Comparing with the given equation, `(E_(a))/(2.303" RT")=(1.25xx10^(4)"K")/(T)`
or `E_(a)=2.303" R "xx1.25xx10^(4)" K"=2.303xx(8.314" JK"^(-1)" mol"^(-1))xx1.25xx10^(4)" K"=239.347" kJ mol"^(-1)`
When `t_(1//2)=256" min",k=(0.693)/(256xx60"s")=4.51xx10^(-5)s^(-1)`
Substituting this value in the given equation,
`log(4.51xx10^(-5))=14.34-(1.25xx10^(4)" K")/(T),i.e.,(-5+0.6542)=14.34-(1.25xx10^(4)"K")/(T)`
or `(1.25xx10^(4)K)/(T)=18.6858" or "T=669" K".`
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