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Two reactions of the same order have equ...

Two reactions of the same order have equal exponential factors but their activation energies differ by `24.9 kj mol^(-1)`. Calcualte the ratio between the rate constants of these reactions at `27^(@)`C (Gas constant `R= 8.3 JK^(-1)mol^(-1))`

Text Solution

Verified by Experts

We know that,
`log _(10) k=log _(10) A-(E)/(2.3030RT)`
So, `log _(10) k_(2)=log_(10)A-(E_(2))/(2.303RT)`
`and log _(10) k_(1)=log_(10)A-(E_(1))/(2.303RT)`
`log_(10)((k_(2))/(k_(1)))=((E_(1)-E_(2)))/(2.303RT)`
`log_(10)((k_(2))/(k_(1)))=(24.9xx1000)/(2.3030xx8.314xx300)`
`(k_(2))/(k_(1))=2.162xx10^(-4)`
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