A catalyst lowered the activation energy byh `25KJmol^(-1)` at `25^(@)C` . By how many times will the rate grow?
Text Solution
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The rate of reaction is related to the activation energy by the following relation: `(K_(2))/(K_(1))=` Antilog `[(DeltaE)/(2.303RT)]` …(i) Given, `DeltaE=25xx10^(3)J` `R=8.314JK^(-1)mol^(-1)` `T=25^(@)C=298K` Substituting all the values in Eq. (i), we get `(K_(2))/(K_(1))=` Antilog `[(25xx10^(3))/(2.303xx8.314xx298)]=24069` `:. K_(2)=K_(1)xx24069` Therefore, the rate increases by `24069` times.
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