1.

A catalyst lowered the activation energy byh `25KJmol^(-1)` at `25^(@)C` . By how many times will the rate grow?

Answer» 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. 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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