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The rate of a reaction gets doubled when...

The rate of a reaction gets doubled when the temperature changes from `7^(@)C` to `17^(@)C`. By what factor will it change for the temperature change from `17^(@)C` to `27^(@)C` ?

A

`1.81`

B

`1.71`

C

`1.91`

D

`1.76`

Text Solution

Verified by Experts

The correct Answer is:
c
`2.303 "log"K_(2)/K_(1)=E_(A)/R[(T_(2)-T_(1))/(T_(1)T_(2))]`
`T=280 K, T=290 K, K_(2)/K_(1)=2`
Find `E_(a)` and then again use:
`2.303"log" K_(2)^(')/K_(1)^(')=E_(A)/R[(T_(2)^(')-T_(1)^('))/(T_(1)^(')T_(2)^('))]`
If `T_(1)^(')=290K, T_(2)^(')=300K,`
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