The velocity constant of a reaction at 290 K was found to be `3.2xx10^(-3)s^(-1)`. When the temperature is raised to 310 K, it will be about

Options

A. `6.4xx10^(-3)`
B. `3.2xx10^(-4)`
C. `9.6xx10^(-3)`
D. `1.28xx10^(-2)`

Answer

D

Explanation

The rates of many reactions are approximately doubled for every `10^(@)C` rise in temperature From 290 K to 310 Ki.e. for `20^(@)C` rise in temperature, the velocity or rate constant =` 2xx2xx3.2xx10^(-3)s^(-1)`
=`12.8xx10^(-3)=1.28xx10^(-2)s^(-1)`