The rate constant of a reaction is given by `k = 2.1 xx 10^(10) exp(-2700//RT)`. It means that

Options

A. log K vs `1//T "will be a curvad line with solpe" =-(2700)/(2.303R)`
B. log k vs `1//T` will be a straight line with intercept on
` "log K axis -log" 2.1xx10^(10)`
C. the number of effective collisions are
`2.1xx1-0^(10) "cm"^(-3)s^(-1)`
D. half-life of the reaction increases with encrease of temperature

Answer

B

Explanation

`k=2.1xx10^(10) "exp" (-2700//RT)`
or log k=log `2.1xx10^(10)-((2700)/(RT))`
i.e. log k vs `1//"T" `will be straight line with entercept on log k axis=log `2.1xx10^(10)`