The rate of reaction becomes 2 times for every `10^@C` rise in temperature. How the rate of reaction will increase when temperature is increased from `30^@C" to" 80^@C` ?

Options

A. 16
B. 32
C. 64
D. 128

Explanation

`(k_(t + 10))/k_t=(r_(t+ 10))/r_1=2`,
For an increase of temperature to `50^@C`, i.e., 5 times, the rate increases by `2^5` times, i.e., 32 times.