The activation energy for most of the reaction is approximately 50 kJ `mol^(-1)` The value of temperature coefficient for such reaction will be

Options

A. `approx 2`
B. `approx 3`
C. 1
D. `gt 4`

Answer

A

Explanation

`log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log.(k_(2))/(k_(1))=(50,000)/(2.303R)((310-300)/(310xx300))`
`log.(k_(2))/(k_(1))=(50,000xx10)/(2.303xx8.314xx310xx300)`
=0.2808
`(k_(2))/(k_(1))`=antilog `0.2808=1.9089~=2`