Given that the temperature coefficient for the saponification of ethylacetate by `NaOH` is `1.75`. Calculate the activation energy.

Options

A. 1.0207 kcal
B. 10.207 kcal/mol1.0207 cal
C. 10.207 cal/mol
D. 149.5 kJ

Answer

B

Explanation

Given, `K_(2)/K_(1) = 1.75`
`T_(1) = 25^(@) C = 25 + 273 = 298` K
`T_(2) = 35^(@) C= 35 + 273 = 308 K`
(Since tem perature coefficient is ratio o f rate constants at `35^(@)` C and `25^(@)` C respectively).
`therefore 2.303 log_(10)K_(2)/K_(1) =E_(a)/R([T_(2)-T_(1)])/(T_(1)T_(2)]`
`therefore 2.303 log_(10) 1.75 =E_(a)/1.987 ([308-298])/(308 xx 298)`
`therefore E_(a) =(2.303 xx 308 xx 298 xx 1.987)/(10) xx log 1.75 cal mol^(-1)`
`E_(a) = 10.207 kcal mol^(-1)`