Question

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

Answer 1

wrong question it cant be calculated

Answer 2

Given,

$K_{2}$ / $K_{1}$ = 1.75

$T_{1}$ = 25°C = 25+273 = 298 K$K_{2} /K_{1}= E_{a} /R [ T_{2} - T_{1} /$

$T_{2}$ = 35°C = 35 + 273 = 308

∵ The temperature coefficient is ratio of rate constants at 35°C and 25° C respectively

2.303 log$K_{2} /K_{1}= E_{a} /R [ T_{2} - T_{1} [tex]/ [tex]T_{1}$×$T_{2}$

⇒2.303 log 1.75 = $E_{a} /1.987$ [308-298/308×298]

$E_{a}$ = 10.207 kcal$mol^{-1}$