Science, asked by jemsey11, 1 year ago

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

Answers

Answered by siddarth17
7
wrong question it cant be calculated
Answered by omegads04
4

Given,

K_{2} / K_{1} = 1.75

T_{1} = 25°C = 25+273 = 298 KK_{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 logK_{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 kcalmol^{-1}

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