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Question: Given that the temperature coefficient for the saponification of ethyl acetate by NaOH is 1.75. Calc...

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

Explanation

Solution

The reactants molecules in a chemical system have initial energy of their own,Er{E_r}. They need to acquire threshold energy, ETh{E_{Th}}, before they react chemically. When the energy of the reactant is less than threshold energy then some extra energy is required to bring about the reaction (effective collision). Therefore, the extra energy supplied to the reactant molecules to attain the threshold energy to undergo chemical reaction is called activation energy, Ea{E_a} .
Activation energy = Threshold energy – Average kinetic energy of the reactant molecules
Ea=EThEr{E_a} = {E_{Th}} - {E_r}.

Complete step by step answer:
In this question the temperature coefficient is given and using this we have to calculate the activation energy for the saponification of ethyl acetate.
Given,
Temperature coefficient (k2k1)=1.75\left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = 1.75
We know that the temperature will rise from 25C{25^ \circ }C to 35C{35^ \circ }C so
T1=25C=298K{T_1} = {25^ \circ }C = 298K
T2=35C=308K{T_2} = {35^ \circ }C = 308K
R=1.98cal.R = 1.98cal.
Now,
Applying Arrhenius equation we will find the activation energy
log(k2k1)=Ea2.303R[T2T1T1T2]\log \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = \dfrac{{{E_a}}}{{2.303\,R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]
log(1.75)=Ea2.303×1.98[308298298×308]\Rightarrow \log \left( {1.75} \right) = \dfrac{{{E_a}}}{{2.303 \times 1.98}}\left[ {\dfrac{{308 - 298}}{{298 \times 308}}} \right]
(We have put the given values in the above equation)
0.2430=Ea2.303×1.98[10298×308]\Rightarrow 0.2430 = \dfrac{{{E_a}}}{{2.303 \times 1.98}}\left[ {\dfrac{{10}}{{298 \times 308}}} \right]
Ea=0.2430×2.303×1.98×298×30810\Rightarrow {E_a} = \dfrac{{0.2430 \times 2.303 \times 1.98 \times 298 \times 308}}{{10}}
Ea=10.207Kcal/mol\Rightarrow {E_a} = 10.207Kcal/mol
Therefore, the value of activation energy for the saponification of ethyl acetate is found to be 10.207Kcal/mol10.207Kcal/mol.

Note: It is to be noted that each reaction at a particular temperature has a definite value of activation energy. The value of activation energy decides the fraction of the total number of effective collisions. If the activation energy is low a large number of molecules will possess this energy. Hence, the number of effective collisions (f represents fraction of the total number of effective collisions) will be more and the rate of the reaction will be high. Thus,
For fast reaction the activation energy is low
For low reaction the activation energy is high
The hydrolysis of an ester molecule with NaOH or KOH to give alcohol and sodium or potassium salts of acid is known as saponification. This process is used in making soaps.