Question

The reversible hydrolysis of ethyl acetate to ethanol and acetic acid has an activation energy of 1 kJ/mol and a pre-exponential factor of 1.5*10^-2 M^xsec^y .The reaction is proceeding at 98.6^oF.The density of water is 1000 g/L and the molecular weight of water is 18.02 g/mol.Assume the reaction is elementary and that reversible reaction is negligible when carried out at 98.6^oF.Below is the chemical structure for ethyl acetate:

Write and balance the equation for the hydrolysis of ethyl acetate to ethanol and acetic acid.

Find the integrated rate law for the loss of ethyl acetate.

What is the concentration of ethyl acetate after 2 minutes if the original concentration was 10mM?

Answer 1

1)Balanced chemical equation for the hydrolysis of ethyl acetate to ethanol and acetic acid:

2) Ester hydrolysis follows first order kinetics and hence Integrated rate law for the loss of ethyl acetate is given as,

d(R)/dt = -k1[R]

For ethyl acetate, d[Ethyl acetate]/dt = -k[Ethyl acetate] The integrated form is, [A] = [A]₀ e^-(K₁t) ...............(1)

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3) We have [A]₀ = 10 mM, [A] = ? after t = 2 min = 120 s, K1 = ? and pre-exponential factor(A) = 1.5 x 10-2 sec^-1 .

T = 98.6 0F = 37 ⁰C = 310.15 K

Ea = 1 kJ/mol = 1000 J/mol.

Arrhenius equation states,

K = A x e^-Ea/RT…………………….(2)

K₁ = (1.5 x 10^-2) x E{ -[1000/(8.314x310.15)] }………….(for first order reaction k1 substituted for k)

K = (1.5 x 10^-2) x 0.6785

K = 1.02 x 10^-2 sec^-1.

Calculation of [A] =?

We have Integrated form of First order reaction,

[A] = [A]₀ e^-k₁^t……

Let us put all known values,

[A] = (10mM)x E[-(1.02 x 10^-2 x 120)

[A] = 10 x 0.294

[A]= 2.94 mM

Hence, concentration of ethyl acetate after 2 min (120 s) will be 2.94 mM.

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