Question

An aqueous solution of ethyl acetate at room temperature has an itial concentration of 0.15 M. Calculate the equilibrium concentrations of ethyl acetate (CH3COOC2H5), ethanol (C2H5OH) and acetic acid if the equilibrium constant at this temperature is Kc= 3.00. Part B. Repeat this calculation for initial ehtyl acetate concentrations of 0.25 M and 0.35 M. Are your answers in agreement with Le Chatelier's Principle?

Answer 1

Pertinent equation for hydrolysis of Ethyl acetate is,

CH3COOC2H5 + H2O <-----------> CH3COOH + HOC2H5

Ethyl acetate Acetic acid Ethanol

Expression for equilibrium constant Kc is given as,

Kc = [CH3COOH][HOC2H5] / [CH3COOC2H5] ..........(1)..([H2O] almost unchanged at equilibrium so not appeared in expression)

Let at equilibrium x M/L of the ethyl acetate get hydrolyzed hence we write the concentrations at equilibrium as,

[CH3COOH] = x M/L

[HOC2H5] = x M/L

[CH3COOC2H5] = (0.15 - x) M/L

Using these concentrations and Kc = 3.00 in above eq.(1) we get,

3 = (x) (x) / (0.15 - x)

3 x (0.15 - x) = x².

x².- 3 x (0.15 - x) = 0.

x².+ 3x - 0.45 = 0

Let us solve this quadratic equation using formula for ax2 + bx + c = 0 as,

x = [-b (b² - 4ac) ^1/2] /2a

comparing our quadratic equation with standard for we get a = 1, b = 3 and c = - 0.45

Hence,

x = [-3 (3² - 4 x 1 x (-0.45)) ^1/2] /2 x1

x = [-3 3.286] /2

x = [-3 + 3.286] /2 or x = [-3 - 3.286] /2 = -ve value hence rulled out.

x = 0.286 / 2

x = 0.143 M/L

Hence equilibrium concentrations of species involved are,

[CH3COOH] = 0.143 M/L

[HOC2H5] = 0.143 M/L

[CH3COOC2H5] = (0.150 - 0.143) = 0.007 M/L

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b) At initial [Ethyl acetate] = 0.25 M

We will have everything same just instead 0.15 we have to take 0.25 M

so,

3 = (x) (x) / (0.25 - x)

3 x (0.25 - x) = x².

x².- 3 x (0.25 - x) = 0.

x².+ 3x - 0.75 = 0

x = [-3 (3² - 4 x 1 x (-0.75)) ^1/2] /2 x1

x = [-3 3.464] /2

x = [-3 + 3.464] /2 or x = [-3 - 3.464] /2 = -ve value hence rulled out.

x = 0.464 / 2

x = 0.232 M/L

Hence equilibrium concentrations of species involved are,

[CH3COOH] = 0.232 M/L

[HOC2H5] = 0.232 M/L

[CH3COOC2H5] = (0.250 - 0.232) = 0.018 M/L

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c) For [Ethyl acetate] = 0.35 M

3 = (x) (x) / (0.35 - x)

3 x (0.35 - x) = x².

x².- 3 x (0.35 - x) = 0.

x².+ 3x - 1.05 = 0

x = [-3 (3² - 4 x 1 x (-1.05)) ^1/2] /2 x1

x = [-3 3.633] /2

x = [-3 + 3.633] /2 or x = [-3 - 3.633] /2 = -ve value hence rulled out.

x = 0.633 / 2

x = 0.317 M/L

Hence equilibrium concentrations of species involved are,

[CH3COOH] = 0.317 M/L

[HOC2H5] = 0.317 M/L

[CH3COOC2H5] = (0.350 - 0.317) = 0.033 M/L

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Le Chateliers principle : Any constrain imposed on equilibrium get nulified by shifting arrow in the direction that nulifies the effect.

From 0.15 to 0.35 M ethyl acetate concentrations we increased concentration of reactatn and by Le Chatelier's principle it sholuld get nulified by lowering or consumption of more and more ethyl acetate it means more and more acetic acid and ethanol should be formed and that is observed from increased values of x i.e.[acetic acid] and [ethanol] 0.007 to 0.017 to 0.033 M/L

Hence, Yes answers are in agreement with Le Chateliers principles.

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