Question

When ethyl alcohol and acetic acid mixed together in equimolecular proportions, equilibrium is attained when two - thirds of the acid and alcohol are consumed. The equilibrium constant of the reaction will be:

• A. 0.4
• B. 40
• C. 0.04
• D. 4

Answer 1

Answer: (A)

The correct option is

D

4

Given that

Equilibrium constant (Kc) is the ratio of the product of the product concentration that is raised to the powers of their coefficient to the products of the reactant concentration that is raised to powers of their coefficient.

Now we considered an equation is

C2H5OH+CH3COOH→CH3COOC2H5+H2O{{C}_{2}}{{H}_{5}}OH+C{{H}_{3}}COOH\to C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$C{H}_{3}COOH+C_2{H}_{5}OH⟶C{H}_{3}COO{C}_2{H}_5+H_{2}O$

The fraction of both ethanol and acetic acid consumed =2/3

The no.of moles of $C_2{H}_{5}OHandC{H}_{3}COOH$ left =$1-\frac{2}{3}=\frac{1}{3}$

The no.of moles of $C{H}_{3}COO{C}_2{H}_{5}and{H}_{2}O$ at equilibrium = 2/3

$K_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}$ = $\frac{\frac{2}{3}\times \frac{2}{3}}{\frac{1}{3}\times \frac{1}{3}}=4$

Therefore,

$Kc=4$