Question

When $3moles$ of ethyl alcohol are mixed with $3moles$ of acetic acid, $2moles$ of ester are formed at equilibrium according to the equation:

$C{H}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons C{H}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$

The value of the equilibrium constant for the reaction is:

• A. $4$
• B. $2/9$
• C. $2$
• D. $4/9$

Answer 1

The correct option is A $4$

Given reaction: When 3 moles of acetic acid are reacting with 3 moles of ethyl alcohol, 2 moles of ester are formed at equilibrium:

$C{H}_{3}COOH\left(l\right)+C_2{H}_{5}OH\rightleftharpoons C{H}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$

3 3 0 0

At any time ‘t’, ‘x’ moles of product forms, so

3-x 3-x x x

At equilibrium, 2 moles of product are formed so $x=2$

1 1 2 2

So equilibrium constant Kc is given by the product of concentrations of products divided by product of concentrations of reactants i.e.$K_c=\frac{2\times 2}{1\times 1}$

Hence equilibrium constant is 4. Option A