Question

For the reaction $A+B\rightleftharpoons C+D$ taking place in a $1L$ vessel; equilibrium concentration of $\left[C\right]=\left[D\right]=0.5M$ if we start with $1$ mole each of $A$ and $B$. Percentage of $A$ converted into $C$ if we start with $2$ moles of $A$ and $1$ mole of $B$, is__________.

• A. $25$%
• B. $40$%
• C. $66.66$%
• D. $33.33$%

Answer 1

The correct option is D $33.33$%

Given

Equilibrium conc. of C and D is 0.5M

Case 1 Conc of A and B be 1M

Case 2 Conc of A be 2 M and B be 1M

Solution

CASE 1

A + B = C + D

t=0 1 1 0 0

t=t 1-0.5 1-0.5 0.5 0.5

$K=\frac{\left[C\right]\ast \left[D\right]}{\left[A\right]\ast \left[B\right]}$

At t=t K=1

CASE 2

Let no. of moles of C and D is x

A + B = C + D

t=0 2 1 0 0

t=t 2-x 1-x x x

$K=\frac{\left[x\right]\ast \left[x\right]}{[2-x]\ast [1-x]}$

Putting K=1

$1=\frac{x^2}{[2-x]\ast [1-x]}$

On solving x=0.667

percentage of A converted=percentage of C formed

0.667/2*100=33.33%

The correct option is D