# Given a reaction $A + B \rightleftharpoons C + D$ taking place in a 1L vessel. If we start with 1 mole each of A and B, equilibrium concentration of [C] = [D] = 0.5 M. 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%

## 1 Answer

Answer: 33.33%

In first case,

A + B $\rightleftharpoons$ C + D
Initial Concentration: [A] = 1 M, [B] = 1 M
Equilibrium Concentration: [A] = (1-0.5) M, [B] = (1-0.5) M, [C] = 0.5M, [D] = 0.5M

Since, [A] = [B] = [C] = [D] = 0.5 M

$\therefore K = \frac{[C][D]}{[A][B]}$ = 1

In second case,

A = B $\rightleftharpoons$ C + D
Initial Concentration: [A] = 2 M, [B] = 1 M
Equilibrium Concentration: [A] = (2-x) M, [B] = (1-x) M, [C] = x M, [D] = x M

$K = \frac{[C][D]}{[A][B]}$ = 1

by putting the concentration values in the equation,
$\therefore \frac{x^2}{(2-x)(1-x)} = 1$
$\Rightarrow x^2 = (2-x)(1-x)$
$\Rightarrow x^2 = -2x - x + 2 + x^2$
$\therefore x = \frac{2}{3}$

$\therefore$ % of A converted to C = $\frac{x}{2} \times 100$
$= \frac{2 \times 100}{3 \times 2}$ = 33.33%

answered Nov 23, 2013

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