$1 m o l e$ of a gas $^{'} A^{'}$ is taken in a vessel of volume $1 L$ . It dissociated according to the reaction $A (g) ⇌ B (g) + C (g)$ at $27$ . Forward and backward reaction rate constants for the reaction are $1.5 \times 10^{- 2}$ and $3 \times 10^{- 2}$ respectively. Find the concentration of $A$ , $B$ and $C$ at equilibrium. Also find $K_{R}$ and $K_{c}$ with proper units.

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Solution

$A (g) ⇌ B (g) + C (g)$
$1$ $0$ $0$ at $t = 0$
$1 - x$ $x$ $x$ at $t = t_{1} s e c$

$K_{c} = \frac{K_{f}}{K_{b}}$
$K_{c} = \frac{1.5 \times 10^{- 2}}{3 \times 10^{- 2}} = .5$

$K_{c} = \frac{[B] [C]}{[A]}$
$K_{c} = \frac{[x] [x]}{[1 - x]}$
$\frac{1}{2} = \frac{x^{2}}{1 - x}$
$2 x^{2} + x - 1$ , equal this binomial thoerem to zero, to get the value of $x$
$x = \frac{1}{2}$

Now the moles at equilibrium,
$f o r A = 1 - x = 1 - .5 = .5$
$f o r B = x = .5$
$f o r C = x = .5$

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A(g)⇌B(g)+C(g) 1 0 0 at t=01−x x x at t=t1sec Kc=KfKb Kc=1.5×10−23×10−2=.5 Kc=[B][C][A] Kc=[x][x][1−x] 12=x21−x 2x2+x−1, equal this binomial thoerem to zero, to get the value of x x=12Now the moles at equilibrium,forA=1−x=1−.5=.5forB=x=.5forC=x=.5