Chemical and Ionic Equilibrium - Result Question 12
12. The equilibrium constant at $298 K$ for a reaction, $A+B \rightleftharpoons C+D$ is $100$ . If the initial concentrations of all the four species were $1 M$ each, then equilibrium concentration of $D$ (in $mol $ $L^{-1}$ ) will be
(2016 Main)
(a) $0.818$
(b) $1.818$
(c) $1.182$
(d) $0.182$
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Answer:
Correct Answer: 12. (b)
Solution:
$K_{\mathrm{eq}}=\frac{[C][D]}{[A][B]}=\frac{(1+x)(1+x)}{(1-x)(1-x)}=\frac{(1+x)^2}{(1-x)^2}$
$ or \hspace{7mm} 100=\left(\frac{1+x}{1-x}\right)^2 \quad \text { or } \quad 10=\frac{1+x}{1-x}$
$\begin{aligned} & or \hspace{7mm}10-10 x=1+x \\ & 10-1=x+10 x \\ & 9=11 x \\ & x=\frac{9}{11}=0.818 \\ & \therefore \quad [D]=1+x=1+0.818=1.818 \\ \end{aligned}$
BETA
