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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Solution:
$$ \begin{aligned} & A+B \rightleftharpoons C+D \ & \begin{array}{lllll} \text { Initially at } t=0 & 1 & 1 & 1 & 1 \end{array} \ & \text { At equilibrium } \begin{array}{llll} 1-x & 1-x & 1+x & 1+x \end{array} \ & K _{eq}=\frac{[C][D]}{[A][B]}=\frac{(1+x)(1+x)}{(1-x)(1-x)}=\frac{(1+x)^{2}}{(1-x)^{2}} \ & 100=\left(\frac{1+x}{1-x}\right)^{2} \quad \text { or } \quad 10=\frac{1+x}{1-x} \ & \text { or } \quad 10-10 x=1+x \ & 10-1=x+10 x \ & 9=11 x \ & x=\frac{9}{11}=0.818 \ & \therefore \quad[D]=1+x=1+0.818=1.818 \end{aligned} $$
