Question

The equilibrium constant $K _{ c }$ at $298 \,K$ for the reaction $A + B \rightleftharpoons C + D$ is $100$ . Starting with an equimolar solution with concentrations of $A , B , C$ and $D$ all equal to $1\, M$, the equilibrium concentration of $D$ is___ $\times 10^{-2}\,M$. (Nearest integer)

Answer 1

First check direction of reversible reaction.
Since $Q _{ c }=\frac{[ C ][ D ]}{[ A ][ B ]}=1< K _{\text {eq. }}$
$ \Rightarrow $ reaction will

Now $: K _{ eq }=100=\frac{(1+ x )(1+ x )}{(1- x )(1- x )}$
$\Rightarrow 100=\left(\frac{1+x}{1-x}\right)^{2}$
(i) $10=\left(\frac{1+x}{1-x}\right)$
$\Rightarrow 10-10 x=1+x$
$\Rightarrow 11 x=9$
$\Rightarrow x=\frac{9}{11}$
(ii) $-10=\frac{1+x}{1-x}$
$\Rightarrow -10+10 x=1+x$
$\Rightarrow -9 x=-11$
$\Rightarrow x=\frac{11}{9}$
$\rightarrow$ ' $x$ ' cannot be more than one, therefore not valid.
therefore equation concretion of $( D )=1+ x$
$=1+\frac{9}{11}=\frac{20}{11}$
$=1.8181=181.81 \times 10^{-2}$
$\simeq 182 \times 10^{-2}$