$1 \mathrm{M}\quad 1 \mathrm{M} \quad 1 \mathrm{M} \quad1 \mathrm{M}$
First check direction of reversible reaction.
Since $\mathrm{Q}_{\mathrm{c}}=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}=1<\mathrm{K}_{\mathrm{eq} .} \Rightarrow$ reaction will move in forward direction to attain equilibrium state.
$\Rightarrow \mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}: \mathrm{K}_{\mathrm{eq}}=100$
$to \quad 1\quad 1\quad\quad 1\quad 1$
$\mathrm{t}_{\text {eq. }} \, 1-\mathrm{x} \, 1-\mathrm{x} \, 1+\mathrm{x} \, 1+\mathrm{x}$
Now $: \mathrm{K}_{\mathrm{eq}}=100=\frac{(1+\mathrm{x})(1+\mathrm{x})}{(1-\mathrm{x})(1-\mathrm{x})}$
$\Rightarrow 100=\left(\frac{1+x}{1-x}\right)^{2}$
$(i)$ $10=\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$
$\Rightarrow 10-10 \mathrm{x}=1+\mathrm{x}$
$\Rightarrow 11 \mathrm{x}=9$
$\Rightarrow \mathrm{x}=\frac{9}{11}$
$(ii)$ $-10=\frac{1+\mathrm{x}}{1-\mathrm{x}}$
$\Rightarrow-10+10 \mathrm{x}=1+\mathrm{x}$
$\Rightarrow-9 x=-11$
$\Rightarrow x=\frac{11}{9}$
$\rightarrow$ $'x'$ cannot be more than one, therefore not valid. therefore equation concretion of $(D)=1+x$
$=1+\frac{9}{11}=\frac{20}{11}$
$=1.8181=181.81 \times 10^{-2}$
$\simeq 182 \times 10^{-2}$
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