b
\({A}+{B} \rightleftharpoons {C}+{D}: {K}_{{eq}}=100\)

\(1 {M}\quad 1 {M} \quad 1 {M} \quad1 {M}\)

First check direction of reversible reaction.

Since \({Q}_{{c}}=\frac{[{C}][{D}]}{[{A}][{B}]}=1<{K}_{{eq} .} \Rightarrow\) reaction will move in forward direction to attain equilibrium state.

\(\Rightarrow {A}+{B} \rightleftharpoons {C}+{D}: {K}_{{eq}}=100\)

\(to \quad 1\quad 1\quad\quad 1\quad 1\)

\({t}_{\text {eq. }} \, 1-{x} \, 1-{x} \, 1+{x} \, 1+{x}\)

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 \((D)=1+x\)

\(=1+\frac{9}{11}=\frac{20}{11}\)

\(=1.8181=181.81 \times 10^{-2}\)

\(\simeq 182 \times 10^{-2}\)