The equilibrium constant at 298 K for a reaction $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$ is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration ofD (in ${\mathrm{molL}}^{-1}$) will be:

$\begin{array}{l}\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D};\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{K}_{\mathrm{eq}}=100\\ 1\mathrm{M}-x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\mathrm{M}-x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\mathrm{M}+x\mathit{ }1\mathrm{M}+x\end{array}$

$K_{\mathrm{eq}}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}=\frac{(1\mathrm{M}+x{)}^2}{(1\mathrm{M}-x{)}^2}=100$

This gives $\frac{1\mathrm{M}+x}{1\mathrm{M}-x}=10$ i.e. $1\mathrm{M}+x=10(1\mathrm{M}-x)$

or $11x=9\mathrm{M}$ or $x=(9/11)\mathrm{M}=0.818\mathrm{M}$

Equilibrium concentration of D is $[\mathrm{D}{]}_{\mathrm{eq}}=1\mathrm{M}+x=1\mathrm{M}+0.818\mathrm{M}=1.818\mathrm{M}$