The equilibrium constant for the reaction; ${\mathrm{H}}_{2( \mathrm{g})}+{\mathrm{CO}}_{2( \mathrm{g})}\iff {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}$is 16 at $1000°\mathrm{C}$. If 1.0 mole of ${\mathrm{H}}_2$ and 1.0 mole of ${\mathrm{CO}}_2$ are placed in one litre flask, the final equilibrium concentration of $\mathrm{CO}$ at $1000°\mathrm{C}$ is

The equilibrium constant for the reaction; ${\mathrm{H}}_{2( \mathrm{g})}+{\mathrm{CO}}_{2( \mathrm{g})}\iff {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}$ is 16 at $1000°\mathrm{C}$.

${\mathrm{H}}_{2( \mathrm{g})}+{\mathrm{CO}}_{2( \mathrm{g})}\iff {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}$

1                1                   0                0      at t=0 sec

1-$\alpha$       1-$\alpha$                     $\alpha$             $\alpha$        at t=eq

$K=\frac{\left[H_{2}O\right]\left[CO\right]}{\left[H_2\right]\left[C{O}_2\right]}$

$=\frac{\alpha ·\alpha }{(1-\alpha )(1-\alpha )}=16$

$=\sqrt{\frac{{\alpha }^2}{(1-\alpha {)}^2}}=\sqrt{16}$

$=\frac{\alpha }{1-\alpha }=4$

$\Rightarrow \alpha =4-4\alpha .$

$5\alpha =4$

$\alpha =\frac{4}{5}$ .

conc. of $CO=\alpha =\frac{4}{5}=0.8M$.

Hence the correct option is (A) $0.8\mathrm{M}$