The value of $K_{c} = 4.24$ at $800 \, K$ for the reaction,
$CO_{(g)} + H_{2}O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$
Calculate equilibrium concentrations of $CO_{2}$,$H_{2}$,$CO$ and $H_{2}O$ at $800 \, K$,if only $CO$ and $H_{2}O$ are present initially at concentrations of $0.10 \, M$ each.

(N/A) For the reaction,
$CO_{(g)} + H_{2}O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$
Initial concentration:
$CO: 0.1 \, M, H_{2}O: 0.1 \, M, CO_{2}: 0, H_{2}: 0$
Let $x \, M$ be the concentration of each product formed at equilibrium.
At equilibrium:
$[CO] = (0.1 - x) \, M, [H_{2}O] = (0.1 - x) \, M, [CO_{2}] = x \, M, [H_{2}] = x \, M$
Equilibrium constant expression:
$K_{c} = \frac{[CO_{2}][H_{2}]}{[CO][H_{2}O]} = \frac{x^{2}}{(0.1 - x)^{2}} = 4.24$
Taking square root on both sides:
$\frac{x}{0.1 - x} = \sqrt{4.24} \approx 2.059$
$x = 2.059(0.1 - x)$
$x = 0.2059 - 2.059x$
$3.059x = 0.2059$
$x = \frac{0.2059}{3.059} \approx 0.0673 \, M$
Equilibrium concentrations:
$[CO_{2}] = [H_{2}] = 0.0673 \, M$
$[CO] = [H_{2}O] = 0.1 - 0.0673 = 0.0327 \, M$