Answer
$ P_{ CO } = 0.713 \space atm$
$ P_{ CO_2 } = 0.287 \space atm$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& P_{ CO }& P_{ CO_2 }\\
Initial& 1.00 & 0 \\
Change& -x& +x\\
Equilibrium& 1.00 -x& x\\
\end{vmatrix}$$
- The exponent of each pressure is equal to its balance coefficient.
$$K_P = \frac{P_{Products}}{P_{Reactants}} = \frac{P_{ CO_2 }}{P_{ CO }}$$
2. At equilibrium, these are the concentrations of each compound:
$ P_{ CO } = 1.00 - x$
$ P_{ CO_2 } = x$
$$0.403 = \frac{(x)}{(1.00 - x)}$$
3. Solve for x:
$0.403(1.00 -x )$
$0.403 - 0.403x = x$
$0.403 = 1.403x$
$$x = \frac{0.403}{1.403} = 0.287$$
$ P_{ CO } = 1.00 - 0.287 = 0.713$
$ P_{ CO_2 } = x = 0.287 $
