The value of $K_{p}$ for the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is $3.0$ at $1000 \ K$. If initially $P_{CO_{2}} = 0.48 \ bar$ and $P_{CO} = 0 \ bar$ and pure graphite is present,calculate the equilibrium partial pressures of $CO$ and $CO_{2}$.

(A) For the reaction,let $x$ be the decrease in pressure of $CO_{2}$.
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
Initial pressure: $0.48 \ bar$ and $0 \ bar$.
At equilibrium: $(0.48 - x) \ bar$ and $2x \ bar$.
$K_{p} = \frac{p_{CO}^{2}}{p_{CO_{2}}} = 3.0$.
Substituting the values: $\frac{(2x)^{2}}{0.48 - x} = 3.0$.
$4x^{2} = 3(0.48 - x) \implies 4x^{2} + 3x - 1.44 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{9 - 4(4)(-1.44)}}{8} = \frac{-3 \pm \sqrt{9 + 23.04}}{8} = \frac{-3 \pm 5.66}{8}$.
Taking the positive root: $x = \frac{2.66}{8} = 0.3325 \approx 0.33$.
Equilibrium partial pressures are:
$p_{CO} = 2x = 2 \times 0.33 = 0.66 \ bar$.
$p_{CO_{2}} = 0.48 - 0.33 = 0.15 \ bar$.