Answer
$$K_p = 7.09 \times 10^{-3}$$
Work Step by Step
1. At equilibrium, these are the partial pressures we get:
$P_{NH_3} = 0 + 2x$
$P_{CO_2} = 0 + x$
2. According to the problem, $P_{NH_3} + P_{CO_2} = 0.363 \space atm$. Substituting:
$2x + x = 0.363$
$3x = 0.363$
$x = 0.121$
$P_{NH_3} = 2(0.121) = 0.242 \space atm$
$P_{CO_2} = 0.121 \space atm$
3. Write the expression for $K_p$ and evaluate it:
$$K_p = \frac{[Products]}{[Reactants]} = P_{NH_3}^2P_{CO_2}$$
$$K_p = (0.242)^2(0.121) = 7.09 \times 10^{-3}$$
