## Chemistry (4th Edition)

$$K_{p}= 7.09 \times 10^{-3}$$
Since we started with only the solid: $P_{NH_3} = 2 P_{CO_2}$ at equilibrium. $P_{CO_2} + P_{NH_3} = 0.363$ atm $P_{CO_2} + 2P_{CO_2} = 0.363$ atm $3P_{CO_2} = 0.363$ atm $P_{CO_2} = 0.121$ atm $P_{NH_3} = 0.242$ atm $$K_{p} = P_{CO_2}P_{NH_3}^2 = (0.121)(0.242)^2 = 7.09 \times 10^{-3}$$