# Equilibrium pressure

Ammonium carbamate dissociates as follows:

$$\ce{NH2COONH4(s)<=> 2NH3(g) + CO2(g)}$$

The value of $$K_p$$ for this reaction is found to be equal to $$\pu{2.92 \times 10^-5 atm^3}$$. If one mole of ammonium carbamate is heated in a sealed container, the total pressure developed in the container is:

• (A) $$\pu{0.0194 atm}$$
• (B) $$\pu{0.0388 atm}$$
• (C) $$\pu{0.0582 atm}$$
• (D) $$\pu{0.0667 atm}$$

I first tried considering initial moles as $$1$$ and taking the moles used at equilibrium as $$x$$. I was able to form an equation with equilibrium constant, $$x$$, total pressure(with the help of fact that partial pressure = mole fraction * total pressure; and also that equilibrium cost = (product of partial pressure of products raised to their stoichiometric coefficients) / (partial pressure of reactants raised to their coefficients).

But I am getting only one equation while two unknowns: $$x$$ and total pressure.

$$K_p=p_{\ce{NH3}}^2p_{\ce{CO2}}$$
Since $$p_{\ce{NH3}}=2p_{\ce{CO2}}$$
$$K_p=4p_{\ce{CO2}}^3$$
and $$p_{\ce{CO2}}=\pu{0.0194 atm}$$. This is one third of the total pressure, which must be $$\pu{0.0582 atm}$$.