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. 




This problem is relatively easy to solve if you remember that the vapor pressure of the products is independent of the amount of solid, provided there is enough solid for it to be in equilibrium with the gases. Therefore:


$$K\_p=p\_{\ce{NH3}}^2p\_{\ce{CO2}}$$


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 $\pu{0.0582 atm}$.


Answer: C


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