[sasha.w]

I've been studying for 4 days straight, like 12 hours a day, so I promise you I'm putting effort here.  Did I do this question correctly? It's shown up on two of the past exams but in different formats.

The decomposition of ammonium carbamate (NH4CO2NH2) is described in the reaction below:

NH4CO2NH2 (s) 2NH3 (g) + CO2 (g)           endothermic reaction

A 9.407 g sample of NH4CO2NH2 (s) is placed in an evacuated (ie. initial pressure is 0 atm) 4.000 L vessel at 24°C. After equilibrium has been established, the total pressure of the system is 0.318 atm. Some NH4CO2NH2 (s) remains in the vessel.

a) What is the equilibrium partial pressure of NH3 (g) and CO2 (g), respectively? on the other exam that has this same type of question, it says compute the final equilbrium partial pressure of H2, this is essentially the same method correct? Should I make an ice table?

P(NH3) = 2x P(CO2)
3x P(CO2) = 0.318 atm
P(CO2)= (0.318)(2)= 0.636 atm

b) Calculate the KP for the reaction.

kp= P(NH3)^2 P(CO2)= (0.636)^2(0.318) = 1.2x10^-1

c) Calculate the number of moles of NH4CO2NH2 (s) initially

d) Calculate the number of moles of NH3 (g) at equilibrium

e) What mole fraction of NH4CO2NH2 (s) has been decomposed to NH3 (g)? Report answer in percentage (%).

[sasha.w]

This is the second question I'm pretty sure I did correctly. Here's a link to the picture so it's easier for you guys to follow.

https://imgur.com/R2KCpiQ

Thanks

[Borek]

What is the equilibrium partial pressure of NH3 (g) and CO2 (g), respectively?

You are given TOTAL pressure and the ratio of amounts of gases is given by teh decomposition stoichiometry.

P(NH3) = 2x P(CO2)

3x P(CO2) = 0.318 atm[/quote]

OK

P(CO2)= (0.318)(2)= 0.636 atm

Sanity check: partial pressure of the gas can't be higher than total pressure.