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Chemistry 11 Online
OpenStudy (anonymous):

URGENT! Consider the following decomposition reaction of ammonium carbonate ((NH4)2CO3): (NH4)2CO3 (s)  2 NH3 (g) + CO2 (g) + H2O (g) In one experiment at 25.0°C, a sample of pure (NH4)2CO3 is placed in an evacuated 2.00 L vessel. At equilibrium, the total pressure is found to be 0.8944 atm. (a) Determine the equilibrium partial pressure of each gaseous species. (b) Determine the mass of ammonium carbonate ((NH4)2CO3) that must have reacted in order to achieve equilibrium. (c) Determine the value of Kp assuming that the reaction temperature is 25.0°C. I can find a but I am stuck at b

OpenStudy (anonymous):

use PV=nRT to find the moles and then convert back

OpenStudy (anonymous):

but wait Im stuck with the ATM at equilibrium, the solid doesnt affect equilibrium at all, so how can I find it with the equilibrium of Pressure Don't I need concentration?

OpenStudy (anonymous):

So you have pressure at EQ right?

OpenStudy (anonymous):

4x=.8944

OpenStudy (anonymous):

Yeah, the total, and I can use basic mole ratio to find individual pressure

OpenStudy (anonymous):

solve for x then use the pressure for lets say CO2 put that into PV=nRT then solve for n then convert over

OpenStudy (anonymous):

So I need to find n for one product than just use stoichiometry ?

OpenStudy (anonymous):

yeah it should work.

OpenStudy (anonymous):

let me work it out for you really fast

OpenStudy (anonymous):

Alright ill try it and post if it works

OpenStudy (anonymous):

(.2236)(2)/(298*.08206) = .0183*96g/mol = 1.76g

OpenStudy (anonymous):

I cant beleive I didnt think of that, and for C , Q goes to infinity no? How to I possible use my Equilibrium constant equation?

OpenStudy (anonymous):

How do I possibly*

OpenStudy (anonymous):

for C? all you do is [NH3]^2[CO2][H2O]=Kp x=.2236 (2*.2236)^2(.2236)*(.2236)=.001

OpenStudy (anonymous):

Not for PNH3 not [NH3] for each

OpenStudy (anonymous):

Oh I thought you had to use ICE and Q to find the direction of the reaction :/ Apparently you don't thanks a lot!

OpenStudy (anonymous):

Q is to judge where you are in respect to K if you have a larger Q you know you have far too many products so you need to get more reactants so the reaction goes to the left, if you have a small Q you know you have too many reactants and you need to go to products. you as in the reaction lol.

OpenStudy (anonymous):

Ah I get it its just to judge the direction of the reaction BEFORE equilibrium but here we already reached it! Thanks a lot!

OpenStudy (anonymous):

Yep you got it! good luck!

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