Finding partial pressures from Kp and Kc in equilibrium

[~angel~]

Just a few questions I'm having trouble with.

1.Ammonia is formed from nitrogen and hydrogen according to the following equlibrium reaction:

N2(g)+3H2(g)-->2NH3(g)

The plant operates close to 700K, at which Kp is 1.00*10^-4 atm-2 and employs the stoiciometric ratio 1:3 of N22. At equilibrium, the partial pressure of NH3 is 50 atm. Calculate the partial pressures of the each reactant.

2.Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 250c, Kc=1.58*10^-8 M^3 for the following equilibrium:

NH2COONH4 (s) <---> 2NH3 (g) + CO2 (g)

If 7.81g of NH2COONH4 is introduced into a 0.500L evacuated container, what is the total partial pressure inside the container at equilibrium at 250c?

Thanks.

 

[WhirlwindMonk]

1. Kp = (Partial Pressure NH3)^2 / ((PP N2) * (PP H2)^3)

2. Kc = [NH3]^2 * [CO2]

That should get you started.

 

[~angel~]

For 1., I got to [N2]*

^3=2.5*10^7 then I just got lost. I know the ratio is 1:3, but that cubed hydrogen has really put me off.

 

[WhirlwindMonk]

for ever x atm of N2 there are 3x atm of H2. So (x)(3x)^3=2.5*10^7

 

[~angel~]

So you get 3x^4=2.5*10^7
x= 53.73?

If that's what you meant, apparently it's the wrong answer for N2.

 

[WhirlwindMonk]

it would be 27x^4=2.5*10^7, you forgot to cube the 3.

 

[~angel~]

lol...thanks for that.