At a certain temperature, the Kp for the decomposition of H2S is 0.832
H2S (g) <-- --> H2 (g) + S (g)
Initially, only H2S is present at a pressue of 0.121 atm in a closed container. what is the total pressure in the container at equilibrium?
Please tell me how to do it and confirm the right answer please. Thanks
Jerry? Sarah? We prefer that you not change screen names.
............H2S ==> H2 + S
initial....0.121 atm.0...0
change.....-2p.....p....+p
equil.....0.121-2p..p....p
Kp = pH2*pS/pH2S
Substitute into the Kp expression from the ICE chart above and solve for p, then 0.121-2p.
Add the partial pressures to find total pressure.
I got p= 0.058447. Is this right. Also what do I do afterwards. I'm confused :(
I obtained 0.584 atm also. You solved for p and that gives you the partial pressure of H2 and S. To find partial pressure of the H2S, that is 0.121-p. Evaluate that, then add the partial pressure of each gas to find the total pressure in the container. The problem asks total P.
okay when i plug it into 1.121-p...i get .537..i get how to do that. But i don't get when u say "add the partial pressure of each gas to find the total pressure in the container. " What do I add to .537. What partial pressure are u talking about? Sorry I'm still confused.
DrBob222 please help?
Note 1. I did NOT obtain 0.584; I obtained 0.05844 which is essentially what you obtained. I just made a typo.
Note 2. that pH2S at the beginning is 0.121 and not 1.121 as you typed. I assume this is just a typo also.
p = partial pressure.
pH2 = 0.05845 atm
pS = 0.05845 atm
pH2S = 0.121-2(0.05845) = 0.00410 and not 0.537.
Ptotal = pH2 + pS + pH2S
Ptotal = 0.05845 + 0.05845 + 0.00410 = ?
and round to 3 significant figures.