At a certain temperature, the Kp for the decomposition of H2S is 0.846. Hâ€‹2â€‹S(g) <---> â€‹Hâ€‹2 (g)...

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At a certain temperature, the Kp for the decomposition of H2S is0.846.
H_â€‹2â€‹S(g) <---> â€‹H_â€‹2 (g) â€‹+ S(g)â€‹Â Â
Initially, only H2S is present at a pressure of 0.161 atm in aclosed container. What is the total pressure in the container atequilibrium? (in atm)

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3.9 Ratings (695 Votes)

Dear friend your answer is

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â H₂SÂ Â Â ==>Â Â Â Â Â H₂Â Â Â Â +Â Â Â Â Â S
initial....Â Â Â Â Â Â Â 0.161 atm.Â Â Â Â Â Â 0...Â Â Â Â Â Â Â Â Â Â Â 0
change.....Â Â Â Â Â -2p.....Â Â Â Â Â Â Â Â Â Â p....Â Â Â Â Â Â Â Â p
equil.....Â Â Â Â Â Â Â 0.161-2p..Â Â Â Â Â Â Â p....Â Â Â Â Â Â Â Â Â Â p

Kp = pH2*pS/pH2S
Substitute into the Kp expression from the ICE chart above and solve for p through quadratic equation

0.846 = pH2*pS/pH2S = (p)(p)/(0.161-p)
Solve for p = 0.1905 so
pH2 = 0.1905
pS = 0.1905

1 mol H2S gas decomposes to 2 mols that the pressure should increase ( 2X 0.161) = 0.322
pH2S = 0.322 - 0.1905 = 0.1315
Ptotal = 0.1905 + 0.1905+ 0.1315 = 0.5125atmÂ Â

Thank you, all the best.

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