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In: Chemistry

At a certain temperature, the Kp for the decomposition of H2S is 0.846. H​2​S(g) <---> ​H​2...

At a certain temperature, the Kp for the decomposition of H2S is 0.846.

H​2​S(g) <---> ​H​2 (g) ​+ S (g)​  

Initially, only H2S is present at a pressure of 0.161 atm in a closed container. What is the total pressure in the container at equilibrium? (in atm)

Solutions

Expert Solution

Dear friend your answer is

                   H2S    ==>      H2     +      S
initial....        0.161 atm.       0...            0
change.....      -2p.....           p....           p
equil.....        0.161-2p..        p....           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 = 0.5125atm  

Thank you, all the best.

queen_honey_blossom answered 2 years ago
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