Question
In: Chemistry
At a certain temperature, the Kp for the decomposition of H2S is 0.846. H2S(g) <---> H2...
At a certain temperature, the Kp for the decomposition of H2S is 0.846.
H2S(g) <---> H2 (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 agoRelated Solutions
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