Question
At a certain temperature, the Kp for the decomposition of H2S is 0.791.
H2S<--> H2 + S
Initially, only H2S is present at a pressure of 0.111 atm in a closed container. What is the total pressure in the container at equilibrium?
Please Help!
H2S<--> H2 + S
Initially, only H2S is present at a pressure of 0.111 atm in a closed container. What is the total pressure in the container at equilibrium?
Please Help!
Answers
We need the phases.
Is that H2S(g) ==> H2(g) + S(g) or S(s)
I assume H2S(g) ==> H2(g) + S(g)
I........0.111.......0.......0
C.........-x.........x.......x
E.......0.111-x.....x........x
Kp = 0.791 = p(H2)*p(S)/p(H2S)
Solve for x = p(H2) = p(S)
p(H2S) = 0.111-x
Then Ptotal = p(H2) + p(S) + p(H2S)
Is that H2S(g) ==> H2(g) + S(g) or S(s)
I assume H2S(g) ==> H2(g) + S(g)
I........0.111.......0.......0
C.........-x.........x.......x
E.......0.111-x.....x........x
Kp = 0.791 = p(H2)*p(S)/p(H2S)
Solve for x = p(H2) = p(S)
p(H2S) = 0.111-x
Then Ptotal = p(H2) + p(S) + p(H2S)
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