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# How do I solve this equilibrium problem?

2H2S (g)==> 2H2 (g)+S2 (g)

When heated, hydrogen sulfide has decomposes to the equation above. A 3.40 g sample of H2S (g) is introduced into an evacuated rigid 1.25L container. The sealed container is heated to 483K.

A. Write the equilibrium constant Kc for the reaction above

B. Calculate the equilibrium concentration  for H2(g) and H2S (g)

C. Calculate the partial pressure of S2(g) in the container at equilibrium at 438K.

D. For the reaction H2 (g)+1/2 S2 (g)==>H2S (g) at 483K, calculate the value of the equilibrium constant Kc

### 1 Answer by Expert Tutors

J.R. S. | Ph.D. in Biochemistry--University Professor--Chemistry TutorPh.D. in Biochemistry--University Profes...
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2H2S (g)==> 2H2 (g)+S2 (g)

A.  Kc = [H2]2[S2]/[H2S]2

B.  3.4g H2S x 1 mole/34 g = 0.1 moles/1.25 L = 0.08 M H2S
...........2H2S (g)==> 2H2 (g)+S2 (g)
I..........0.08................0..........0........
C.......-x....................+x.........+1/2x....
E....0.08-x...............x.............0.5x......
Without knowing the equilibrium concentration of at least one of the species, or without knowing the value of Kc, one cannot carry this calculation any further.

D.  If you have additional information for B and C above, then you can solve for Kc, and in part D, the value of Kc will be 1/sqrt of that Kc since the reaction is reversed and halved.