a. mols H2S = grams/molar mass = 3.40/34 = 0.1
Surely I don't need to write the Kc expression for you.
b.
..........2H2S ==> 2H2 + S2
initial..0.1........0.....0
change....2x.......2x.....x
equil...0.1-2x.....2x.....x
From the problem, (S2) = x = 0.0372.
Therefore, (H2) is 2x
(H2S) = 0.1-2x
You can do the arithmetic.
c. Substitute the values of concns into Kc and solve for Kc.
d. The problem asks for the original Kc but half of it and reversed. Take square root to take care of the 1/2 part and take the reciprocal to take care of the reversed part.
e. Use PV = nRT.
Post your work if you get stuck.
Need help in AP chemistry on Equilibrium
When heated, hydrogen sulfide gas decomposes according to the equation
2 H2S(g) ⇄ 2 H2(g) + S2(g)
A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed
container is heated to 483 K, and 3.72 x 10
-2
mol of S2(g) is present at equilibrium.
(a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction
represented above.
(b) Calculate the equilibrium concentration, in mol L
-1
, of the following gases in the container
at 483 K.
(i) H2(g)
(ii) H2S (g)
(c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K.
(d) For the reaction H2(g) + ½ S2(g) ↔ H2S(g) at 483 K, calculate the value of the equilibrium
constant, Kc.
(e) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K
2 answers
Note that I dropped a - sign here on the change of -2x. The equil value of 0.1-2x is correct as written.
b.
..........2H2S ==> 2H2 + S2
initial..0.1........0.....0
change....2x.......2x.....x
equil...0.1-2x.....2x.....x
b.
..........2H2S ==> 2H2 + S2
initial..0.1........0.....0
change....2x.......2x.....x
equil...0.1-2x.....2x.....x