Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A...

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Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A...

Consider the reaction for the decomposition of hydrogen disulfide:

2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C

A 0.500 L reaction vessel initially contains 0.175 mol of H2S and 6.25×10−2 mol of H2 at 800∘C.

Find the equilibrium concentration of [S2].

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Answer 1

Answer #1

Volume of the vessel = 0.500 L

Initial concentration of H₂S (g) = 0.175/0.500 = 0.35 M

Initial concentration of H₂ (g) = 6.25×10⁻² /0.500 = 0.125 M

2H₂S(g) 2H₂(g) + S₂(g)

Initial concentration 0.35 0.125 0

Change in concentration - 2X 2X X

Equilibrium concentration (0.35 - 2X) (0.125+2X) X

Equilibrium constant,

K_c = [H₂]²[S₂]/[H₂S]²

or, 1.67 x 10^-7 = (0.125+2X)²(X)/(0.35 - 2X)²

or, 1.67 x 10^-7 = (0.125)²(X)/(0.35 )² [(0.35 - 2X) 0.35 as 2X << 0.35 and (0.125+2X) 0.125 as 2X << 0.125]

or, 0.015625X = 0.0000000204575

or, X = 1.31 x 10^-6 M

Therefore, the equilibrium concentration of S₂(g), [S₂] = 1.31 x 10^-6 M

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Answer 2

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Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g) Kc = 1.67×10^-7 at 800∘C The...

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