2 h2s(l) ⇌ 2h2(g) + s2(g) (Kc at 700c is 9.30x10^-8) A) Write the equilibrium expression...

2 h2s(l) ⇌ 2h2(g) + s2(g) (Kc at 700c is 9.30x10^-8)

A) Write the equilibrium expression for this reaction

B) If 0.45 mol of H2S gas is placed in a 3 L container at 700C what is the concentration, in mol/L, of hydrogen gas when equilibrium is reached.

Solutions

Expert Solution

A ) equilibrium expression Kc = [H2]^2 [S2]

pure liquid doesnot take in the part of equilibrium expression

molarity of H2S = 0.45 / 3 = 0.15 M

2H2S (l) <---------------------> 2H2 (g) + S2 (g)

0.15 0 0 ---------------> iniitla

0.15-2x 2x x ------------------> equilibrium

Kc = [H2]^2 [S2]

Kc = (2x)^2 (x)

Kc = 4 x^3

9.30 x 10^-8 = 4 x^3

x = 2.85 x 10^-3

equilibrium concentrations :

hydrogen equilibrium concentration = 2x = 5.71 x 10^-3 M

sulfur equilibrium concentration = x = 2.85 x 10^-3 M

queen_honey_blossom answered 1 year ago

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