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Hydrogen Sulfide decomposes according to the following reaction for which Kc= 9.30 x 10^-8 at 700...

Hydrogen Sulfide decomposes according to the following reaction for which Kc= 9.30 x 10^-8 at 700 degrees C.

2H2S (g) makes 2H2 (g) + S2(g)

If .43 mol H2S is placed in a 3 L container what is the equilibrium concentration of H2 (g) at 700 degrees C?

Solutions

Expert Solution

moles of H2S = 0.43

volume = 3 L

concentration = moles / volume = 0.43 / 3

                       = 0.143 M

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

0.143                                0             0    ------------ initial

0.143 - 2x                         2x            x    ------------ at equlibrium

Kc = [H2]^2[S2] / [H2S]^2

9.30 x 10^-8 = (2x)^2 (x) / (0.143 - 2x)

4x^3 + 1.86 x 10^-7x - 1.33 x 10^-8 = 0

by solving this equation we get

x = 1.482 x 10^-3

Equilibrium concentration of H2 = 2x = 2 x 1.482 x 10^-3

                                                   = 2.96 x 10^-3 M

Equilibrium concentration of [H2] = 2.96 x 10^-3 M

queen_honey_blossom answered 3 months ago
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