Question

At 500˚C the equilibrium constant for the reaction

2 H2S(g) ⇌ S2(g) + 2 H2(g) is Kc = 1.67 x 10–6

If 0.350 mol of H2S were placed in a 1.00 L container and allowed to come to equilibrium at that temperature, calculate the concentrations (in mol/L) of of hydrogen gas at equilibrium.

Answer 1

To find the equilibrium concentration of hydrogen gas in the reaction 2H2S(g) ⇌ S2(g) + 2H2(g) with Kc = 1.67 x 10−6, one must use the equilibrium constant expression, initial concentrations, the stoichiometry of the reaction, and solving for the change in concentration x.

Step-by-step explanation:

To calculate the concentration of hydrogen gas (H2) at equilibrium for the given reaction 2H2S(g) ⇌ S2(g) + 2H2(g), with the equilibrium constant (Kc) being 1.67 x 10−6, you must apply the concept of chemical equilibrium and the reaction quotient. Start by writing the expression for the equilibrium constant:

Kc = [S2][H2]² / [H2S]²

Let's denote the change in concentration of H2S that reacts as x. Since the reaction involves the decomposition of 2 moles of H2S to form 1 mole of S2 and 2 moles of H2, the change in concentration of S2 and H2 will be +x and +2x, respectively.

Initial concentrations (M): [H2S] = 0.350, [S2] = 0, [H2] = 0

Change in concentrations (M): [H2S] = -2x, [S2] = +x, [H2] = +2x

Equilibrium concentrations (M): [H2S] = 0.350 - 2x, [S2] = 0 + x, [H2] = 0 + 2x

Now, plug these expressions into the equilibrium constant expression and solve for x:

Kc = (x)(2x)^2 / (0.350 - 2x)²

To find the value of x, this equation must be solved. However, as Kc is very small, the assumption can be made that 2x is much smaller than 0.350, allowing the simplification of the equation and solving for x.

Once x is known, compute the equilibrium concentration of H2 by multiplying 2x by the volume of the container, which is 1.00 L.