Question

For the reactions system 2H₂(g) + S₂(g) → 2H₂S(g), a 1.00 liter vessel is found to contain 0.50 moles of H₂, 0.020 moles of S₂, and 68.5 moles of H₂S. Calculate the numerical value of the Keq of this system.

K =
Are the products or reactants favored?

Answer 1

The equilibrium constant (Keq) for the system 2H₂(g) + S₂(g) → 2H₂S(g) is 938450, indicating that the products are highly favored over the reactants in this reaction.

Step-by-step explanation:

The reaction 2H₂(g) + S₂(g) → 2H₂S(g) at equilibrium in a 1.00-liter vessel with 0.50 moles of H₂, 0.020 moles of S₂, and 68.5 moles of H₂S, allows us to calculate the equilibrium constant (Keq) using concentrations in moles per liter (M). The equilibrium concentrations of H₂, S₂, and H₂S are 0.50 M, 0.020 M, and 68.5 M, respectively.

The expression for Keq for this reaction is:

Keq = [H₂S]^2 / ([H₂]^2 × [S₂])

By plugging in the equilibrium concentrations, we get:

Keq = (68.5^2) / (0.50^2 × 0.020) = (4692.25) / (0.50 × 0.50 × 0.020) = 4692.25 / 0.005 = 938450

Thus, the numerical value of Keq is 938450.

In this system, because Keq is significantly greater than 1, this implies that the products are highly favored over the reactants