Answer:

Given:
Temperature (T) = 90 °C = 363 K
Number of moles of H2 (nH2) = 0.2 mol
Number of moles of S (nS) = 1 mol
Volume of the vessel (V) = 1 L
Equilibrium constant (K) = 6.8 × 10^-2

Calculating the initial pressure of H2 and S:
Using the ideal gas equation, we can calculate the initial pressure of H2 and S.

PV = nRT
For H2: P(H2) = n(H2)RT/V = (0.2 mol × 0.0821 L atm/mol K × 363 K)/1 L = 5.96 atm
For S: P(S) = n(S)RT/V = (1 mol × 0.0821 L atm/mol K × 363 K)/1 L = 29.2 atm

Calculating the equilibrium pressure of H2S:
H2(g) + S(s) ⇌ H2S(g)
At equilibrium, the mole ratio of H2, S, and H2S is 1:1:1 (stoichiometry of the balanced chemical equation). Let x be the equilibrium partial pressure of H2S. Then, the equilibrium partial pressure of H2 and S will also be x.

Using the equilibrium constant expression, we can write:

K = [H2S] / [H2][S]
K = x / (x × x)

Solving for x, we get:

x = √(K × P(H2) × P(S))
x = √(6.8 × 10^-2 × 5.96 atm × 29.2 atm) = 2.56 atm

Therefore, the partial pressure of H2S at equilibrium is 2.56 atm.

Explanation:
The given equilibrium is an example of a gas-solid equilibrium. At high temperatures and pressures, solid sulfur reacts with gaseous hydrogen to form gaseous hydrogen sulfide. The equilibrium constant (K) tells us the degree of completion of the reaction at equilibrium. In this case, K is less than 1, indicating that the reaction is not complete and the reactants are favored at equilibrium.

Using the ideal gas equation, we can calculate the initial pressure of H2 and S. Since the volume of the vessel is constant throughout the reaction, the total pressure inside the vessel is the sum of the partial pressures of the gases.

At equilibrium, the mole ratio of H2, S, and H2S is 1:1:1, which means that the partial pressures of all three gases are equal. Using the equilibrium constant expression, we can relate the partial pressures of H2, S, and H2S at equilibrium. Solving for x, we get the partial pressure of H2S at equilibrium.