A 1.0-L reaction vessel at \(90^{\circ} \mathrm{C}\) contains \(8.00 \mathrm{~g}\) of sulfur, hydrogen, and hydrogen sulfide gases with partial pressures of \(6.0 \mathrm{~atm}\) and \(0.40 \mathrm{~atm}\), respectively, at equilibrium: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ (a) Calculate \(K\) for the reaction at equilibrium. (b) The mass of sulfur is increased to \(10.0\) grams. What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished? (c) The pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is increased to \(1.0 \mathrm{~atm}\). What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished?

Given the initial partial pressures of hydrogen gas (H_2) and hydrogen sulfide (H_2S) at equilibrium, we are asked to calculate the equilibrium constant K for the given reaction. First, we need to write an equilibrium expression involving the partial pressures: $$K_p = \frac{P_{H_2S}}{P_{H_2}}$$ Now, we can use the given values to find the equilibrium constant K: $$K_p = \frac{0.40\,\mathrm{atm}}{6.0\,\mathrm{atm}} = 0.0667$$

We are given that the mass of sulfur is increased from 8.00 g to 10.00 g. Since H_2S is formed from both H_2 and S, the additional S will push the reaction towards the formation of H_2S. Let (x) be the increase in the partial pressure of H_2S at the new equilibrium. Applying stoichiometry and using the initial values: $$P_{H_2S, new} = 0.40\,\mathrm{atm} + x$$ $$P_{H_2, new} = 6.0\,\mathrm{atm} - x$$ Next, we can write a new equilibrium expression using these values: $$K_p = \frac{P_{H_2S, new}}{P_{H_2, new}} = \frac{0.40 + x}{6.0 - x} = 0.0667$$ Now, solve for x: $$x = 0.363\,\mathrm{atm}$$ Substituting x back into our expressions for the new partial pressures, we get: $$P_{H_2S, new} = 0.763\,\mathrm{atm}$$ $$P_{H_2, new} = 5.637\,\mathrm{atm}$$ These are the new equilibrium partial pressures of H_2 and H_2S after increasing the mass of sulfur.

We are given that the pressure of H_2S is increased to 1.0 atm. Increasing the pressure of H_2S will push the reaction towards the formation of H_2 and S. Let (y) be the increase in partial pressure of H_2 at the new equilibrium. Applying stoichiometry and using the initial values: $$P_{H_2S, new} = 1.0\,\mathrm{atm} - y$$ $$P_{H_2, new} = 6.0\,\mathrm{atm} + y$$ Write an equilibrium expression with these new values: $$K_p = \frac{P_{H_2S, new}}{P_{H_2, new}} = \frac{1.0 - y}{6.0 + y} = 0.0667$$ Now, solve for y: $$y = -0.2\,\mathrm{atm}$$ Note that a negative value for y is not physically meaningful because we originally increased the pressure of H_2S. This means that adding H_2S has not affected the equilibrium, and the partial pressures of H_2 and H_2S remain unchanged. Therefore, the partial pressures of H_2 and H_2S when equilibrium is reestablished are the same as initially given: $$P_{H_2} = 6.0\,\mathrm{atm}$$ $$P_{H_2S} = 0.40\,\mathrm{atm}$$