Final answer:
To calculate the equilibrium pressures in the decomposition of H₂S, we establish an ICE table, apply the equilibrium constant Kp = 2.2 x 10^-6, and solve for the change 'x'. Given the assumption that the change is negligible, the equilibrium pressures are approximated as [H₂S] = 0.810 atm, [H₂] = 0.014 atm, [S₂] = 0.0072 atm.
Step-by-step explanation:
According to the reaction H₂S(g) ⇌ 2H₂(g) + S₂(g), we can set up an ICE (Initial, Change, Equilibrium) table and use the equilibrium constant to solve for the pressures of H₂, S₂, and the remaining H₂S at equilibrium.Given the Kp value and initial pressure, we will assume a change of 'x' atm for H₂S decomposing into 2x atm for H₂ and x atm for S₂. Plugging these values into the equilibrium expression, Kp = (2x)^2 * x / (0.824 - x), we can solve for 'x'. However, since the question states the change in pressure of H₂S is neglected, we can approximate the value of 'x' without involving complex calculations.
Thus, the final equilibrium pressures are: [H₂S] = 0.810 atm, [H₂] = 0.014 atm, [S₂] = 0.0072 atm. This confirms that the assumption of neglecting the change is valid as it's less than 2% difference from the initial pressure of H₂S, maintaining the integrity of the small approximation.