Dissociation of Ammonium Hydrogen Sulfide in a Sealed Flask
Ammonium hydrogen sulfide, NH4HS(s), dissociates according to NH4HS(s) ⇌ NH3(g) + H2S(g). At 298 K the equilibrium constant is Kp = 0.108, with partial pressures in atm. A 1.00 L rigid flask at 298 K is charged with 0.200 atm of NH3 and 0.100 atm of H2S, and 0.0100 mol of solid NH4HS is added. The flask is then sealed. Use R = 0.0821 L·atm·mol⁻¹·K⁻¹. Assume ideal gases, constant temperature, and that the solid, if present, occupies negligible volume. In a second experiment the same gas pressures are used but only 0.0050 mol of solid NH4HS is added. The flask initially contains no other gases. You may treat the solid as having unit activity.
1. For the first experiment, determine the direction in which the system adjusts and calculate the equilibrium partial pressures, assuming solid remains at equilibrium.
Read the worked answer
Reaction quotient before any adjustment: Q = P(NH3)P(H2S) = 0.200×0.100 = 0.0200. Since Q < Kp = 0.108, the forward reaction is favoured: more solid decomposes, increasing both partial pressures. Let x be the increase in each partial pressure in atm.
Then (0.200 + x)(0.100 + x) = 0.108.
Expand: x² + 0.300x + 0.0200 − 0.108 = 0, so x² + 0.300x − 0.0880 = 0.
x = [−0.300 + √(0.0900 + 0.352)]/2 = [−0.300 + 0.6648]/2 = 0.1824.
So P(NH3) = 0.382 atm and P(H2S) = 0.282 atm.
Check the solid: x = 0.182 atm corresponds to n = xV/RT = 0.182/(0.0821×298) = 7.44×10⁻³ mol decomposed. This is less than the 0.0100 mol added, so solid remains and the equilibrium assumption is valid.
2. For the second experiment, decide whether equilibrium with solid can be reached. If not, calculate the final partial pressures and the amount of solid remaining.
Read the worked answer
In the second experiment only 0.0050 mol of solid is present. If all of it decomposed, it would add 0.0050 mol of each gas.
Initial moles: n(NH3) = 0.200/(0.0821×298) = 8.18×10⁻³ mol; n(H2S) = 0.100/(0.0821×298) = 4.09×10⁻³ mol.
After complete decomposition: n(NH3) = 1.318×10⁻² mol and n(H2S) = 9.09×10⁻³ mol.
The corresponding pressures are P = nRT/V: P(NH3) = 1.318×10⁻²×24.46 = 0.322 atm; P(H2S) = 9.09×10⁻³×24.46 = 0.222 atm.
Their product is 0.322×0.222 = 0.0715, still below Kp = 0.108. Therefore the solid disappears before the gas pressures can reach the equilibrium product. After that no solid remains, so the system cannot establish the NH4HS(s) equilibrium; the reaction stops with P(NH3) = 0.322 atm and P(H2S) = 0.222 atm, and zero solid remains. Q is then 0.0715 < Kp.