Q4. Only 0.25 mol of CaCO3 is heated to 1100 K in an evacuated 20.0 L vessel (Kp = pCO2 = 2.00 bar at 1100 K). Is equilibrium reached?
Even if all of it decomposed, pCO2 = nRT/V = 0.25 × 0.0831 × 1100/20.0, about 1.14 bar, less than 2.00 bar. So all the CaCO3 decomposes, no CaCO3 is left, and the final state is a phase-exhausted boundary state rather than two-solid coexistence. The equality pCO2 = 2.00 bar requires both solids; the equilibrium constant remains defined.
Q5. Solid NH4HS is put in a flask that already holds NH3 at 0.20 bar; Kp = 0.11 bar2 for NH4HS(s) ⇌ NH3(g) + H2S(g). Find the partial pressures and total pressure at equilibrium, assuming ideal gases at fixed temperature and volume and enough solid remains.
If p bar of each gas forms, p(0.20 + p) = 0.11, so p2 + 0.20p − 0.11 = 0 and p = (−0.20 + √0.48)/2 ≈ 0.246 bar. So pH2S ≈ 0.246 bar, pNH3 ≈ 0.446 bar, and the total ≈ 0.69 bar. The added NH3 holds back the decomposition.
Q6. Write Kp for H2O(l) ⇌ H2O(g). What familiar quantity is it?
Kp = pH2O, since the liquid is left out. It is the vapour pressure of water at that temperature, which is why the vapour pressure is fixed at a given temperature.
Q7. Write Kc for Zn(s) + Cu2+(aq) ⇌ Zn2+(aq) + Cu(s), and say why each species is in or out.
Kc = [Zn2+]/[Cu2+]. The metals are pure solids, with constant concentrations, so they are left out; the ions are dissolved and their concentrations change, so they stay in.
Q8. Repeat NCERT Problem 6.6 (Kp = 3.0 at 1000 K for CO2(g) + C(s) ⇌ 2CO(g)) with an initial CO2 pressure of 1.00 bar.
(2x)2/(1.00 − x) = 3.0 gives 4x2 + 3x − 3.0 = 0, so x = [−3 + √(9 + 48)]/8 = (−3 + 7.55)/8, about 0.569. pCO ≈ 1.14 bar and pCO2 ≈ 0.43 bar. Check: 1.142/0.43 ≈ 3.0.