Q4. For which of these is Kp = Kc? (a) N2(g) + O2(g) ⇌ 2NO(g); (b) 2SO2(g) + O2(g) ⇌ 2SO3(g); (c) CO(g) + H2O(g) ⇌ CO2(g) + H2(g); (d) PCl5(g) ⇌ PCl3(g) + Cl2(g).
(a) and (c): in both Δn = 0 (2 − 2), so (RT)0 = 1. For (b) Δn = −1 and for (d) Δn = +1.
Q5. Repeat NCERT Problem 6.4 (Kc = 4.24 at 800 K for CO + H2O ⇌ CO2 + H2), but start with 0.20 M each of CO and H2O.
x/(0.20 − x) = √4.24 = 2.059, so x = 0.4118/3.059, about 0.1346 M (2 × 0.0673). [CO2] = [H2] ≈ 0.1346 M; [CO] = [H2O] ≈ 0.0654 M. Doubling both starting concentrations doubles every equilibrium concentration here, because Δn = 0.
Q6. Write Kc for Fe3+(aq) + SCN−(aq) ⇌ Fe(SCN)2+(aq) and give its unit.
Kc = [Fe(SCN)2+]/[Fe3+][SCN−]; one concentration on top and two underneath, so the unit is (mol L−1)−1 = L mol−1.
Q7. In NCERT Table 6.5, Kp for N2O4 ⇌ 2NO2 rises from 47.9 at 400 K to 1700 at 500 K, while Kp for N2 + 3H2 ⇌ 2NH3 falls from 41 to 3.6 × 10−2. What does that suggest about the two reactions?
A constant that grows with temperature belongs to an endothermic forward reaction (splitting N2O4 absorbs heat); one that shrinks belongs to an exothermic forward reaction (forming ammonia gives out heat). This is taken up under factors affecting equilibria (NCERT Section 6.8).
Q8. A student uses R = 8.314 J mol−1 K−1 to convert Kc = 3.75 × 10−6 mol L−1 to Kp for 2NOCl ⇌ 2NO + Cl2 at 1069 K. What does she get, and why is it wrong?
3.75 × 10−6 × 8.314 × 1069, about 3.33 × 10−2, a hundred times the correct 3.33 × 10−4 bar. With Kc in mol L−1 and Kp in bar, R must be 0.0831 bar L mol−1 K−1. Since 1 L bar = 10−3 m3 × 105 Pa = 100 J, the number 8.314 is 100 times 0.0831, so it makes Kp 100 times too big (in general 100Δn; here Δn = 1). The same slip would explain NCERT’s printed 0.033.