Q4. For H2(g) + I2(g) ⇌ 2HI(g), Kc = 57.0 at 700 K. A flask holds [H2] = [I2] = 0.050 M and [HI] = 0.50 M. Is it at equilibrium? If not, which way does it move?
Qc = (0.50)2/(0.050 × 0.050) = 0.25/0.0025 = 100. Since 100 > 57.0, it is not at equilibrium; the reaction goes backward, HI forming H2 and I2.
Q5. Starting from 0.20 M HI only at 700 K, find the equilibrium concentrations (Kc = 57.0 for H2 + I2 ⇌ 2HI).
If [H2] = [I2] = x, then [HI] = 0.20 − 2x and (0.20 − 2x)2/x2 = 57.0. Taking square roots, (0.20 − 2x)/x = 7.55, so x = 0.20/9.55, about 0.021 M. [H2] = [I2] ≈ 0.021 M, [HI] ≈ 0.158 M.
Q6. At 380 K, Kc = 1.80 for PCl5 ⇌ PCl3 + Cl2. If 1.00 mol of PCl5 is put in a 1 L vessel, what fraction dissociates? Compare with the 3.00 mol of NCERT Problem 6.9.
x2/(1.00 − x) = 1.80 gives x2 + 1.80x − 1.80 = 0 and x = [−1.80 + √(3.24 + 7.20)]/2 = (−1.80 + 3.231)/2 ≈ 0.716. So about 72% dissociates, against 1.59/3.00 = 53% in Problem 6.9: a less concentrated sample dissociates more.
Q7. Paper burning in air has an enormous K at room temperature, yet a book lasts for centuries. How?
K only says where equilibrium lies, not how fast it is reached. The reaction is very slow at room temperature (its activation energy is high), so the paper and the oxygen stay far from equilibrium; a flame starts it.
Q8. NCERT Table 6.5 gives Kp = 47.9 for N2O4 ⇌ 2NO2 at 400 K. With the same 13.8 g of N2O4 in 1 L at 400 K as in Example 2 (4.98 bar at the start), what total pressure is reached?
4x2/(4.98 − x) = 47.9 gives 4x2 + 47.9x − 238.5 = 0, so x = (−47.9 + 78.2)/8 ≈ 3.78 bar. Then pN2O4 ≈ 1.20 bar, pNO2 ≈ 7.57 bar, and the total ≈ 8.76 bar, not the 9.15 bar of NCERT Problem 6.8.