Question

6. If 0.54 mol of H2S is placed in a 3.0 L container, what is the equilibrium concentration of H2 at 710 C?

6. In the decomposition of hydrogen sulfide: 2H2S(9) + 2H2(g) + S2(g) K = 9.90 x 10-8 at 710°C If 0.54 mol H2S is placed in a 3.0 liter container, what is the equilibrium concentration of H2(g) at 710 °C? Can the small x approximation be used here? Justify it (see p. 768 of text). (1.9 x 103 M] 2H₂Sigg I Hreg) + S2 (g) 0.84 m) Initical 3. 0L * 0.18 change 1 - 2x Equilibrium 016- 2x k. [Hz]?[s] [H2S]² = (2x)² (+) - = 99x108 (0.18-2x)

Answer 1

Ź 22 (9) + S2 (9) 2 H₂S (9) 0.18 - 2x 0.18-21 +2x +x 20 a (2x)2(x) Ke = (0.18-2x) 2 9.90 x 108 - 4x3 (6. 18-242 Most trusted stationery u 7 10.18-21 = x <<<< 0.18 0.18 4,23 _ 2.20 x 10-9 x= 9.29 X 16-4m. (he) = 2x = (1.9 x 103 m