Question

At a certain temperature, the K for the decompositon of H2S is 0.759.

H2S(g) <---> H2(g) + S(g)

Initally, only H2S is present at a pressure of 0.103 bar in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Hasig) Halg) + Sly) at t- 0 0103 bas 0 0 at ta teguy 0.103-4 2 since, given that K= 0.759 and we know that X doesn't change without change in temp. so, K= Keque and, 727999999 Кеч - Бир Бер) B (H2S) (9) - 0.759 = x.e 10.103-) 0.759 - x2 x2 0.103-x 3) & % 2 - 0.759 (0.103-) x² +0.759x – 0.078 = 0 = -0.759 + 120.759)2+ 4-4-60-078) 2 = -0.759 + 10.576 +0.312 2 : -0.7591 10.888

9) x = -0.759 + Jo.888 - 9.759 fo.942 2 - x = -0.759to.942 a 0.759 -0.942 2 2 x = 8.0916 since total preouse at Equi = Prosest Praight Psig) = 0.103-x tx tre = 0.103 +x total pressuse at Egry = 0.103+0.0916 . Protal = 0.1946 bar. ) Any