Question

At a certain temperature, the ????p for the decomposition of H2S is 0.776. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.200 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

0.365 atm is the total pressure in the container at equilibrium.

Step-by-step explanation:

Given , equilibrium constant of the reaction in terms of partial pressure:
`K_p=0.776`

Initial Partial pressure of the hydrogen sulfide gas =
`p_(H_2S)=0.200 atm`

Let the partial pressure of hydrogen gas and sulfur gas at equilibrium be p.

`H_2S(g)\rightleftharpoons H_2(g)+S(g)`

Initially 0.200 atm

At eq'm 0.200-p p p

`K_p=(p_(H_2)p_(s))/(p_(H_2S))`

`0.776=(p* p)/((0.200-p))`

Solving the equation we will get two values of 'p' from which we will ignore the negative value.

p = 0.165 atm

Partial pressure of the hydrogen sulfide gas at equilibrium=
`p_(H_2S)=0.200 atm-p=0.200 atm-0.165 atm =0.035 atm`

Total pressure in the container at equilibrium :

0.200-p + p + p =0.200 atm +p = 0.365 atm