Question

At a certain temperature, the Kp for the decomposition of H2S is 0.842.
H2S(g) image from custom entry tool H2(g) + S(g)
Initially, only H2S is present at a pressure of 0.104 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Answer:

0.1976 atm is the total pressure in the container at equilibrium.

Explanation:

The value of $K_p$ for the given reaction = 0.842

Partial pressure of the hydrogen sulfide = 0.104 atm

Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm

Partial pressure of the hydrogen gas at equilibrium = x

Partial pressure of the sulfur gas at equilibrium = x

$H_2S(g)\rightleftharpoons 2H_2(g)+S(g)$

Initially

0.104 atm                      0  0

At equilibrium

(0.104 -2x)                      2x   x

The expression of  $K_p$ is given by ;

$K_p=\frac{x\times x}{(0.104 -x) }$

$0.842 =\frac{x^2}{(0.104-x)}$

Solving for x:

x = 0.0936 atm

Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm

= (0.104-0.0936 ) atm = 0.0104 atm

Partial pressure of the hydrogen gas at equilibrium = x = 0.0936 atm

Partial pressure of the sulfur gas at equilibrium = x = 0.0936 atm

Total pressure in the container will be sum of all the partial pressure of the hydrogen sulfide gas, hydrogen gas and sulfur gas at the equilibrium.

P = 0.0104 atm + 0.0936 atm + 0.0936 atm = 0.1683 atm

0.1976 atm is the total pressure in the container at equilibrium.