Question

At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.746 0.746 . H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) H2S(g)↽−−⇀H2(g)+S(g) Initially, only H 2 S H2S is present at a pressure of 0.240 0.240 bar in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Answer: The total pressure of container at equilibrium is 0.431 bar

Step-by-step explanation:

We are given:

Pressure of hydrogen sulfide = 0.240 bar

The given chemical equation follows:

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

Initial: 0.240

At eqllm: 0.240-x x x

The expression of
`K_p` for above equation follows:

`K_p=(p_(H_2)* p_S)/(p_(H_2S))`

We are given:

`K_p=0.746`

Putting values in above expression, we get:

`0.746=(x* x)/(0.240-x)\\\\x=0.191,-0.940`

Neglecting the negative value of 'x' because pressure cannot be negative.

So, the equilibrium pressure of hydrogen gas = x = 0.191 bar

The equilibrium pressure of sulfur gas = x = 0.191 bar

The equilibrium pressure of hydrogen sulfide gas = (0.240 - x) = (0.240 - 0.191) = 0.049 bar

Total pressure of the container at equilibrium =
`p_(H_2)+p_(S)+p_(H_2S)`

Total pressure of the container at equilibrium = 0.191 + 0.191 + 0.049 = 0.431 bar

Hence, the total pressure of container at equilibrium is 0.431 bar