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

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:

$K_p=\frac{p_{H_2}\times p_S}{p_{H_2S}}$

We are given:

$K_p=0.746$

Putting values in above expression, we get:

$0.746=\frac{x\times x}{0.240-x}\\\\x=0.191,-0.940$

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