Question

At a certain temperature, the K p for the decomposition of H 2 S is 0.859 . H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) Initially, only H 2 S is present at a pressure of 0.245 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Answer:

P = 0.444 atm

Explanation:

The reaction of decomposition of H₂S is:

                    H₂S(g)   ⇄  H₂(g)  + S(g)      

Initial            0.245          0           0

Equilibrium  0.245-x       x            x

The equilibrium constant (Kp) of the above reaction is the following:

$K_{p} = \frac{P_{H_{2}}*P_{S}}{P_{H_{2}S}}$     (1)

Knowing that initally, only H₂S is present at P = 0.245, the Kp of equation (1) at equilibrium is:

$0.859 = \frac{x*x}{0.245 - x}$  

$x^{2} -0.859*0.245 + 0.859x = 0$

Using the quadratic formula we get x₁ = 0.199 and x₂ = -1.058.

Taking only the possitive number we have that:

$P_{H_{2}} = P_{S} = x = 0.199 atm$      

$P_{H_{2}S} = (0.245 - 0.199) atm = 0.046 atm$      

Therefore, the total pressure in the container at equilibrium is:

$P_{T} = P_{H_{2}} + P_{S} + P_{H_{2}S} = (0.199*2 + 0.046) atm = 0.444 atm$

I hope it helps you!