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:
[tex] K_{p} = \frac{P_{H_{2}}*P_{S}}{P_{H_{2}S}} [/tex] (1)
Knowing that initally, only H₂S is present at P = 0.245, the Kp of equation (1) at equilibrium is:
[tex] 0.859 = \frac{x*x}{0.245 - x} [/tex]
[tex] x^{2} -0.859*0.245 + 0.859x = 0 [/tex]
Using the quadratic formula we get x₁ = 0.199 and x₂ = -1.058.
Taking only the possitive number we have that:
[tex]P_{H_{2}} = P_{S} = x = 0.199 atm [/tex]
[tex]P_{H_{2}S} = (0.245 - 0.199) atm = 0.046 atm [/tex]
Therefore, the total pressure in the container at equilibrium is:
[tex] P_{T} = P_{H_{2}} + P_{S} + P_{H_{2}S} = (0.199*2 + 0.046) atm = 0.444 atm [/tex]
I hope it helps you!
