Question

At a certain temperature, the Kp for the decomposition of H2S is 0.813.
H2S (Two wa arrow) H2(g) + S(g)
Initially, only H2S is present at a pressure of 0.280 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Initially:
H2S = 0.28
H2(g) = 0
S(g) = 0
At equilibrium:
H2S = 0.28 - x
H2(g) = x
S(g) = x

Kp = Products/Reactants = ([x][x])/(0.28-x)
x^2 +0.813x-0.22764 = 0
x =(-b plus or minus sqrt(b^2 -4ac))/2a
x = 0.220303199 or approximately 0.220

What is the total pressure in the container at equilibrium?
Pressure of the reactants + the products
so
x + x + (0.28-x)
0.220 + 0.220 + (0.280 - 0.220)
= 0.5
This is the total pressure at equilibrium.
I hope this helps Janavi.

Answer 2

Thank you sooo much Karan!