Question

At a certain temperature, the Kp for the decomposition of H2S is 0.842.

Initially, only H2S is present at a pressure of 0.126 atm in a closed container. What is the total pressure in the container at equilibrium?

H2S (g) <---> H2(g) + S(g)

Answer 1

Answer – Given, Kp = 0.842 , P of H₂S = 0.126 atm

Now we need to put ICE chart -

      H₂S (g) <---> H₂(g) + S(g)      

I    0.126               0           0

C    -x                +x         +x

E   0.126-x         +x        +x

Kp = P (H₂ (g)) P (S (g)) / P (H₂S(g))

0.842= x *x / (0.126-x)

0.842 (0.126-x) = x²

0.106 -0.842x = x²

x² + 0.842x – 0.106 = 0

a= 1, b = 0.842 , c = -0.106

we know the quadratic formula

x = -b+/-√b²-4ac / 2a

by placing the value in it and calculate x value

x= - 0.953 and x = 0.111

so, x = 0.111, because we consider lowest and positive value from both x values.

so at equilibrium concentrations

P of HI =0.126-x

             = 0.126 - 0.111

          = 0.015 atm

x = P of H₂ = P of S = 0.111 atm

so total pressure at equilibrium = 0.015 +0.111 +0.111

                                                   = 0.237 atm