homework question

Question

At a certain temperature,the kp for the decomposition of H2S is 0.893

H2S(g)↽−−⇀H2(g)+S(g)

Initially, only H2S is present at a pressure of 0.146 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

An easy way to view this is via an ICE table:

H₂S(g) <===> H₂(g) + S(g)

0.146.................0.............0........Initial

-x.....................+x............+x.......Change

0.146-x.............x..............x........Equilibrium

Kp = (H₂)(S) / (H₂S)

0.893 = (x)(x) / 0.146-x

x² = 0.130 - 0.893x

x² + 0.893x - 0.130 = 0

x = 0.127 atm

So, at equilibrium, the following partial pressures will be present:

H₂S = 0.146 - 0.127 = 0.019 atm

H₂ = 0.127 atm

S = 0.127 atm

Total pressure @ equilibrium = 0.019 atm + 0.127 atm + 0.127 atm = 0.273 atm