Question

At a certain temperature, the ????p for the decomposition of H2S is 0.841. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.259 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Answer : The total pressure in the container at equilibrium is, 0.4667 atm

Solution : Given,

Initial pressure of
`H_2S` = 0.259 atm

Equilibrium constant,
`K_p` = 0.841

The given equilibrium reaction is,

`H_2S(g)\rightleftharpoons H_2(g)+S(g)`

Initially 0.259 0 0

At equilibrium (0.259 - x) x x

Let the partial pressure of
`H_2` and
`S` will be, 'x'

The expression of
`K_p` will be,

`K_p=((p_(H_2))(p_(S)))/(p_(H_2S))`

Now put all the values of partial pressure, we get

`0.841=((x)* (x))/((0.259-x))`

By solving the term x, we get

`x=0.2077atm`

The partial pressure of
`H_2` and
`S` = x = 0.2077 atm

Total pressure in the container at equilibrium =
`0.259-x+x+x=0.259+x=0.259+0.2077=0.4667atm`

Therefore, the total pressure in the container at equilibrium is, 0.4667 atm