Question

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is charged with 1.42 bar of N 2 and 2.87 bar of H 2 , what will the equilibrium partial pressures in the mixture be?

Answer 1

Answer : The partial pressure of
`N_2,H_2\text{ and }NH_3` at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of
`N_2` = 1.42 bar

Initial pressure of
`H_2` = 2.87 bar

`K_p` = 0.036

The given equilibrium reaction is,

`N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)`

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of
`K_p` will be,

`K_p=((p_(NH_3))^2)/((p_(N_2))(p_(H_2))^3)`

Now put all the values of partial pressure, we get

`0.036=((2x)^2)/((1.42-x)* (2.87-3x)^3)`

By solving the term x, we get

`x=0.287\text{ and }3.889`

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of
`NH_3` at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of
`N_2` at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of
`H_2` at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of
`N_2` + Partial pressure of
`H_2` + Partial pressure of
`NH_3`

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar