Question

At 850 K, the value of the equilibrium constant Kp for the ammonia synthesis reaction:

N2(g) + H2(g) <---> N2H2(g)

is 0.1650. If a vessel contains an initial reaction mixture in which [N2] = 0.0150 M, [H2] = 0.0200 M, and [N2H2] = 0.000150 M, what will the [N2H2] be when equilibrium is reached?

Answer 1

we know that

Kp = Kc (RT)^dn

given reaction is

N2 + H2 ---> N2H2

here

dn = 1 - 1 - 1 = -1

so

Kp = Kc (RT)^-1

Kc = Kp (RT)

Kc = 0.165 x (0.0821 x 850)

Kc = 11.514525

now

initially

[N2] = 0.015

[H2] = 0.02

[N2H2] = 0.00015

now

N2 + H2 ---> N2H2

using ICE table

at equilibrium

[N2] = 0.015 - y

[H2] = 0.02 - y

[N2H2] = 0.00015 + y

Kc = [N2H2] / [N2] [H2]

11.514525 = [0.00015 + y] / [0.015 - y] [0.02 - y]

solving

we get

y = 0.0024

so

at equilibrium

[N2H2] = 0.00015 + y = 0.00015 + 0.0024 = 0.00255 M