Question

Consider the reaction.

N2(g) + 3H2(g) <-> 2NH3(g)

At equilibrium, the concentrations of the different species are as follows.

[NH3] = 0.105 M
[N2] = 1.1 M
[H2] = 1.50 M

What is the equilibrium constant for the reaction at this temperature?

a. 0.0030
b. 0.030
c. 34
d. 340

Answer 1

Answer : The correct option is, (a) 0.0030

Solution : Given,

Concentration of
`NH_3` = 0.105 M

Concentration of
`N_2` = 1.1 M

Concentration of
`H_2` = 1.50 M

The balanced equilibrium reaction is,

`N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)`

The expression for equilibrium constant of the reaction will be,

`K_(eq)=([NH_3]^2)/([N_2][H_2]^3)`

Now put all the given values in this expression, we get

`K_(eq)=([NH_3]^2)/([N_2][H_2]^3)`

`K_(eq)=((0.105)^2)/((1.1)* (1.50)^3)\\\\K_(eq)=2.96* 10^(-3)=3* 10^(-3)=0.003`

Therefore, the equilibrium constant for the reaction at this temperature will be, 0.0030

Answer 2

N2(g) + 3H2(g) <-> 2NH3(g)

Constant equilibrium, Ke = [NH3]eq^2 / { [N2]eq * [H2]eq^3 }


[NH3]eq = 0.105 M
[N2]eq = 1.1 M
[H2]eq = 1.50 M

Ke = 0.105^2 / [ 1.1 * (1.50)^3 } = 0.00297 = 0.0030

Answer: option a. 0.0030