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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

Respuesta :

Edufirst
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

Answer : The correct option is, (a) 0.0030

Solution : Given,

Concentration of [tex]NH_3[/tex] = 0.105 M

Concentration of [tex]N_2[/tex] = 1.1 M

Concentration of [tex]H_2[/tex] = 1.50 M

The balanced equilibrium reaction is,

[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]

The expression for equilibrium constant of the reaction will be,

[tex]K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]

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

[tex]K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]

[tex]K_{eq}=\frac{(0.105)^2}{(1.1)\times (1.50)^3}\\\\K_{eq}=2.96\times 10^{-3}=3\times 10^{-3}=0.003[/tex]

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

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