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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?

A0.0030
B0.030
C34
D340

Verified Answer :

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 \(NH_3\) = 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,

\(K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}\)

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

\(K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}\)

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

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

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