What is Keq for the reaction N2 + 3H2 2NH3 if the equilibrium concentrations are [NH3] = .250 M, [N2] = .590 M, and [H2] = .750 M?

After thorough researching, the Keq for the reaction N2 + 3H2 2NH3 if the equilibrium concentrations are [NH3] = .250 M, [N2] = .590 M, and [H2] = .750 M is 1.33. The correct answer to the following given statement above is 1.33

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BarrettArcher BarrettArcher · Answer 2 · 2019-04-18T09:12:30+02:00

Answer : The value of [tex]K_{eq}[/tex] for the given reaction is, 0.252

Explanation : Given,

Concentration of [tex]NH_3[/tex] = 0.250 M

Concentration of [tex]N_2[/tex] = 0.590 M

Concentration of [tex]H_2[/tex] = 0.750 M

The given balanced equilibrium reaction is,

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

The expression for equilibrium constant for this 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 the value of [tex]K_{eq}[/tex]

[tex]K_{eq}=\frac{(0.25)^2}{(0.59)\times (0.75)^3}[/tex]

[tex]K_{eq}=0.252[/tex]

Therefore, the value of [tex]K_{eq}[/tex] for the given reaction is, 0.252