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scoray [572]
3 years ago
5

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

M?
Chemistry
2 answers:
OverLord2011 [107]3 years ago
6 0

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

Explanation : Given,

Concentration of NH_3 = 0.250 M

Concentration of N_2 = 0.590 M

Concentration of H_2 = 0.750 M

The given balanced equilibrium reaction is,

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

The expression for equilibrium constant for this 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 the value of K_{eq}

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

K_{eq}=0.252

Therefore, the value of K_{eq} for the given reaction is, 0.252

Lena [83]3 years ago
3 0
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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