Question

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

Answer 1

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

Answer 2

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