Question

What is Keq for the reaction N2 + 3H2 2NH3 if the equilibrium concentrations are [NH3] = 3 M, [N2] = 1 M, and [H2] = 2 M?

Answer 1

Answer : The correct option is, (D) $k_{eq}=1.125$

Explanation : Given,

Concentration of $NH_3$ = 3 M

Concentration of $N_2$ = 1 M

Concentration of $H_2$ = 2 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 $K_{eq}$

$K_{eq}=\frac{(3)^2}{(1)\times (2)^3}$

$K_{eq}=1.125$

Therefore, the value of $K_{eq}$ for the given reaction is, 1.125

Answer 2

Answer;

= 1.125 mol^-2 dm^6

Explanation;

Kc = ([NH3]^2)/([N2]*[H2]^3)  

if the equilibrium concentrations are [NH3] = 3 M, [N2] = 1 M, and [H2] = 2 M

      = (3²)/(1²)(2²)

      = 9/8  

      = 1.125 mol^-2 dm^6