Question

Consider the reaction.

Answer 1

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 2

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