Question

what is $k_{eq}$ for the reaction $n_2 + 3h_2
ightleftharpoons 2nh_3$ if the equilibrium concentrations are $nh_3=3 m$, $n_2=1 m$, and $h_2=2 m$?

a. $k_{eq}=4.5$
b. $k_{eq}=1.5$
c. $k_{eq}=1.125$
d. $k_{eq}=0.75$

Explanation:

Step1: Write equilibrium - constant expression

For the reaction $N_2 + 3H_2
ightleftharpoons2NH_3$, the equilibrium - constant expression $K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}$.

Step2: Substitute the given concentrations

Substitute $[NH_3]=3M$, $[N_2]=1M$, and $[H_2]=2M$ into the expression:
$K_{eq}=\frac{(3)^2}{(1)\times(2)^3}=\frac{9}{1\times8}=\frac{9}{8} = 1.125$.

Answer:

C. $K_{eq}=1.125$