Question

question 2 (4 points)
consider the following reaction:
n₂(g) + 3h₂(g) ⇌ 2nh₃(g) k = 1.67
initially, 4.65 mol/l nitrogen and 8.55 mol/l hydrogen were placed into a container and allowed to reach equilibrium. at equilibrium there are 4.05 mol/l of hydrogen.
what are the concentrations of nitrogen and ammonia at equilibrium?
n₂ = 2.15 m
nh₃ = 2 m
n₂ = 15 m
nh₃ = 3 m
n₂ = 3.15 m
nh₃ = 3 m
n₂ = 3 m
nh₃ = 3.15 m

Explanation:

Step1: Calculate the change in hydrogen concentration

The initial concentration of hydrogen $[H_2]_0 = 8.55$ mol/L and the equilibrium concentration $[H_2]_e=4.05$ mol/L. The change in hydrogen concentration $\Delta[H_2]=[H_2]_0 - [H_2]_e=8.55 - 4.05 = 4.5$ mol/L.

Step2: Determine the change in nitrogen and ammonia concentrations based on stoichiometry

From the reaction $N_2(g)+3H_2(g)
ightleftharpoons 2NH_3(g)$, the mole - ratio of $N_2$ to $H_2$ is 1:3 and of $NH_3$ to $H_2$ is 2:3.
The change in nitrogen concentration $\Delta[N_2]=\frac{1}{3}\Delta[H_2]$. Since $\Delta[H_2]=4.5$ mol/L, $\Delta[N_2]=\frac{4.5}{3}=1.5$ mol/L.
The initial concentration of nitrogen $[N_2]_0 = 4.65$ mol/L. So the equilibrium concentration of nitrogen $[N_2]_e=[N_2]_0-\Delta[N_2]=4.65 - 1.5=3.15$ mol/L.
The change in ammonia concentration $\Delta[NH_3]=\frac{2}{3}\Delta[H_2]$. Since $\Delta[H_2]=4.5$ mol/L, $\Delta[NH_3]=\frac{2\times4.5}{3}=3$ mol/L. Since the initial concentration of ammonia is 0, the equilibrium concentration of ammonia $[NH_3]_e = 3$ mol/L.

Answer:

C. $[N_2]=3.15$ M, $[NH_3]=3$ M