Question

The reaction of nitrogen gas and oxygen gas to form nitrogen monoxide gas is shown below. If the measured concentrations of all three chemicals at some point in time are: [N2] = 0.80 M


[O2] = 0.050 M

[NO] = 0.10 M

Which statement is TRUE about the reaction at this point in time? N2(g) + O2(g) ⇄ 2 NO(g) K = 0.10


The reaction is at equilibrium.

The reverse reaction is occurring at a faster rate than the forward reaction.

The forward reaction is occurring at a faster rate than the reverse reaction.

This set of concentration values is impossible because the concentrations of N2 and O2 must be the same.

Answer 1

Answer:

The reverse reaction is occurring at a faster rate than the forward reaction.

Explanation:

When the reaction:

aA + bB ⇔ cC + dD

has not reached the balance, it is possible to calculate:

$Qc=\frac{[C]^{c}*[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is called the reaction quotient, and the concentrations expressed in it are not the equilibrium concentrations, but other concentrations given at a time of the reaction.

Comparing Qc with Kc allows to find out the status and evolution of the system:

• If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
• If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
• If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In the case of the reaction:

N₂ (g) + O₂ (g) ⇄ 2 NO (g)

$Qc=\frac{[NO]^{2} }{[N_{2} ]*[O_{2} ] }$

Being:

• [N₂] = 0.80 M
• [O₂] = 0.050 M
• [NO] = 0.10 M

and replacing:

$Qc=\frac{0.10^{2} }{0.80*0.050 }$

you get:

Qc= 0.25

Being Kc=0.10, Qc>Kc.  Then the system is not in equilibrium and will evolve to the left to increase the concentration of reagents. So, the reverse reaction is occurring at a faster rate than the forward reaction.