Chemistry, 02.10.2019 00:30, XxKaitlynnxX

At equilibrium, the concentrations in this system were found to be [n2]=[o2]=0.100 m and [no]=0.400 m. n2(g)+o2(g)↽−−⇀2no(g) if more no is added, bringing its concentration to 0.700 m, what will the final concentration of no be after equilibrium is re‑established?

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answered: queenkimm26

The final concentration of NO after equilibrium is re‑established will be:

[NO] = 0.600 M

Explanation:

We know that for a generic reaction aA (g) + bB (g) ⇄ cC (g) + dD (g), where a, b, c, and d are stoichiometric coefficients and A, B, C, and D are reagents and products, the equilibrium constant must be written as:

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

*brackets [ ], mean molar concentration

Then, as we know the equilibrium concentrations of reagents and products in the reaction $N_{2} (g) + O_{2} (g)$ ⇄ 2 NO (g), the equilibrium constant could be calculated as well:

$K = \frac{[NO]^{2}}{[O_{2}].[N_{2}]}$

$K = \frac{(0.400 M)^{2}}{(0.100 M).(0.100 M)} = 16$

Equilibrium constants only varies with temperature changes. So, K found before could be used with other equilibrium concentrations, supposing no temperature changes.

Let´s analyze what happens if more NO is added:

As more product is added, to re-establish the equilibrium, the reaction will consume NO to form reagents.This will happen in stoichiometric quantities, so supposing 2x the quantity of NO consumed, x amount of nitrogen and x of oxygen will be produced.

$N_{2} (g) + O_{2} (g)$ ⇄ 2 NO (g)

Initial situation: $[N_{2}] = 0.100 M; [O_{2}] = 0.100 M; [NO] = 0.700 M$

Equilibrium situation: $[N_{2}] = 0.100 M + x; [O_{2}] = 0.100 M + x; [NO] = 0.700 M - 2x$

Now, we can re-write equilibrium constan K with this new concentrations:

$K = \frac{[NO]^{2}}{[O_{2}].[N_{2}]} = 16$

$K = \frac{(0.700 M - 2x)^{2}}{(0.100 M + x).(0.100 M + x)} = 16$

Next step is to solve this equation to find x value.

$\frac{(0.700 M - 2x)^{2}}{(0.100 M + x).(0.100 M + x)} = 16$ $\frac{(0.700 M - 2x)^{2}}{(0.100 M + x)^{2}} = 16$

Solving the binomial squares $(a+b)^{2} = a^{2} + 2.a.b + b^{2}$

$\frac{4x^{2}-2.8x+0.49}{x^{2}+0.2x+0.01} = 16$ $4x^{2}-2.8x+0.49 = 16x^{2}+3.2x+0.16$

Grouping terms:

$12x^{2}+6x-0.33 = 0$

The quadratic formula of the roots of the general quadratic equation $ax^{2} +bx+c$ is:

$x_{1} =\frac{-b+\sqrt[]{4.a.c} }{2.a} \\x_{2} =\frac{-b-\sqrt[]{4.a.c} }{2.a}$

Then, for our problem:

$x_{1} =0.050 M\\x_{2} = -0.550 M$

Only one of this answers is correct. To know which one, let´s look wheather equilibrium concentrations are possible or not.

Case $x_{1} =0.050 M$

Equilibrium situation: $[N_{2}] = 0.100 M + x = 0.150 M; [O_{2}] = 0.100 M + x = 0.150 M; [NO] = 0.700 M - 2x = 0.600 M$

As all concentrations are possitive, it´s a possible situation. Let´s analyze the other to confirm.

Case $x_{2} =-0.550 M$

Equilibrium situation: $[N_{2}] = 0.100 M + x = -0.450 M; [O_{2}] = 0.100 M + x = -0.450 M; [NO] = 0.700 M - 2x = 1.800 M$

$x_{2}$ isn´t a possible answer because it results in negative equilibrium concentrations of the reagents. Then Case $x_{1} =0.050 M$ is the correct one.