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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Answer from: 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.

The answer is: The final concentration of NO after equilibrium is re‑established will be [NO] = 0.600 M

Answer from: Quest

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –

6.022

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,

will have a gram-molecular weight of 352g.

Answer from: Quest

explanation:

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