Respuesta :
Answer:
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:
[tex]K = \frac{[C]^{c}.[D]^{d}}{[A]^{a}.[B]^{b}}[/tex]
*brackets [ ], mean molar concentration
Then, as we know the equilibrium concentrations of reagents and products in the reaction [tex]N_{2} (g) + O_{2} (g)[/tex] ⇄ 2 NO (g), the equilibrium constant could be calculated as well:
[tex]K = \frac{[NO]^{2}}{[O_{2}].[N_{2}]}[/tex]
[tex]K = \frac{(0.400 M)^{2}}{(0.100 M).(0.100 M)} = 16[/tex]
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.
[tex]N_{2} (g) + O_{2} (g)[/tex] ⇄ 2 NO (g)
Initial situation: [tex][N_{2}] = 0.100 M; [O_{2}] = 0.100 M; [NO] = 0.700 M[/tex]
Equilibrium situation: [tex][N_{2}] = 0.100 M + x; [O_{2}] = 0.100 M + x; [NO] = 0.700 M - 2x[/tex]
Now, we can re-write equilibrium constan K with this new concentrations:
[tex]K = \frac{[NO]^{2}}{[O_{2}].[N_{2}]} = 16[/tex]
[tex]K = \frac{(0.700 M - 2x)^{2}}{(0.100 M + x).(0.100 M + x)} = 16[/tex]
Next step is to solve this equation to find x value.
- [tex]\frac{(0.700 M - 2x)^{2}}{(0.100 M + x).(0.100 M + x)} = 16[/tex]
- [tex]\frac{(0.700 M - 2x)^{2}}{(0.100 M + x)^{2}} = 16[/tex]
Solving the binomial squares [tex](a+b)^{2} = a^{2} + 2.a.b + b^{2}[/tex]
- [tex]\frac{4x^{2}-2.8x+0.49}{x^{2}+0.2x+0.01} = 16[/tex]
- [tex]4x^{2}-2.8x+0.49 = 16x^{2}+3.2x+0.16[/tex]
Grouping terms:
- [tex]12x^{2}+6x-0.33 = 0[/tex]
The quadratic formula of the roots of the general quadratic equation [tex]ax^{2} +bx+c[/tex] is:
[tex]x_{1} =\frac{-b+\sqrt[]{4.a.c} }{2.a} \\x_{2} =\frac{-b-\sqrt[]{4.a.c} }{2.a}[/tex]
Then, for our problem:
- [tex]x_{1} =0.050 M\\x_{2} = -0.550 M[/tex]
Only one of this answers is correct. To know which one, let´s look wheather equilibrium concentrations are possible or not.
- Case [tex]x_{1} =0.050 M[/tex]
Equilibrium situation: [tex][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[/tex]
As all concentrations are possitive, it´s a possible situation. Let´s analyze the other to confirm.
- Case [tex]x_{2} =-0.550 M[/tex]
Equilibrium situation: [tex][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[/tex]
[tex]x_{2}[/tex] isn´t a possible answer because it results in negative equilibrium concentrations of the reagents. Then Case [tex]x_{1} =0.050 M[/tex] is the correct one.
The answer is: The final concentration of NO after equilibrium is re‑established will be [NO] = 0.600 M
The final concentration of NO after equilibrium is re-established is 0.6
The given equation for the chemical reaction is given as:
[tex]\mathbf{N_{2(g)} + O_{2(g)} \to 2NO_{(g)}}}[/tex]
The I.C.E table can be constructed as:
N₂ + O₂ ↔ 2[NO]
Initial 0.1 0.1 0.70
Concentration +x +x -2x
Equilibrium 0.1 +x 0.1 + x 0.70 - 2x
According to the equilibrium constant, the equilibrium constant for the I.C.E table is:
[tex]\mathbf{K _c= \dfrac{ (0.7-2x)^2 }{( 0.1+x )(0.1+x) } }[/tex]
where;
- equilibrium constant [tex]\mathbf{K_c}[/tex] = 16
[tex]\mathbf{16= \dfrac{ (0.7-2x)^2 }{( 0.1+x )(0.1+x) } }[/tex]
By solving for (x), x = 0.05
Therefore, we can conclude that the final concentration of NO = (0.70 -2x)
= 0.70 - 2(0.05)
= 0.6
Learn more about the equilibrium constant here:
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