Answer :
There is an error in the first sentence of the question; the right format is:
Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO2.
It should be NO2 and not NO.
Answer:
The equilibrium molarity of NO = 0.21695 m
Explanation:
Given that :
the volume = 500 mL = 0.500 m
number of moles of [tex]NO_3 = 1.9 \ mol[/tex]
number of moles of [tex]NO_2 = 1.6 \ mol[/tex]
Then we can calculate for their respectively concentrations as :
[tex][NO_3] = \frac{number \ of \ moles}{volume}[/tex]
[tex][NO_3] = \frac{1.9}{0.500}[/tex]
[tex][NO_3] = 3.8 \ M[/tex]
[tex][NO_2] = \frac{number \ of \ moles}{volume}[/tex]
[tex][NO_2] = \frac{}{} \frac{1.6}{0.500}[/tex]
[tex][NO_2] = 3.2 \ M[/tex]
The chemical reaction can be written as:
[tex]NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}[/tex]
The ICE table is as follows;
[tex]NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}[/tex]
Initial 3.8 - 3.2
Change +x x -2x
Equilibrium 3.8+x +x 3.2 - 2x
[tex]K_c=\frac{[NO_2]^2}{[NO_3][NO]}[/tex]
[tex]where \ K_c = 8.33[/tex]
[tex]8.33 = \frac{(3.2-2x)^2}{(3.8+2x)x} \\ \\ 8.33 = \frac{(3.2-2x)^2}{(3.8x+2x^2)}[/tex]
[tex]8.33(3.8x + 2x^2) = (3.2-2x)^2 \\ \\ 31.654x + 16.66x^2 = (3.2-2x)(3.2-2x) \\ \\ 31.654x + 16.66x^2 = 10.24 - 12.8x +4x^2 \\ \\ 10.24 - 44.454x -12.66 x^2 = 0 \\ \\ 12.66x^2 +44.454x -10.24 = 0[/tex]
Using quadratic formula;
[tex]\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}[/tex]
= [tex]\frac{-(44.454) + \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)} \ \ OR \ \ \frac{-(44.454) - \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)}[/tex]
= 0.21695 OR -3.7283
Going by the positive value;
x = 0.21695
[tex][NO_3] = 3.8 +x = 3.8 + 0.21695[/tex]
= 4.01695 m
[NO] = x = 0.21695 m
[tex][NO_2] = 3.2 +x = 3.2 + 0.21695[/tex]
= 3.41695 m