Question

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO. The following reaction becomes possible: NO3gNOg 2NO2g The equilibrium constant K for this reaction is 8.33 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places

Answer 1

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 $NO_3 = 1.9 \ mol$

number of moles of $NO_2 = 1.6 \ mol$

Then we can calculate for their respectively concentrations as :

$[NO_3] = \frac{number \ of \ moles}{volume}$

$[NO_3] = \frac{1.9}{0.500}$

$[NO_3] = 3.8 \ M$

$[NO_2] = \frac{number \ of \ moles}{volume}$

$[NO_2] = \frac{}{} \frac{1.6}{0.500}$

$[NO_2] = 3.2 \ M$

The chemical reaction can be written as:

$NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}$

The ICE table is as follows;

                    $NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}$

Initial              3.8         -               3.2

Change          +x          x               -2x

Equilibrium     3.8+x    +x              3.2 - 2x

$K_c=\frac{[NO_2]^2}{[NO_3][NO]}$

$where \ K_c = 8.33$

$8.33 = \frac{(3.2-2x)^2}{(3.8+2x)x} \\ \\ 8.33 = \frac{(3.2-2x)^2}{(3.8x+2x^2)}$

$8.33(3.8x + 2x^2) = (3.2-2x)^2 \\ \\ 31.654x + 16.66x^2 = (3.2-2x)(3.2-2x) \\ \\ 31.654x + 16.66x^2 = 10.24 - 12.8x +4x^2 \\ \\ 10.24 - 44.454x -12.66 x^2 = 0 \\ \\ 12.66x^2 +44.454x -10.24 = 0$

Using quadratic formula;

$\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}$

= $\frac{-(44.454) + \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)} \ \ OR \ \ \frac{-(44.454) - \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)}$

= 0.21695 OR -3.7283

Going by the positive value;

x = 0.21695

$[NO_3] = 3.8 +x = 3.8 + 0.21695$

= 4.01695 m

[NO] = x  = 0.21695 m

$[NO_2] = 3.2 +x = 3.2 + 0.21695$

= 3.41695 m