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

Step-by-step 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] = (number \ of \ moles)/(volume)`

`[NO_3] = (1.9)/(0.500)`

`[NO_3] = 3.8 \ M`

`[NO_2] = (number \ of \ moles)/(volume)`

`[NO_2] = \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=([NO_2]^2)/([NO_3][NO])`

`where \ K_c = 8.33`

`8.33 = ((3.2-2x)^2)/((3.8+2x)x) \\ \\ 8.33 = ((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;

`(-b\pm √((b)^2-4ac) )/(2a)`

=
`(-(44.454) + √((44.454)^2-4(12.66)(-10.24)) )/(2(12.66)) \ \ OR \ \ (-(44.454) - √((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