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

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.

The equilibrium molarity of NO = 0.21695 m

Explanation:

Given that :

the volume = 500 mL = 0.500 m

number of moles of

number of moles of

Then we can calculate for their respectively concentrations as :

The chemical reaction can be written as:

The ICE table is as follows;

Initial              3.8         -               3.2

Change          +x          x               -2x

Equilibrium     3.8+x    +x              3.2 - 2x

Using quadratic formula;

=

= 0.21695 OR -3.7283

Going by the positive value;

x = 0.21695

= 4.01695 m

[NO] = x  = 0.21695 m

= 3.41695 m

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