Question

Suppose a 500.mL flask is filled with 0.20mol of NO3 and 1.9mol of NO . The following reaction becomes possible: +NO3 (g) + NO (g) ---> 2NO2 (g) The equilibrium constant K for this reaction is 5.74 at the temperature of the flask. Calculate the equilibrium molarity of NO3 . Round your answer to two decimal places.

Answer 1

Sol . As Volume of flask = 500 mL = 0.5 L

intiial moles of NO3 = 0.20 mol

So , initial molarity of NO3 = initial moles of NO3 / Volume of flask = 0.20 / 0.5 = 0.4 M

Now , initial moles of NO = 1.9 mol  

So , initial molarity of NO = initial moles of NO / Volume of flask = 1.9 / 0.5 = 3.8 M

Now , Reaction :

NO3(g) + NO(g) <----> 2NO2(g)

Initial 0.4 3.8 0

Change - x - x +2x

Equilibrium 0.4 - x 3.8 - x 2x

As Kc = [NO2]² / ( [NO3] [NO] )

5.74 = (2x)² / ( ( 0.4 - x )(3.8 - x ) )

5.74 ( 1.52 - 4.2x + x² ) = 4x²

8.7248 - 24.108x + 5.74x² = 4x²

or , 1.74x² - 24.108x + 8.7248 = 0  

or , x² - 13.855x + 5.014 = 0  

Solving this quadratic equation ,  

x = ( - ( -13.855) +- ( -13.855 × -13.855 - 4 × 1 × 5.014 )^1/2 ) / 2

x = ( 13.855 +- 13.111 ) / 2  

So , x = ( 13.855 + 13.111 ) / 2 = 13.483

or , x = ( 13.855 - 13.111 ) / 2 = 0.372

But x = 13.483 is so high than the initial molarity  of NO3 and NO , which is not posssible .

Therefore , x = 0.372

Now , Equilibrium Molarity of NO3 = 0.4 - x  

= 0.4 - 0.372 =   0.028 M