Question

Suppose a

250.mL

flask is filled with

0.60mol

of

NO3

and

1.1mol

of

NO2

. The following reaction becomes possible:

+NO3gNOg

2NO2g

The equilibrium constant

K

for this reaction is

0.467

at the temperature of the flask.

Calculate the equilibrium molarity of

NO

. Round your answer to two decimal places.

M

Answer 1

Calculating initial Molarity : Number of moles of NO3 = 0.6 mol Volume of NO3 = 250 ml Volume in Liters of NO3 = 0.25 L Molarity of NO3 = ? Number of moles Molarity of NO3 = Volume in Liters 0.6 mol Molarity of NO3 = 0.25 L Molarity of NO3 = 2.4 M Calculating initial Molarity : Number of moles of NO2 = 1.1 mol Volume of NO2 = 250 mL Volume in Liters of NO2 = 0.25 L Molarity of NO2 = ? Number of moles Molarity of NO2 = Volume in Liters 1.1 mol Molarity of NO2 = 0.25 L Molarity of NO2 = 4.4 M

Construct ICA table : Reaction NO: (g) + NO (g) + Initial(M) 2.40 0 .00 Change (M) -X -X Equilibrium(M) 2.40-X -X 2 NO2 (g) 4.4 +2x 4.4+2x Equilibrium expression: K - [ NO2 12 equilibrium constant K = [ NO3 ] [ NO ] 0.467 = [ 4.4 +2x 1? [ 2.40 -> ] [-x] Calculating x Value: 3.533 x + 18.721 x + 19.36 = 0 By solving the above quadratic equation, we get x = -1.408 The equilibrium molarity of NO = -X = 1.41 M