Question

Suppose a 250.mL flask is filled with 1.7mol of O2 and 1.6mol of NO. The following reaction becomes possible: +N2g + O2g <-->2NOg The equilibrium constant K for this reaction is 0.628 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. *** must calculate for M first by dividing the moles by Liters before placing in the ICE table.

Answer 1

Volume of flask = 250 ml = 0.250 L

Molarity of O2 = 1.7 mol / 0.250 L = 6.8 mol / L = 6.8 M

Molarity of NO = 1.6 mol / 0.250 L = 6.4 M

Given reaction N2(g) + O2(g) <------> 2 NO(g)

Initial concentration 0 6.8 6.4

Change in concentration -x -x. +2x

Equilibrium concentration - x. 6.8 -x. 6.4+2x

K = [NO]2 / {[N2][O2]}

K = (6.4+2x)2 / (-x)(6.8-x)

0.628 = (40.96 + 4x2 + 25.6 x)/ (-6.8x +x2)

0.628(-6.8x + x2) = 40.96 + 4x2 + 25.6 x

-4.27x + 0.628 x2 = 40.96 + 4x2+25.6 x

3.372x2+ 29.87 x + 40.96 = 0

This is quadratic equation in x, after solving we get x = -1.69

Equilibrium molarity of NO = 6.4+2x = 6.4 + 2(-1.69) = 3.02 M