Question

3. Nitrogen and oxygen react to produce nitric oxide according to the following equation: N2 (g) + O2(g) → 2 NO (g) The equilibrium constant for this reaction is 1.70 x 10-3. Suppose that 0.180 mol N2 and 0.650 mol O2 are mixed in a 4.00-L reaction vessel. What will be the concentrations of N2, 02, and NO when equilibrium is established? (Hint: assume that the amounts of N2 and O2 that react are small-less than 10% of the starting amounts - and check your assumption when you have solved the equation.)

Answer 1

Sol.

Initial moles of N2 = 0.180 mol

Volume of vessel = 4 L

So , Initial concentration of N2  

= 0.180 mol / 4 L = 0.045 M

Also , Initial moles of O2 = 0.650 mol

Initial concentration of O2  

= 0.650 mol / 4 L = 0.1625 M

Reaction : N2(g) + O2(g) <----> 2NO(g)

Initial 0.045 0.1625 0

change - x - x + 2x

equilibrium 0.045 - x 0.1625 - x 2x

So , Kc = [NO]² / ( [N2] × [O2] )

1.70 × 10^-3 = (2x)² / ( ( 0.045 - x ) ( 0.1625 - x ) )

Assuming that x is very small , so ,  

0.045 - x = 0.045

and 0.1625 - x = 0.1625

Therefore ,

1.70 × 10^-3 = 4x² / ( 0.045 × 0.1625 )

x² = 0.0000031078

x = 0.00176

Equilibrium Concentrations are :

[N2] = 0.045 - x = 0.045 - 0.00176 =   0.04324   M  

[O2] = 0.1625 - x = 0.1625 - 0.00176 =   0.16074 M  

[NO] = 2x = 2 × 0.00176 =   0.00352 M