Question

The reaction 2NO(g) + O2(g) ⇌ 2NO2(g) has Kp = 1.65 × 1011 at 25°C. To a 1.00 L flask, 1.03 g of NO and 532 mL of O2 measured at 29°C and 772 torr are mixed. When the mixture comes to equilibrium at 25°C, what is the concentration of NO(g)? If needed, use “E” for scientific notation. Do not enter units as part of your answer. Report your answer to the correct number of significant figures.

Answer 1

2NO(g) + O2(g) <------------> 2NO2(g)                Kp = 1.65 × 10^11

moles of NO = 1.03 / 30 = 0.0343 mol

concentration of NO = 0.0343 M

volume of O2 = 532 mL = 0.532 L

pressure = 772 torr = 1.016 atm

temperrature = 29 oC = 302 K

P V = n R T

1.016 x 0.532 = n x 0.0821 x 302

n = 0.0218

moles of O2 = 0.0218

concentration of O2 = 0.0218 M

Kp = Kc (RT)^Dn

1.65 x 10^11 = Kc x (0.0821 x 302)^-1

Kc = 4.09 x 10^12

2 NO (g)   + O2 (g)    <--------------------> 2 NO2 (g)

0.0343         0.0218                                      0

0.0343 - 2x   0.0218 - x                               2x

Kc = (2x)^2 / (0.0343-2x)^2(0.0218 - x )

4.09 x 10^12 = 4x^2 / (0.0343-2x)^2(0.0218 - x )

x = 0.0163

concentration of NO = 0.0343 - 2x

concentration of NO = 0.00170 M