Question

To study the following reaction at 20C, a mixture of NO(g), NO2(g) and H2O(g) was prepared in a 10.0L glass reaction vessel.

H2O(g) + NO2(g) + NO(g) <=> 2HNO2 (g)

For NO, NO2 and HNO2, the initial concentrations were as follows: [NO]=2.59 x 10^-3 M, [NO2]=2.59 x10^-3 and [HNO2]=0M. The inital partial pressure of H2O(g) was 17.5 torr. When equilibrium was reached, the HNO2 concentration was 4.0x10^-4M.

a.) From this data, calculate the value of the equilibrium constant, Kc.

b.) What is the value of the Kp equilibrium constant at 20C?

Answer 1

pH2O = 17.5 torr

760 torr = 1 atm

17.5 torr = 0.023 atm

PV = nRT

0.023*10 = n*0.0821*(273+20)

n = 0.00956

Molarity of H2O = n / V = 0.000956

H2O(g) + NO2(g) + NO(g) <=> 2HNO2 (g)

0.000956 0.00259 0.00259 0

0.000956-x 0.00259-x 0.00259-x 2x

At equilibrium

2x = 0.0004

x = 0.0002

[HNO2] = 0.0004

[H2O] = 0.000956 - 0.0004 = 0.000556

[NO2] = [NO] = 0.00259 - 0.0004 = 0.00219

Kc = [HNO2]^2 / [H2O] [NO2] [NO]

Kc = 6

b]

Kp = Kc RT^(delta n)

delta n = No of products - No of reactants = 2 - 3 = -1

Kp = Kc / RT = 6 / (0.0821*293) = 0.2494

Kp = 0.25