Question

1. At a particular temperature, K 0.267 for the reaction SOg+NO SO2+NO) If all four gases had initial concentrations of 0.800 M, calculate the equilibrium concentrati 0.8004 267-(0.800tx ons of the gases. (6 pts) (0.800-xX0.800- NO2 S0a No S0ろ 00 80m Wom -X +X 0.8 00t 0.8 004 O.800-x10.800 -X 2. Consider the following reaction: 2 NOIBr 2 NOBrig) Kp = 28.4 at 298 K In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br) is 126 torr. What is the partial pressure of NOBr in this mixture? (5 pts) 108x 2- 216 Br= 126 3. Consider the following process

Answer 1

1) Kc = 0.267

Reaction Quotient (Q) = 0.8*0.8/0.8*0.8 = 1

It means, Q > Kc

Therefore, reaction will proceed in backward direction i.e reactant side.

Kc for backward reaction = 1/0.267 = 3.7453

New reaction will be like,

SO2 + NO2 ----------> SO3 + NO , Kc = 3.7453

0.8 0.8 0.8 0.8 (at t= 0)

0.8-x 0.8-x 0.8+x 0.8+x (at equilibrium)

Now,

Kc = (0.8+x)^2 / (0.8-x)^2

3.7453 = [0.8+x / 0.8-x]^2

0.8+x / 0.8-x = 1.9353

0.8+x = 1.54824 - 1.9353x

2.9353x = 0.74824

x = 0.2549

Hence, below will be equilibrium concentration:

[SO2] = [NO2] = 0.8-x = 0.8 - 0.2549 = 0.5451 M

[SO3] = [NO] = 0.8+x = 0.8+0.2549 = 1.0549 M .... Answer

2)

2NO + Br2 ---------> 2NOBr , Kp = 28.4

Kp = (Pnobr)^2 / (Pno)^2 * (Pbr)

28.4 = x^2 / (108)^2 * (126)

x = 6460.53 torr

Hence, partial pressure of NOBr = x = 6460.53 torr .... Answer