Question

For the reaction Br₂ (g) + Cl₂ (g) → 2 BrCl Kp = 1.11 x 10^-4 at 150K.
a) If 1.0 atm each of bromine and chlorine gas are placed in an otherwise empty 1 L container and allowed to equilibrate at a temperature of 150K, what will be the partial pressures at equilibrium of all three gases?
Br₂  atm
Cl₂  atm
BrCl  atm
b) suppose 0.2 atm each of all three gases are placed in a selaed container and allowed to equilibrate at 150K. What will be the partial pressure of BrCl at equiibrium?
atm

For the reaction N₂O₄ (g) → 2 NO₂ Kc = 0.513 at a temperature of 500K.
a) If the concentration of NO₂ is 0.1 M at equilibrium, what is the equilibrium concentration of N₂O₄ ?  M
b) If the equilibrium concentration of N₂O₄ is 0.0085 M, what is equilibrium concentration of NO₂?  M

Answer 1

a)

Br₂ (g) + Cl₂ (g) ------------------> 2 BrCl                    Kp = 1.11 x 10^-4

1 atm     1 atm                           0

1 - x    1- x    2x

Kp = P²BrCl / P_Br2 x P_Cl2

1.11 x 10^-4 = (2x)^2 / (1 - x)^2

2x / 1 - x = 0.01054

2x = 0.01054 - 0.01054 x

x = 0.00524

equilibrium partial pressures:

P_{Br2 =} 0.995 atm

P_Cl2 = 0.995 atm

P_BrCl = 0.0105 atm

b)

Br₂ (g)   + Cl₂ (g) ------------------> 2 BrCl                    Kp = 1.11 x 10^-4

0.2 atm     0.2 atm                            0.2

0.2 - x    0.2- x    0.2 + 2x

Kp = (0.2 + 2x)^2 / (0.2 - x)^2

1.11 x 10^-4 = (0.2 + 2x)^2 / (0.2 - x)^2

0.2 + 2x / 0.2 - x = 0.01054

0.2 + 2x = 2.107 x 10^-3 - 0.01054 x

2.01054 x = - 0.1979

x = - 0.0984

partial pressure of BrCl = 0.00314 atm