09022018

Question 1

Manipulating equilibrium expressions

Answer 1

• When a reversible chemical equation is manipulated, it is also necessary to make appropriate changes in the equilibrium expression and the equilibrium constant.
• Table 15.2

Question 2

Example:
The following reactions have the indicated equilibrium constants at 100C:
(1) 2NOBr (g) <> 2NO (g) + Br2 (g); K_c=0.014
(2) Br2 (g) + Cl2 (g) <> 2BrCl(g); K_c=7.2
Determine the value of K_c for the following reactions at 100C:
A. 2NO (g) + Br2 (g) <> 2NOBr (g)
B. 4NOBr(g) <> 4NO (g) + 2Br2 (g)
C. 2NOBr (g) + Cl2 <> 2NO (g) + 2BrCl (g)
D. NOBr (g) <> NO (g) + 1/2 Br2 (g)

Answer 2

A. K = [NOBr]^2 / ([NO]^2 * [Br2]) = 1/0.014 = 71
B. K = ([NO]^4 * [Br2]^2) / [NOBr]^4 = (0.014)^2 = 2.0 * 10^-4
C. K = { ([NO]^2 * [Br2]) / [NOBr]^2 }^1/2 = (0.014)^1/2 = 0.12
D. K = ([NO] * [Br2]^1/2) / NOBr = (0.14)*.

Question 3

Gaseous equilibria

Answer 3

• When an equilibrium expression contains only gases, we can write an alternate form of the expression in which the concentrations of gases are expressed as partial pressures (atm). Thus, for the equilibrium:
  N2O4 (g) <> 2NO2 (g)
  We can write as:
  K_c = [NO2]^2 / [N2O4]. OR K_P = (P_NO2)^2 / P_N2O4
• The relationship btw K_c and K_P can be expressed as:
  K_P = K_C[(0.08206 Latm/Kmol) * T]^dn
  Where dn = moles of gaseous products - moles of gaseous reactants

Question 4

Example:
Write K_P expressions for
A. PCl3 (g) + Cl2 (g) <> PCl5 (g)
B. O2 (g) + 2H2 (g) <> 2H2O (l)
C. F2 (g) + H2 (g) <> 2HF (g)

Answer 4

A. K_P = (P_PCl5) / {(P_PCl3) * (P_Cl2)}
B. K_P = 1 / {(P_O2) * (P_H2)^2
C. K_P = (P_HF)^2 / (P_F2) * (P_H2)

Question 5

Example:
The equilibrium constant, K_c, for the reaction
N2O4 (g) <> 2NO2 (g)
Is 4.63 * 10^-3 at 25C. What is the value of K_P at this temperature.

Answer 5

K_P = {K_c * (0.08206 Latm/Kmol)*T}^dn
> dn = 2-1
> K_P = 4.63 * 10^-3 * 0.08206 * 298 = 0.113

Question 6

Using Q and K to predict the direction of reaction

Answer 6

• The equilibrium expression may be used to predict th direction of a reaction and to calculate equilibrium concentrations
• Predictions are made based on comparisons btw Q_c and K_c
• There are 3 possibilities:
  1. QK: The ratio of initial concentrations of products to reactants is too large. To reach equilibrium products must be converted to reactants. The system proceeds in the reverse direction.

Question 7

Example:
At 375C, the equilibrium constant for the reaction
N2 (g) + 3H2 (g) <> 2NH3 (g)
Is 1.2. At the state of a reaction, the concentrations of N2, H2, and NH3 are 0.071M, 9.210^-3M, and 1.8310^-4M, respectively. Determine whether the system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.

Answer 7

Q_c = [NH3]^2 / ([N2] * [H2]^3)
= (1.8310^-4)^2 / {(0.071)(9.2*10^-3)} = 0.61
> Q_c < K_c
> Forward direction

Question 8

PCl5 (g) (decomposes)
PCl5 (g) <> PCl3 (g) + Cl2 (g)
0.686 atm, 63% decomposition rate

Answer 8

K_P = {(P_PCl3)*(P_Cl2)} / (P_PCl5)

           P_PCl5    |    P_PCl3    |    P_Cl2
   1  |  0.686     |       0          |      0
   c  |      -X        |        X         |      X                  X = 0.63*0.686 = 0.432
   €  |0.686-0.432 = 0.254| 0.432 | 0.432|

K_P = {(0.432)*(0.432)} / 0.254 = 0.74 atm
K_c = K_P /(RT)^dn = 0.74 / (0.08206 L*atm/mol*K)