Le Chatelier's Principle and Equilibrium Calculations

Le Chatelier's Principle

_______________________ states:

If an outside influence upsets an equilibrium, the system undergoes a change in the direction that counteracts the disturbance and if possible, restores the system to equilibrium

Q > K: shifts to reactants

Q = K: no change

Q < K: shifts to products

If Q > K, the reaction shifts to _____________

If Q = K, there is ________________

If Q < K, the reaction shifts to ____________

Reaction shifts to products

PCl3(g) + Cl2(g) ⇄ PCl5(g)

How is the reaction effected when adding Cl2?

Reaction shifts to reactants

PCl3(g) + Cl2(g) ⇄ PCl5(g)

How is the reaction effected when adding PCl5?

Reaction shifts to reactants

PCl3(g) + Cl2(g) ⇄ PCl5(g)

How is the reaction effected when removing PCl3?

Exothermic

- (△H°) change in temperature of heat is a negative value (-83 NJ/mol)

- heat is the product

Endothermic

- (△H°) change in temperature of heat is a positive value (83 NJ/mol)

- heat is the reactant

How does an exothermic reaction differ from an endothermic reaction?

Heat as product: produced

Heat as reactant: absorbed

If heat in a reaction is a product, the heat is being _____________

If heat in a reaction is a reactant, the heat is being ______________

Exothermic

- Increase in temperature shifts the reaction to reactants

Endothermic

- Increase in temperature shifts the reaction to products

What happens if (△H°) temperature of heat is increased in an exothermic reaction vs an endothermic reaction?

Exothermic

- Decrease in temperature shifts the reaction to Products

Endothermic

- Decrease in temperature shifts the reaction to reactants

What happens if (△H°) temperature of heat is decreased in an exothermic reaction vs an endothermic reaction?

Reaction shifts to the side with less gaseous molecules/moles

Change in Vol/pressure of gases equilibrium:

What side does the reaction shift to when volume decreases while pressure increases?

Reaction shifts to the side with more gaseous molecules/moles

Change in Vol/pressure of gases equilibrium:

What side does the reaction shift to when volume increases while pressure decreases?

No change, both sides of the equation have the same amount of moles

If volume increases while pressure decreases, what side does this reaction shift to?

Br2(g) + Cl2(g) ⇄ 2BrCl(g)

Initials (M) + Change (M) = Equilibrium (M)

What does ICE stand for

KC = [HI]2/ [H2][I2]

1. Solve for Initial(M)

[H2] = [I2] → 0.200 mol/2.00L = 0.100M

2. ICE

H2(g) + I2(g) ⇄ 2HI(g)

I(M) 0.100 0.100 0

C(M) - 0.080 - 0.080 + 2(0.080)

E(M) 0.020 0.020 0.160

3. Solve for KC

Kc = (0.160)^2/ (0.020)^2 = 64

Calculate K from equilibrium concentrations/pressures:

For the reaction below, we initially have 0.200 mol of each reactant in a 2.00L flask. I2 concentration drops to 0.020M when it reaches equilibrium. Solve for Kc

H2(g) + I2(g) ⇄ 2HI(g)

1. Solve for I(M)

PCl3 M = 0.200 mol/1.00L = 0.200M

Cl2 M = 0.100 mol/1.00L = 0.100M

2. ICE

PCl3(g) + Cl2(g) ⇄ PCl5(g)

I(M) 0.200 0.100 0

C(M) - 0.080 - 0.080 + 0.080

E(M) 0.120 0.020 0.080

3. Solve for KC

Kc = (0.080)/ (0.120)(0.020) = 33

Calculate K from equilibrium concentrations/pressures:

0.200 mol PCl3 and 0.100 mol Cl2 in 1.00L. Solve for Kc

E(M) PCl3 = 0.120 mol

PCl3(g) + Cl2(g) ⇄ PCl5(g)

1. Solve for I(M)

N2O4 M = 0.300mol/ 2.00L = 0.150 M

2. ICE

N2O(g) ⇄ 2NO2(g)

I(M) 0.150 0

C(M) - x + 2x

E(M) 0.150 - x 2x

3. Solve for x by plugging into KC equation

4.50 = (NO2)2 / (N2O4)

4.50 = (2x)^2 / (0.150 - x)

4x2 / (0150 - x)

Ax^2 + bc + c = 0 → 4.50 (0.150 - x) = 4x^2

4x^2 + 4.50x - 0.675 = 0

4. Plug into quadratic equation →

a = 4

b = 4.5

c = - 0.625

x = 0.134

4. Solve for E(M) by plugging in x-value

NO2 M = 2(0.134) = 0.268M

N2O4 M = 0.150 - 0.134 = 0.016 M

Calculate equilibrium concentrations/pressures from K:

If KC = 4.50 for N2O(g) ⇄ 2NO2(g) and 0.300mol N2O4 is initially in a 2.00L container, what would be the equilibrium for both gases?

negligible

If change in concentration (x) is small relative to initial concentration, x is ______________

not negligible

If initial concentration = 0, x is _________________

- 500

- 5%

Initial Concentration Threshold:

Initial concentration should be at least _______ or more greater than K

value of x should be less than _____ of initial concentration

1. Determine if I(M) is 500 > K

(0.100/7.3 x 10^-18) 500 > K

2. ICE

2H2O(g) ⇄ 2H2(g) + O2(g)

I(M) 0.100 0 0

C(M) - 2x + 2x + x

E(M) 0.100 - 2x 2x x

3. Solve for x and plug into KC equation

7.3 x 10^-18 = (2x)^2(x) / (0.100)^2

x = 2.6 x 10^-7

4. Solve for E(M) by plugging in x-value

H2O M = 0.100 - 2(2.6 x 10^-7) = 0.100 M

H2 M = 2(2.6 x 10^-7) = 5.2 x 10^-7 M

O2 M = 2.6 x 10^-7 M

Solve for E(M) of H2O, H2 and O2

2H2O(g) ⇄ 2H2(g) + O2(g)

KC = 7.3 x 10^-18 at 1000°C

H2O I(M) = 0.100M

1. Solve for I(M)

N2O M = 0.200mol/ 4.00L = 0.050M

NO2 M = 0.400mol/ 4.00L = 0.100M

2. ICE

N2O(g) + NO2(g) ⇄ 3NO(g)

I(M) 0.050 0.100 0

C(M) - x - x + 3x

E(M) 0.050 - x 0.100 - x 3x

3. Solve for x and plug in to solve for E(M) NO

1.4 x 10^-10 = (3x3)/ (0.050)(0.100)

x = 3.0 x 10^-5

NO M = 3(3.0 x 10^-5) = 9.0 x 10^-5 M

Solve for E(M) of NO

N2O(g) + NO2(g) ⇄ 3NO(g)

KC = 1.4 x 10^-10 at 200°C