Class Demo with hall• Chemical Equilibrium – when the rate of the

forward and reverse reactions are equal• Dynamic – Reactions at eq never stop• Equilibrium DOES NOT mean that the amount of

reactants and products are equal. They have reached an unchanging ratio

Definition

Equilibrium

Equilibrium GIDefinition

Cis Trans

1. NaCl (s) Na+(aq) + Cl-(aq)[unsaturated]

2. NaCl (s) Na+(aq) + Cl-(aq) [saturated]

Definition

N2O4(g) 2NO2(g)

Clear BrownCold Hot

1. Starting with all reactants (N2O4)

N2O4(g) 2NO2(g)

2. Starting with all products (NO2)

Graphs

1. Kc = Eq. Constant involving molarity

a. Molarity = [ ]b. Example = [0.50 M]

2. Kp = Eq. Constant involving pressure

a. Atmospheresb. We live at about 1 atm

3. Generic ExampleaA + bB cC + dD

Eq. Constants

2O3(g) 3O2(g)

2NO(g) + Cl2(g) 2NOCl(g)

H2(g) + I2(g) 2HI(g)

Eq. Constants

Heterogeneous Equilibrium1. More than one state is present2. Exclude solids and liquids from K. (not

considered to have a molarity or pressure)

SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)

Pb(NO3)2(aq)+Na2SO4(aq)PbSO4(s)

+2NaNO3(aq)

Ba2+(aq) + SO42-(aq) BaSO4(s)

Heterogeneous Eq.

1. Exclude liquid water (often the solvent)2. Keep gaseous water3. Examples

CO2(g) + H2(g) CO(g) + H2O(l)

3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

A Note About Water

1. An exampleCO(g) + Cl2(g) COCl2(g)

Kc = 4.6 X 109

2. RulesK>>1 Favors the productsK<<1 Favors the reactantsK~1 Reactants ~ Products

1. Does the following reaction favor the products or reactants?

N2(g) + O2(g) 2NO(g)Kc = 1 X 10-30

2. For the following reaction, Kc = 794 at 298 K and Kc = 54 at 700 K. Should you heat or cool the mixture to promote the formation of HI?

H2(g) + I2(g) 2HI(g)

1. Kc= 2.5 X 10-30 for N2(g) + O2(g) 2NO(g) calculate Kc for:

2NO(g) N2(g) + O2(g)

2. Calculate Kc for ½ N2(g) + ½ O2(g) NO(g)

3. The Kc for N2(g) + 3H2(g) 2NH3(g) is 4.43 X10-3. Calculate Kc for:

2N2(g) + 6H2(g) 4NH3(g)

Converting Between Kc and Kp

Kp = Kc (RT)n

R = 0.0821 L atm/mol KT = Kelvin Temperature n = change in number of moles of gas

Calculate Kp for the following reaction at 300 oC:

N2(g) + 3H2(g) 2NH3(g)

Kc = 9.60

ANS: 0.00434 (4.34 X 10-3)

Calculate Kp for the following reaction at 1000 K:

2 SO3(g) 2SO2(g) + O2(g)

Kc = 4.08 X 10-3

ANS: 0.335

1. A mixture is allowed to reach eq.. At eq., the vessel contained 0.1207 M H2, 0.0402 M N2, and 0.00272 M NH3. Calculate the equilibrium constant.

N2(g) + 3H2(g) 2NH3(g)

(Ans: 0.105)

2. At eq., a vessel contained 0.00106 M NO2Cl, 0.0108 M NO2, and 0.00538 M Cl2. Calculate the equilibrium constant.

NO2Cl(g) NO2(g) + Cl2(g)

(Ans: 0.558)

3. A mixture of 0.00500 mol of H2 and 0.0100 mol of I2 is placed in a 5.00 L flask and allowed to reach eq.. At eq., the mixture is found to be [HI] = 0.00187 M. Calculate Kc.

H2(g) + I2(g) HI(g)

(Ans: 51)

4. A vessel is charged with 0.00609 M SO3. At eq., the SO3 concentration had dropped to 0.00244 M SO3. What is the value of Kc?

SO3(g) SO2(g) + O2(g)

(Ans: 0.0041)

5. 4.00 mol of HI was placed in a 5.00 L flask and allowed to decompose. At eq. It was found that the vessel contained 0.442 mol of I2. What is the value of Kc?

HI(g) H2(g) + I2(g)

(Ans: 0.020)

6. An eq. mixture of gases is analyzed. The partial pressure of nitrogen is 0.432 atm and the partial pressure of hydrogen is 0.928 atm. If Kp is 1.45 X 10-5, what is the partial pressure of ammonia?

N2(g) + 3H2(g) 2NH3(g)

7. Consider the following equilibrium:PCl5(g) PCl3(g) + Cl2(g)

a. At equilibrium, the partial pressure of PCl5 and PCl3 are measured to be 0.860 atm and 0.350 atm, respectively. If Kp = 0.497, what is the partial pressure of Cl2? (1.22 atm)

b. Suppose at equilibrium the partial pressure of PCl5 is 2.00 atm. Calculate the partial pressure of PCl3 and Cl2 . Assume Kp is still 0.497 and only PCl5 was initially in the flask. (0.997 atm)

1. 0 = ax2 + bx + c

2. x = -b + \/ b2 – 4ac2a

3. 2x2 + 4x = 1

1. A gas cylinder is charged with 1.66 atm of PCl5 and allowed to reach eq.. If the Kp= 0.497, what are the pressures of all the gases at equilibrium?

PCl5(g) PCl3(g) + Cl2(g)

(Ans: 0.97 atm, 0.693 atm)

2. A 1.000 L flask is filled with 1.000 mol of H2 and 2.000 mol of I2. The Kc = 50.5. What are the concentrations of all the gases at equilibrium?

H2(g) + I2(g) 2HI(g)

(Ans: 0.065 M, 1.065 M, 1.87 M)

Do I always need the quadratic, or can I cheat?

The equilibrium constant for the following reaction is 2400.

2NO(g) N2(g) + O2(g)

If the initial concentration of NO is 0.157 M, calculate the equilibrium concentrations of NO, N2 and O2.

(Ans: 0.0016 M, 0.0777 M, 0.0777 M)

Equilibrium

1. Reaction Quotient2. Calculated the same as K, but using initial

concentrations

3. Q < Kshifts to productsQ = Kat equilibriumQ > Kshifts to reactants

Q: Reaction Quotient

1. If you introduce 0.0200 mol of HI, 0.0100 mol of H2 and 0.0300 mol of I2 in a 2.00 L flask, which way will the reaction proceed to reach equilibrium?

H2(g) + I2(g) 2HI(g) Kc = 51

(Ans: Q = 1.3)

2. Predict which way the following reaction will proceed as it reaches eq. Assume that you start with [SO3] = 0.002 M, [SO2] = 0.005 M and [O2] = 0.03M.

2SO3(g) 2SO2(g) + O2(g)

Kc = 0.0041

(Ans: Q = 0.2)

3. Predict which way the following reaction will proceed as it reaches eq. Assume that you start with [NH3] = 0.002 M, [N2] = 0.005 M and no H2. Kc= 0.105

N2(g) + 3H2(g) 2NH3(g)

Blue Bottle Demo5 grams KOH3 grams Dextrose250 mL of water1 drop methylene blue

LeChatelier’s Principle

• Definition – If a system at eq. Is disturbed, it will shift to relieve that disturbance

LeChatelier’s Principle

If a system at eq. is disturbed, it

will shift to relieve that disturbance

Le’Chatelier’s Principle

N2(g) + 3H2(g) 2NH3(g)

• Add N2

• Add NH3

• Remove NH3 as it forms• Remove H2

N.B. Does NOT apply to solids and liquids. They do not appear in the K.

LiCl(s) Li+(aq) + Cl-(aq)

Disturbing and K

• Adding products = K increases (TEMPORARILY)• Adding reactants = K decreases

(TEMPORARILY)

N2(g) + 3H2(g) 2NH3(g)

Kc = [NH3]2 [N2][H2]3

N2(g) + 3H2(g) 2NH3(g)

• Identify the # of moles of gas on either side.• Show piston drawing

1. Increase the volume of the container2. Decrease the volume of the container

N2(g) + 3H2(g) 2NH3(g)

1. Increase the pressure of the system2. Decrease the pressure of the system

Soda example (CO2(aq) CO2(g))

N.B. Adding a noble or inert gas has no effect on the eq. Pressure change without a volume change.

• Endothermic Reactions – absorb heat from the surroundings– Heat is added (reactants)– Cooking is an example

H +

• Exothermic Reactions – Release heat– Give off heat (products)– Fire is an example

H -

CO(g) + 3H2(g) CH4(g) + H2O(g)

H = -206 kJ/mol

1. Heat the system2. Cool the system

1. Catalyst2. Examples

a. Enzymesb. Vitaminsc. Catalytic convertor

3. No effect on the position of equilibrium

Example 1N2O4(g) 2NO2(g)

H = 58 kJ/mol

a. Add N2O4

b. Remove NO2 as it formsc. Increase the total pressured. Increase the total volumee. Cool the solutionf. Add a catalyst

Example 2.

2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)H = -37 kJ/mol

a. Add PbSb. Remove SO2 as it formsc. Add O2d. Increase volumee. Decrease the pressuref. Heat the flask

Example 3.Given the following eqn., how could you promote the

formation of PCl3 and Cl2?

PCl5(g) PCl3(g) + Cl2(g)H = 88 kJ/mol

Example 4.How could you promote the formation of CH4?

CO(g) + 3H2(g) CH4(g) + H2O(g) H = -206 kJ/mol

SINGLE ARROWReaction goes to completion

HCl + NaOH NaCl + H2O

ICEnd

Na+ Cl-

H2O

Double ARROWReaction goes to equilibrium (all species present)

N2 + 3H2 2NH3

ICEquilibrium

N2 H2 K = [NH3]2 ratio

NH3 [N2][H2]3

HC2H3O2(aq) + H2O(l) ↔ H3O+(aq) + C2H3O2-(aq)

Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq)

Co(H2O)62+(aq) + 4Cl-(aq) + 50 kJ/mol ↔ CoCl4

2-(aq) + 6H2O(l)

(pink) (blue)

NaCl(s) ↔ Na+(aq) + Cl-(aq)

14.[O2]3/[O3]2 1/[Cl2]2

[C2H6]2[O2]/[C2H4]2[H2O]2 [CH4]/[H2]2

[Cl2]2/[HCl]4[O2]16. a) Products b) Reactants 18. Kc = 85820a) 1.35 X 105 b) H2S favored22a) 13.3 b) 0.274 c) 0.034926. Kp = 1/PSO2

Na2O is a solid, no molarity or pressure28. Kc = 10.530. 66.8, Products favored32. a) 0.14 M, 0.020 M, 0.40 M b) 58

38. a) 0.0013(R) b) Reactants c) 1.1 X10-5(P) 40.0.0535 atm42a. 0.0362 g I2 b) 0.018 g SO2

44. 0.13 M46.0.011 M48a. 0.0432 M PH3 and BCl3

52 a) increase(P) b) decrease(R) c) decrease(R)d) decrease(R) e) no change f) decrease(R)

54 a) Endothermic b) more moles of gas in product56 a) -90.7 kJ b) exothermic

c) Increase pressure 59. Kp = 24.7 Kc = 0.00367

60. 0.71

Take-Home Pretest1.1.09 g NaOH2.20.43.43524.1675.Products6.Q = 0.637, moves to products7.[H2] = 0.312 M, [Cl2] = 0.012 M, [HCl] = 0.376 M

Equilibrium

1.82 = 2m log 1.82 = log 2m log 1.82 = m log 2

0.260 = 0.301 mm = 0.864