IIIIIIIVV Ch. 18 – Reversible Reactions & Equilibrium pp. 549 – 559.

I II III IV V

Ch. 18 – Reversible Reactions & Equilibrium

pp. 549 – 559

A. Reversible Reactions

Reactions can be reversible Conversion of reactants to products

and products to reactants happens simultaneously

Forward reaction: 2SO2(g) + O2(g) → 2SO3(g)

Reverse reaction: 2SO2(g) + O2(g) ← 2SO3(g)

A. Reversible Reactions Chemical equilibrium – when the rates of the

forward and reverse reaction are equal

Physical Equilibrium: vapor-liquid phase or solution equilibrium

No net change occurs in the actual amounts of the components of the system (reaction will never complete)

dynamic state

Concentration at equilibrium is temperature dependent

A. Reversible Reactions

Equilibrium position – made up of relative concentration of reactants and products at equilibrium

A B

Different sized arrows indicate favored direction of a reaction

Catalyst speeds up forward and reverse reactions equally

decreases time it takes to reach equilibrium

B. Equilibrium Constant

Law of Mass Action – expression of equilibrium condition using equilibrium constant, Keq

jA + kB ↔ lC + mD

kj

ml

eq BA

DCK

][][

][][

Brackets indicate concentrations at equilibrium (*pure solids and liquids do not affect equilibrium values!*)


Example

4NH3(aq) + 7O2(g) ↔ 4NO2(aq) + 6H2O(g) Arrows indicate coefficients

72

43

62

42

][][

][][

ONH

OHNOKeq


Write the balanced equation and the equilibrium expression for the production of ammonia

N2 (g) + 3H2 (g)↔ 2NH3 (g)

322

23

]][[

][

HN

NHKeq

C. Evaluating Equilibrium Constant

N2 (g) + 3H2 (g)↔ 2NH3 (g) Keq > 1

Products favored at equilibrium Keq < 1

Reactants favored at equilibrium Keq = 1

Large amount of both products and reactants at equilibrium

322

23

]][[

][

HN

NHKeq

Example

Solve for Keq in N2 (g) + 3H2 (g)↔ 2NH3 (g) if [N2] = .625, [H2] = 4.0 and [NH3] = 2.0

Plug into Keq expression

Keq=[2.0]2

[.625] [4.0]3Keq= .10

•Are the reactants or products favored in this reaction?

•Reactants


Write the balanced equation and the equilibrium expression for:

Zn (s) + 2 Ag+ (aq)↔ Zn2+ (aq) + 2 Ag (s)

2

2

][

][

Ag

ZnKeq

Example Solve for Keq in Zn (s) + 2

Ag+ (aq)↔ Zn2+ (aq) + 2 Ag (s)

if [Zn] = 109 M, [Ag+] = 2.0 and [Zn2+] = 24M Plug into Keq expression

Keq=[24]

[2.0]2Keq= 6.0

Reaction Equilibrium Concentrations in a problem, may not always be at

equilibrium! If you know initial molarities of reactants you can calculate

the ending molarites of all substances if we know the Keq value.

Can also calculate the Keq if you know the initial molarities and one end molarity.

To help us with this we make a RICE chart! R = Reaction I = Initial C = Change E = Equilibrium

Calculating Equilibrium Constant

Hydrogen & Iodine react to form hydrogen iodide. There are initially 2.00 moles of both hydrogen and iodine present in a 1.0 L flask. After the reaction has reached equilibrium, there are 1.20 moles of hydrogen remaining. Calculate the Keq for the reaction.

1. First write out the reaction: H2 + I2 2 HI

Reaction [H2] [I2] [2 HI]

Initial

Change

Equilibrium


2. Then fill in the amounts given in molarity.

2 moles H2 / 1 L = 2.0 M H2

2 moles I2 / 1 L = 2.0 M I2

1.2 moles H2 / 1 L = 1.2 M H2

For now… the initial amount of products is always zeroReaction [H2] [I2] [2 HI]

Initial 2.00 2.00 0

Change

Equilibrium 1.20

Calculating Equilibrium constant

3. Figure out the change (subtraction) You are given 1 ending amount,1.20 moles of H2

Change: Initial – Equilibrium = .80 moles of H2

4. Use the coefficients of the equation to figure out the rest of the ‘change’ row. (mole ratio)


Initial 2.00 2.00 0

Change 0.80 0.80 1.60

Equilibrium 1.20

Calculating Equilibrium constant

5. Find the Equilibrium concentration by subtracting on the reactants side But adding on the products side

6. Substitute values into the Keq expression


Initial 2.00 2.00 0

Change 0.80 0.80 1.60

Equilibrium 1.20 1.20 1.60

78.1)20.1)(20.1(

)60.1(

]][[

][ 2

22

2

MM

M

IH

HIKeq

Example 2: When 2.50 moles of nitrogen and 3.40 moles of

hydrogen are mixed in a .500 liter container, they react to form ammonia. If the concentration of nitrogen is found to be 3.50 M when the reaction reaches equilibrium, what is the value of the equilibrium constant?

1. First write out the reaction: N2 + 3 H2 2 NH3

Reaction [N2] [3 H2] [2 NH3]

Initial

Change

Equilibrium


2. Then fill in the amounts given in molarity.

2.5 moles N2 / 0.500 L = 5.00 M N2

3.40 moles H2 / 0.500 L = 6.80 M H2

Initial amount of products is zero 3. Calculate the Change 4. Use mole ratio for rest of the Change row

Reaction [N2] [3H2] [2 NH3]

Initial 5.00 6.80 0

Change 1.50 4.50 3.00

Equilibrium 3.50


5. Find the Equilibrium concentration by subtracting on the reactants side But adding on the products side

6. Substitue values in Keq expression

Reaction [N2] [3H2] [2 NH3]

Initial 5.00 6.80 0

Change 1.50 4.50 3.00

Equilibrium 3.50 2.30 3.00

211.0)30.2)(50.3(

)00.3(

]][[

][3

2

322

23

MM

M

HN

NHKeq

Solubility Equilibrium Special constant for the dissociation of solids that

dissolve in water. Ionic solids dissociate into their cations & anions

NaCl(s) Na+1 (aq) + Cl-1 (aq)

Ions formed carry an electrical current (electrolytes) Solubility

The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve at a specific Temperature

H2O

Solubility All ionic solids are soluble up to a point… so

they may not be completely soluble Soluble: if concentration of solution is at least

0.1 M @ room temperature Slightly Soluble: if concentration of solution is

between 0.0001 M and 0.1 M @ room temp. Insoluble: if concentration of solution is less

than .0001 M @ room temperature

Solubility Product Constant (Ksp)

This is the product of the equilibrium concentrations of ions in a saturated solution of a salt

*There is no denominator in solubility product equilibrium constant.

Ex: AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = [Ag+][Cl-]

H2O

Solubility Product Equilibrium Example: The solubility of AgCl in pure water is 1.3

x10-5 M . Calculate the Ksp using the expression above.

The mole ratio of AgCl to both Ag+ and Cl-1 is 1:1, the solubility of each ions is equal to the solubility of AgCl.

Ksp = (1.3x10-5 ) (1.3x10-5 )

= 1.69 x10-10 (.000000000169) Is AgCl considered soluble, insoluble or slightly

soluble? insoluble

Example Calculate the solubility of CaF2 in grams/L if Ksp = 4.0 x 10-8 . 1. Write Balanced Reaction: CaF2 (s) Ca2+ (aq) + 2 F-1 (aq) 2. Write the solubility expression:

Ksp = [Ca2+][F-1]2

4.0x10-8 = [Ca2+][F-1]2

There are 2 unknowns so we have to write in terms using the mole ratios. For every one mole of Ca2+ there are 2 moles of F-1

Assign a variable x for solubility of Ca2+

H2O

Example [Ca2+] = x and [F-] = 2x Substitute the values into the equilibrium

expression & solve for x 4.0x10-8 = [x] [2x]2

4.0x10-8 = [x] [4x2] 4.0x10-8 = 4x3

1.0x10-8 = x3

√ 1.0x10-8 = √ x3 X = 2.2 x 10-3 M

44

3 3

Example

X = 2.2 x 10-3 M We have assigned x as the solubility

of the Ca2+ ion which is equal to the solubility of the salt, CaF2.

Now convert from mol/L to g/L

2.2x10-3 mol CaF2 L

78.1 g CaF2

1 mole CaF2

= 0.017 g/L of CaF2

E. Factors Affecting Equilibrium Lechâtelier’s Principle

If a stress is applied to a system in dynamic equilibrium, the system will change in a way to relieve the stress

Reaction shifts towards products or reactants in order to relieve stress

Stresses that upset equilibrium include changes in:- Concentration- Temperature- Pressure

E. Factors Affecting Equilibrium

Concentration in reactants causes forward rate to

increase, then increase in reverse rate and equilibrium is re-established, but is shifted in direction of products

H2CO3 (aq) CO2 (aq) + H2O (l)

<1% add H2CO3 >99%

in products has reverse effect in concentration shifts equilibrium

toward that substance

E. Factors Affecting Equilibrium Temperature

shifts away from side that contains the heat

Think of heat as a reactant or product, same as before with addition or removal

Remove heat (cool)

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

‘ Add heat

Temperature

Exothermic Reactions: give off heat Heat in kJ or Joules is a product

Endothermic Reactions: take in heat Heat in kJ or Joules is a reactant

Equilibrium is temperature dependent, Keq can only be affect overall by this change T!

E. Factors Affecting Equilibrium Pressure

Only affects gaseous solution with unequal number of moles of reactants and products

Increase shifts towards side w/ least amount of gas

Decrease shifts towards side w/ most amount of gas

Decrease pressure

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

Increase pressure

Example

N2 (g) + 3 H2 (g) 2NH3 (g) + 92 kJ

•What direction is the shift if you reduce the pressure?

•How would cooling the system shift the reaction?

•How would adding N2(g) shift the system?

•What direction is the shift if you remove NH3 ?

•left

•right

•right

•right

I. Rates of ReactionReaction Energy and Reaction Kinetics

Collision Theory

• Reaction rate depends on the collisions between reacting particles.

• The particles collide and make new substances.

• Successful collisions occur if the particles...

– collide with each other

– have the correct orientation

– have enough kinetic energy to break bonds

Collision Theory

Particle Orientation

Required Orientation

Successful Collision Unsuccessful Collisions

Activation EnergyActivation Energy (Ea)

– minimum energy required for a reaction to occur

– Activated Complex: the transitional structure in a collision that exists while old bonds are breaking and new bonds are being formed.

A. Collision TheoryActivation Energy (Ea)

minimum energy required for a reaction to occur

ActivationEnergy

Activation Energy Cont…

Activation Energy:- depends on reactants- is always positive

- low Ea = fast reaction rate- takes less energy for the reaction to start.

• Reaction Rate: the change in concentration of reactants per unit time as a reaction proceeds.

E

a

Factors Affecting Rxn Rate1. Nature of Reactants

- substances vary greatly in their tendencies to react.- bonds are broken and other bonds are formed in reactions.- the rate of reaction depends on the particular reactants and the bonds involved.

2. Surface Area– high SA = fast rxn rate– more opportunities for collisions– Increase surface area by…

• using smaller particles – if we make the pieces of the reactants smaller, we increase the number of particles on the surface which can react.

• dissolving in water – gases & dissolved particles can mix & collide freely. Reactions happen rapidly.

Factors Affecting Rxn Rate Cont.

3. Concentration

- high conc = fast rxn rate

- more opportunities for collisions because there are more particles in the same volume that can react.

There are less red particles in the same volume so there is less chance of a collision

There are more red particles in the same volume so there is more chance of a collision so the reaction goes faster

Factors Affecting Rxn Rate Cont.4. Temperature

- high temp. = fast rxn. rate

- high KE

- when we increase the temperature, we give the

particles energy

- this makes the particles move faster

- so there are more opportunities for collision

- it is easier to reach activation energy

Factors Affecting Rxn Rate Cont.5. Catalyst

– substance that increases rxn rate without being consumed in the rxn

– lowers the activation energy

II. Energy DiagramsCh. 17 – Reaction Energy and

Reaction Kinetics

Terms and Symbols• Reactants: the chemicals you start with in a

reaction.

• Products: the chemicals formed during the reaction.

• Heat of reaction (E orHrxn): the difference in energy between reagents and products (units are Joules).

• Transition State: the highest point on the energy diagram, representing the point at which the reaction is half-completed.

• Activation Energy (Ea): the amount of energy required for the reaction to take place. The higher the Ea, the slower the reaction (units are Joules).

Energy Diagrams

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Reactants

E

Products

Forward Rxn

(exothermic)

Reverse Rxn

(endothermic)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Activated Complex (transition state)

Ea’

Ea

En

erg

y (E

) in

kJ/

mo

l

Course of Reaction (time)

• Show relationship between time and energy during the course of a chemical reaction.

Endothermic Reaction• A reaction in which heat is absorbed

• Products have higher potential energy than the reactants.

• The pink curve shows the uncatalyzed reaction. The blue curve shows what happens when a catalyst is present.

• The energies and amounts of the products and reactants stays the same, and the E stays the same. The catalyst just allows the reaction to reach equilibrium faster.

Course of Reaction

Ene

rgy

Exothermic Reactions• Reactions in which heat is released.

• Products have lower potential energy than the reactants.• The blue curve shows the uncatalyzed reaction. The red

curve shows what happens when a catalyst is present.• Again, nothing changes but the amount of time it takes for the

reaction to reach equilibrium.• Exothermic rxns are referred to as “spontaneous” because

they can proceed to products without outside intervention.

Course of Reaction

Ene

rgy

Formulas

*Eforward = Eproducts – Ereactants

*Ereverse = Ereactants – E products

* Ea = energy of activated complex – energy of reactants

* Ea’ = energy of activated complex – energy of products

Practice #1• For the energy diagram provided, label the reactants,

products, E, Ea, and Ea’. Also, determine the values of E for the forward and reverse reactions, and the values of Ea and Ea’.

80

60

40

20

0

-20

Forward Reverse

Ene

rgy

(kJ/

mol

)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

EEa

Ea’

reactants

products

Practice #1 Continued*Eforward = Eproducts – Ereactants = 55 kJ/mol – (-20 kJ/mol)

= 75 kJ/mol

* Ereverse = Ereactants – Eproducts = -20 kJ/mol – 55 kJ/mol

= -75 kJ/mol

80

60

40

20

0

-20

Forward Reverse

Ene

rgy

(kJ/

mol

)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

EEa

Ea’

reactants

products

Practice #1 Continued* Ea = energy of activated complex – energy of reactants

= 80 kJ/mol – (-20 kJ/mol) = 100 kJ/mol

* Ea’ = energy of activated complex – energy of products

= 80 kJ/mol – 55 kJ/mol = 25 kJ/mol

80

60

40

20

0

-20

Forward Reverse

Ene

rgy

(kJ/

mol

)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

EEa

Ea’

reactants

products

Practice #2• Draw and label an energy diagram that depicts the

following reaction. Place the reactants at energy level of 0 kJ/mol. Eforward = -10 kJ/mol and Ea’ = 40 kj/mol

Eforward = Eproducts – Ereactants

-10 kJ/mol = Eproducts – 0 kJ/mol = -10 kJ/mol

Ea’ = activated complex – Eproducts

40 kJ/mol = a.c – (-10 kJ/mol) = 20 kJ/mol 40

30

20

10

0

-10

Ene

rgy

(kJ/

mol

)

Course of Reaction

reactants

products

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - E

Ea Ea’

40

30

20

10

0

-10

40

30

20

10

0

-10

40

30

20

10

0

-10

Ene

rgy

(kJ/

mol

)

40

30

20

10

0

-10

IIIIIIIVV Ch. 18 – Reversible Reactions & Equilibrium pp. 549 – 559.

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Transcript of IIIIIIIVV Ch. 18 – Reversible Reactions & Equilibrium pp. 549 – 559.