Chapter 19 Notes.pptx

BELLWORK

TUESDAY, FEBRUARY 27

TH

Make of list of three chemical changes

that occur slowly and three that occur

quickly.

Chemistry:

Chapter 19

19.1 Rates of Reaction

 Rate – a measure of the speed of any change

that occurs within an interval of time.

 Reaction Rate - change in concentration of a

reactant or product per unit time

 Some reactions go very fast (burning) and

Question:



In your own words, explain what

rate is.



Teacher’s definition: how fast

Collision Theory:



In order to react, two or more particles must

collide with sufficient energy (

called the

activation energy

) and with the proper

molecular orientation. If the colliding

Question:



In your own words, explain

what activation energy is.

 Teacher’s definition: the energy needed

Using the graph below, what is the

energy of each of the following:

 reactants? _______________________  products? _______________________

 activation energy? _______________________  activated complex? _______________________  ΔH? _______________________

80 kJ 20 kJ 100 - 80

100 kJ 20 – 80

= 20 kJ

= - 60 kJ



Activated complex

- (also called

Factors that affect reaction

rate:



Temperature

- Reactions go faster at

higher temperatures. Particles have

more kinetic energy. More colliding

particles have enough energy to

Factors that affect reaction

rate:



Concentration

- Increasing the

concentration of reactants usually

increases the reaction rate. If there are

more particles to collide, there should be

a greater number of collisions that

Factors that affect reaction

rate:



Particle size

-

Increased surface

area increases

reaction rate.

Again, more

Factors that affect reaction

rate:



Catalysts

- A catalyst is a substance

Question:

Charlie is dissolving sugar in tea. Circle

the actions that will cause the sugar to

dissolve more quickly.

-powder the sugar

-chill the tea

Enzyme

Inhibitor



substance that blocks the

Rate Law



equation that is written that

expresses how the reaction rate of a

particular reaction is dependent

upon the concentrations of its

reactants.



For the reaction aA + bB  cC + dD,

the general form of the rate law

would be:



Rate is usually expressed as mol/L∙time.



k is the specific rate constant. It is

constant for a given reaction at a given

temperature. The faster a reaction, the

larger the k value.



[A] and [B] represent the concentrations of

reactants A and B in moles per liter (M).



x and y are the order of the reactant. They

can only be determined by analyzing

experimental data. These exponents are

usually positive integers.

Question:



What happens when

Determining the order of a reactant

(We must use experimental data for this!):



If doubling the initial [ ] of a reactant causes the initial rate to:



double, the reaction is first order in that reactant.



quadruple, the reaction is second order in that reactant.



increase 8 times, the reaction is third order in that reactant.



If doubling the initial [ ] of a reactant does not change the initial rate, the reaction is

Rate = k[A][B]

2



The exponents determine the

order of the reactants.



The sum of the exponents is the

order

of the reaction

.



Rate = k[A][B]2 is first order in A, second order

Question:



How do you determine the

EXAMPLES

2x = 4 [A]

A  B

Trial [A] Rate

1 0.05 3 x 10-4

2 0.10 1.2 x 10

-3

3 0.20 4.8 x 10

-3

Rate Law:

rate = k[A]

EXAMPLES

2x = 4 [A]

2x = 2 [B]

2A + B 2C

Trial [A] [B] Rate

1 0.1 0.2 0.10

2 0.1 0.4 0.20

3 0.2 0.4 0.80

EXAMPLES

2x = 1 [A]

2x = 2 [B]

[B]

1

rate = k

A + B  C

Trial [A] [B] Rate

1 0.10 0.20 1.0 x 10-5

2 0.20 0.20 1.0 x 10-5

3 0.20 0.40 2.0 x 10-5

EXAMPLES

2x = 2 [NH₄+ ]

2x = 2

[NO₂-]

NH₄+ + NO

2-  N2 + 2H2O

Trial [NH₄+] [NO

2-] Rate(mol/

Lmin)

1 0.100 0.005 1.35 x 10-7

2 0.100 0.010 2.70 x 10-7

3 0.200 0.010 5.40 x 10-7

Rate Law: Value of k:

rate = k [NH

₄+

] [NO

2-

]

1 1

rate = k

[NH4+ ] [NO1 2-]1

Pick a trial… plug in values… solve for k

1.35 x 10

-7

= k

[0.100 ][0.005]

1 1

Bellwork

Wednesday, March 9th

The following data were obtained at 55o_C.

(CH₃)₃CBr + OH-  _(CH

3)3COH + Br

- Exp. [(CH

3)3CBr] [OH-] Initial rate (mol/L s)

  1 0.10 0.10 1.0 x 10-3  2 0.20 0.10 2.0 x 10-3  3 0.10 0.20 1.0 x 10-3

 

 Write the rate law for the reaction.

Reaction Mechanisms

 A reaction mechanism is a series of steps

(reactions) that make up the overall reaction. Most reactions do not occur in a single step

but are the sum of multiple steps.

 Rate-limiting or rate-determining step

-slowest step in a reaction. This determines how fast the reaction can go.

 Mechanisms are usually not completely known

because of the short life of their “temporary products”. Possible mechanisms must:

 Agree with the rate law

Question:

 A group of 4 friends are getting ready for a

party:

 Person 1 takes 15 minutes  Person 2 takes 30 minutes  Person 3 takes 55 minutes

 Person 4 takes 1 hour and 30 minutes

 How long will it take them to be ready to go?

19.2 EQUILIBRIUM

 Chemical equilibrium- state where

concentrations of products and reactants remain constant

 equilibrium is dynamic

 any chemical reaction in a closed vessel will

reach equilibrium

 at equilibrium…

jA + kB



lC + mD

[ ] = concentration = mol/L = Molarity

K

_c

= K = K

_eq

= equilibrium constant

(used interchangeably)

K

_eq

=

[Products]

[Reactants]

j k

EXAMPLE

•

Write the K

_eq

expression for the following

reaction:

4 NH

₃

(g) + 7 O

₂

(g)



4 NO

₂

(g) + 6 H

₂

O (g)

K

_eq

=

[Products]

[Reactants]

=

[NO

₂

] [H

₂

O]

[NH

₃

] [O

₂

]

4 6

If we know equilibrium

concentrations, we can calculate

the equilibrium constant, K

_eq

.



K

_eq

changes with temperature (not

with concentration or pressure).



Concentrations of

pure solids and

pure liquids are not included

in

EXAMPLE

 Write the K_eq expression for the following

reaction:

CaCO₃(s)  CaO (s) + CO₂(g)

K

_eq

= [CO

₂

]

Why are CaCO

₃

& CaO left out?

Because



A value for K

_eq

greater than one

means

that the equilibrium is far to the right

(

mostly products

).

Large Numerator



A value for K

_eq

less than one

means

that the equilibrium is far to the left

(

mostly reactants

).

Large

Denominator

Summary



Large K > 1 products are "favored"



K = 1 neither reactants nor products

are favored

EXAMPLES

• _{Analysis of an equilibrium mixture of nitrogen, hydrogen}

and ammonia contained in a 1 L flask at 300º_{C gives the}

following results: hydrogen: 0.15 mol, nitrogen: 0.25 mol, ammonia: 0.10 mol. Calculate K_eq.

3H

₂

(g) + N

₂

(g)



2NH

₃

(g)

K

_eq

=

[H

₂

]

[N

₂

]

[NH

₃

]

2

1 3

[0.15] [0.25]

[0.10]

2 1 3

=

Calculate the value of K_eq for the following reaction at equilibrium.

HC₂H₃O₂ (aq) + H₂O (l)  C₂H₃O₂- _{(aq) + H}

3O+ (aq)

• _{An analysis of the equilibrium mixture gives the following}

results:

[HC₂H₃O₂] = 0.25 M, [C₂H₃O₂-_{] = 0.075 M, and}

[H₃O+_{] = 6.0 X 10}-5 _M

K_eq=

[HC₂H₃O₂]

[H₃O+]

[C₂H₃O₂-]

=

[0.25]

[6.0 X 10-5 ]

[0.075]

K_eq= 1.8 x 10-5 More products or reactants

Bellwork

Thursday, March 9

th

This equation is at equilibrium:

•

CO

_(g)

+ H

₂

O

_(g)

↔ CO

_{2 (g)}

+ H

_{2 (g)}

– If a 1.00L vessel has 2.50 mol CO₂, 2.80 mol H₂O, 5.00 mol CO and 1.22 mol H₂ gas, what is K_eq?

LeChatelier's Principle



When a stress is

applied to a

system, the

equilibrium will

shift in the

direction that

will relieve the

stress.

Changes in

concentration

 An increase in concentration of:

 a reactant will cause equilibrium to shift to

the right to form more products.

 a product will cause equilibrium to shift to

the left to form more reactants.

 A decrease in concentration of:

 a product will cause equilibrium to shift to

the right to form more products.

 a reactant will cause equilibrium to shift to

Question:

 Summarize what happens when you

A + B



C + D

 Remove A or B  Add C or D

Example:

As₄O₆ + 6C  As₄ + 6CO

• _{addition of CO} • _{addition of C}

• addition of As₄O₆

Changes in

temperature

 Treat energy as a product or reactant and

temperature changes work just like changes in concentration!

 An increase in temperature of an exothermic

reaction (H is negative) will cause equilibrium to shift to the left. A decrease in temperature of an exothermic reaction will cause equilibrium to shift to the right.

 An increase in temperature of an endothermic

reaction (H is positive) will cause equilibrium to shift to the right. A decrease in temperature of an endothermic reaction will cause

Question:

 Summarize what happens when you

Example:

N₂ + O₂  2NO

H = 181 kJ (endothermic)

 addition of heat

 lower temperature

Example:

2SO₂ + O₂  2SO₃

H= -198 kJ (exothermic)

 increase temperature  remove heat

Example:

CaCO₃ + 556 kJ  CaO + CO₂

Changes in pressure

 Changes in pressure only affect

equilibrium systems having gaseous products and/or reactants.

 Increasing the pressure of a gaseous system

will cause equilibrium to shift to the side with fewer gas particles.

 Decreasing the pressure of a gaseous



If the system has the same number

of moles of gas on each side,

changes in pressure do not affect

equilibrium.



The addition of

an inert

(nonreactive) gas does not affect

the equilibrium

system since the

partial pressures of the gases

Question:



Summarize what happens

Example:

P₄(s) + 6Cl₂(g)  4PCl₃(l)

 increase container volume

 Shifts to side with more gas

 decrease container volume

 Shifts to side with less gass

 add argon gas

Example:

PCl₃(g) + Cl₂(g)  PCl₅(g)

 a. decrease container volume  b. add helium gas

N/C

PCl₃(g) + 3NH₃(g)  P(NH₂)₃(g) + 3HCl(g)

 increase container volume

Addition of a catalyst



Adding a catalyst does not

affect equilibrium. Catalysts

speed up the forward and

Question:



Explain what a catalyst is in

Example:

CO(g) + H₂O(g)  H₂(g) + CO₂(g)

H = -41kJ

Determine in which direction the reaction will shift when each of these stresses are added.

 temperature is increased

 helium gas is added

 water vapor is added

 gaseous carbon dioxide is removed

 a catalyst is added

+ 41kJ

N/C

Example:

2SO₃(g)  2SO₂(g) + O₂(g)

H = 197 kJ

Determine in which direction the reaction will

shift when each of these stresses are added.

 oxygen gas is added

 pressure is incr. by decr. volume  argon gas is added

 temperature is decreased

 gaseous sulfur dioxide is removed

197kJ +

Example:

Consider the reaction:

2NO₂(g)  N₂(g) + 2O₂(g) which is exothermic

• NO₂ is added • N

2 is removed

• The volume is halved • He (g) is added

• The temperature is increased • A catalyst is added

+ HEAT

N/C

BELLWORK

19.3

Entropy, Enthalpy & Gibbs Free Energy



Spontaneous process

-occurs without outside

intervention

Question:

Entropy, S - a measure of randomness or disorder

 associated with probability (There are more

ways for something to be disorganized than organized.)

 Entropy increases going from a solid to a liquid

to a gas.

 Entropy increases when solutions are formed.  Entropy increases in a reaction when more

atoms or molecules are formed.

 The entropy of a substance increases with

2nd law of

thermodynamics

 In any spontaneous process there is always

an increase in the entropy of the universe.

 Remember that we calculate the change in

enthalpy (heat) by subtracting the reactants’ value from the value of the products.

 They are given in kilojoules/mol. Remember

that the enthalpy of a free element in its

standard state is zero. (Enthalpy values are found on page 316.)

 Ho = H

Question:



What is the difference

EXAMPLE

Calculate the change in enthalpy for the reaction:

2H₂(g) + O₂(g)  2H₂O(l)

Ho = H products - H reactants

[ 2 ( -285.8 ) ] - [ 2 ( 0 ) + ( 0 ) ]

Ho =

Gibbs free energy, G



energy available to do work





G

o

= standard free energy

change



change in free energy that

occurs if the reactants in their

standard states are converted to

products in their standard states



G

o

=



H

o

-T



S

o



A

spontaneous reaction has a

negative



G

. For example,

when ice melts



H is positive

(endothermic),



S is positive and



G = 0 at 0

o

C.

_{Entropy, ΔS}

If...

_{Enthalpy, ΔH Spontaneity}

Positive Positive Yes at high temp

Negative Positive Never spontaneous Positive Negative Always spontaneous

Question:



What happens when



S is