Chapter 14 & 16 Chemical Equilibrium and reaction rates

16.1 Rates of Rxn: Collision Theory

• Amount of time for a reaction to take place can vary greatly

- the measurement of this time is in the form of a rate

-measurement of the speed of any change that occurs within an interval of time.

• ex- iron rusting may be expressed as 0.50 mol/year

Reaction RateReaction Rate

The change in concentration of a reactant or product per unit of time

2 1

2 1

[ ] [ ]A at timet A at timetRate

t t

[ ]ARate

t

Collision Theory

• Reaction rate depends on the collisions between reacting particles.

• Successful collisions occur if the particles...

– collide with each other

– have the correct orientation

– have enough kinetic energy to break bondsCO + NO2 CO2 + NO

Ineffective collision

Effective collisionN O

O

OC

NO

O

CO

Collision Theory

• Collision Theory-atoms, ions and molecules can react to form products when they collide, provided they have enough kinetic energy.

• This minimum amount of KE in order to react is called the activation energy (Ea)

Collision Theory• Activation Energy

– depends on reactants

– low Ea = fast rxn rate

Ea

Endothermic ReactionsEndothermic Reactions

Exothermic ReactionsExothermic Reactions

Factors Affecting Rxn Rate

5 mph “fender bender”

50 mph “high-speed crash”

1. Temperature- • usually increasing temperature increases the reaction rate. • This increases the # of particles having enough KE

Factors Affecting Rxn Rate2. Concentration

– increased concentration = increased rxn rate

– more opportunities for collisions (collision frequency increases)

Factors Affecting Rxn Rate

3. Particle Size– smaller particle size increases the rate.– This increases the amount of total surface

area exposed therefore increasing the collision frequency

Factors Affecting Rxn Rate

4. Catalyst*– substance that increases rxn rate without

being consumed in the rxn– lowers the activation energy

*inhibitor-*inhibitor- a substance that a substance that interferes with the catalyst. This interferes with the catalyst. This can slow down the rate or even can slow down the rate or even stop it completely. stop it completely.

CatalysisCatalysis•Catalyst: A substance that speeds up a reaction without being consumed

•Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

•Homogeneous catalyst: Present in the same phase as the reacting molecules.

•Heterogeneous catalyst: Present in a different phase than the reacting molecules.

Exothermic Reaction with a Exothermic Reaction with a CatalystCatalyst

Endothermic Reaction withEndothermic Reaction witha Catalysta Catalyst

• A catalyst can speed up the rate of a given chemical reaction by

•A increasing the equilibrium constant in favor of products.

•B lowering the activation energy required for the reaction to occur.

•C raising the temperature at which the reaction occurs.

•D increasing the pressure of reactants, thus favoring products.

Which reaction diagram shows the effect of using the appropriate

catalyst in a chemical reaction?

• H2O2 hydrogen peroxide, naturally breaks down into H2O and O2 over time. MnO2, manganese dioxide, can be used to lower the energy of activation needed for this reaction to take place and, thus, increase the rate of reaction. What type of substance is MnO2?

A a catalystB an enhancerC an inhibitorD a reactant

14.1 Chemical Equilibrium14.1 Chemical EquilibriumReversible Reactions:

A chemical reaction in which the products

can react to re-form the reactantsChemical Equilibrium:

• When the rate of the forward reactionequals the rate of the reverse reaction• the concentration of products andreactants remains unchanged

2HgO(s) 2Hg(l) + O2(g)

Arrows going both directions ( ) indicates equilibrium in a chemical equation

Rate Comparison for H2(g) + I2(g) 2HI(g)Equal amounts of H2 and I2, [HI] goes up.

Decline in the rate of the forward system, most reaction has been carried out

The greater number of HI, reverse reaction starts to occur and reverse reaction increases in rate.

At some point, rate of forward and reverse becomes equal

Chemical Equilibria are Dynamic

•In some cases, the equilibrium has a higher concentration of products than reactants.

•This type of equilibrium is also shown by using two arrows.reac tants products

•The forward reaction has a longer arrow to show that the products are favored.

14.2 Systems at Equilibrium• Equilibrium Constants (Keq) is the ratio of product concentration to

reactant concentrationaA + bB cC + dD

Keq = [C]c [D]d K eq = [product]

[A]a [B]b [reactant]

– Keq < 1, reactants favored at equal.– Keq > 1, products favored at equal. – Does not include pure solids and liquids

Write the equilibrium expressions, Kc, for the following reaction:

2 NO2(g) 2 NO(g) + O2(g)

Keq ; Equilibrium constant (No Units)

Kc (concentration); Used for solutions (aq) or gases.

Kp (Pressure); Used for gases only.

Other Equilibrium Expressions

22

22

C ][NO

][O[NO] K

Write a equilibrium expression for PCl5(g) PCl3(g) + Cl2(g)

][PCl

]][Cl[PCl K

5

23eq

For the above reaction, the equilibrium constant is 35. If the concentrations of PCl5 and PCl3 are 0.015 M and 0.78 M respectively. What is the concentration of Cl2?

][PCl

]][Cl[PCl K

5

23eq

M 0.015

]M)[Cl (0.78 53 2 [Cl2] = 0.67 M

An aqueous solution of carbonic acid reacts to reach equilibrium as described below.

–2 3 2 3 3H CO ( ) H O( ) HCO ( ) H O ( ) aq l aq aq

The solution contains the following solution concentrations: carbonic acid, 3.3 × 10−2 M; bicarbonate ion, 1.19 × 10−4 Mhydronium ion, 1.19 × 10−4M. Determine the Keq.

eqK–3 3

2 3

[HCO ][H O ]

[H CO ]

eqK – -4 -43 3

-22 3

-7[HCO ][H O ] (1.19 10 ) (1.19 10 )

[H CO ] (3.3 10 )4.3 10

Which is more favorable? Forward or backward? Backward, LOW Keq

For the system involving N2O4 and NO2 at

equilibrium at a temperature of 100C, the product

concentration of N2O4 is 4.0x10-2 M and the

reactant concentration of NO2 is 1.4x10-1 M.

What is the Keq value for this reaction?2 NO2 N2O4

Keq= 2.0

Calculating equilibrium concentrations from initial concentrations:Steps in calculating equilibrium concentrations:(1) Write equilibrium expression(2) Calculate change in concentration of each substance(3) Calculate equilibrium concentrations

Given the following reaction: A B + C. The initialconcentration of A is 0.260 M and at equilibrium, [C] is 0.160 M.Calculate the Keq.

[A]

[B][C] Keq

M 0.100

M) (0.160

2

Initial (M)

Change (M)

Equilibrium (M)

A B + CKeq = ?

= 0.256

0.260 0 0

x + x + x 0.160 + 0.160 +

0.1600.1600.1600.100

Given the following reaction: H2(g) + I2(g) ↔ 2 HI(g).

At 25°C, 0.100 mol of H2 and 0.100 mol of I2 react in a 1.00

L container. At equilibrium, the [H2] is 0.080 M. Calculate

Keq at this temperature.

]][I[H

[HI] Keq

22

2

2

2

M) (0.080

M) (0.040

Initial (M)

Change (M)

Equilibrium (M)

0.100 0.100 0

x x + 2x

0.020 0.020 + 0.040

0.0400.0800.080

H2 + I2 2 HIM = mol / L

= 0.25

Keq for the equilibrium below is 1.8 × 10−5 at a temperature of 25°C. Calculate when [NH3] = 6.82 × 10−3.

aq l aq aq–3 2 4NH ( ) H O( ) NH ( ) OH ( )

Initial (M)

Change (M)

Equilibrium (M)

6.82 x 10 -3 0 0

x + x + x

xx6.82x10-3 - x

Keq = ? NH3 NH4+ + OH-

eq

xK

– 2–5 4

-33

[NH ][OH ]1.8 10

[NH ] 6.82 10

Disregard, to make the calculation easier!

x2 = (1.8 10−5) (6.82 10−3) = 1.2 × 10−7

- -474[NH ] 1.2 10 3.5 10

14.3 Le Chatelier’s Principle

When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress.

When you take something away from a When you take something away from a system at equilibrium, the system system at equilibrium, the system shiftsshifts in such a way as to in such a way as to replace replace what you’ve taken awaywhat you’ve taken away..

Le Chatelier Translated:

if a stress is applied to a system in a dynamic equilibrium, the system changes to relieve the stress

Factors Affecting Equilibrium: Le Chatelier’s Principle

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

– Concentration:• Increasing O2, will cause the shift (away from

O2)

– Temperature:• Increasing T will cause the shift (away from

heat)

– Pressure: • Increase in P will shift (favors side with less

gas molecules)

LeChatelier Example #1LeChatelier Example #1A closed container of ice and water at equilibrium. The temperature is raised.

Ice + Energy (ΔH) Water

The equilibrium of the system shifts to the _______ to use up the added energy.

right

LeChatelier Example #2LeChatelier Example #2A closed container of N2O4 and NO2 at equilibrium. NO2 is added to the container.

N2O4 (g) + Energy (ΔH) 2 NO2 (g)

The equilibrium of the system shifts to the _______ to use up the added NO2.

left

LeChatelier Example #3LeChatelier Example #3A closed container of water and its vapor at equilibrium. Vapor is removed from the system.

water + Energy (ΔH) vapor

The equilibrium of the system shifts to the _______ to replace the vapor.

right

LeChatelier Example #4LeChatelier Example #4A closed container of N2O4 and NO2 at equilibrium. The pressure is increased.

N2O4 (g) + Energy (ΔH) 2 NO2 (g)

The equilibrium of the system shifts to the _______ to lower the pressure, because there are fewer moles of gas on that side of the equation.

left

6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H = 2820 kJ.

• Some CO2(g) is added and causes . . .

• Shift to right so get more C6H12O6(s) .

• Temperature is raised . . .• Shift to right, more C6H12O6(s) forms.

• Volume is decreased• There are equal moles of gaseous

reactants and products, so no change.

6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H = 2820 kJ

• Some O2(g) is removed . . .• Shift to right, more C6H12O6(s) forms.• Some C6H12O6(s) is removed . . .• No change (glucose is a solid so is not in the

equilibrium expression). No more C6H12O6(s) .• A catalyst is added . . .• No effect on equilibrium, only allows it be attained

more quickly. No more C6H12O6(s) .• Some H2O is removed . . .• No change, water is a liquid. No additional C6H12O6(s)

forms.

• When a reaction is at equilibrium and more reactant is added, which of the following changes is the immediate result? – A The reverse reaction rate remains the

same.– B The forward reaction rate increases.– C The reverse reaction rate decreases.– D The forward reaction rate remains the

same.

• In which of the following reactions involving gases would the forward reaction be favored by an increase in pressure?– A A + B AB– B A + B C + D– C 2A + B C + 2D– D AC A + C

4HCl(g) + O 2(g) 2H2O(l)+ 2Cl2(g) + 113kJ

• Which action will drive the reaction to the right? – A heating the equilibrium mixture – B adding water to the system – C decreasing the oxygen concentration – D increasing the system’s pressure

NO2(g) + CO(g) NO(g) + CO2(g)

• The reaction shown above occurs inside a closed flask. What action will shift the reaction to the left? – A pumping CO gas into the closed flask – B raising the total pressure inside the flask – C increasing the NO concentration in the

flask

16.2 How can rate reaction be explained?

•The rate law describes the way in which reactant concentration affects reaction rate.–A rate law is the expression that shows how the rate of formation of product depends on the concentration. (all other than solvent)

Rate law equation

Consider the equationA + B C + D

rate = k [reactant]x

x = Order of reaction

• can not be determined from the equation.

•MUST be calculated

Rate of reaction = k [A]x[B]y

k = rate constant x = order of reaction with respect to Ay = order of reaction with respect to Bx + y = overall order of reaction

To calculate the rate of reaction,

• Ratio of rates a two different concentrations of A

X

1

2

1

2

A

A

rate

rate

Determine the reaction order and write the rate law equation 2HI(g) → H2(g) + I2(g)

X

1

2

1

2

A

A

rate

rate

X

015.0

030.0

M/s 1.1x10^-3

M/s 4.4x10^-3

M

M

4 = 2x

x = 2

Rate of reaction = k [HI]x

Rate = k [HI]2

The initial rate of decomposition of acetaldehyde, CH3CHO, CH3CHO --> CH4 + CO at 600C was

measured at a series of concentrations with the followingresults:

a) Determine the reaction order.

(b) Write the rate law.

Experiments 1 2 3 4

[CH3CHO] (M) 0.10 0.20 0.30 0.40

rate (mol/L s) 0.085 0.34 0.76 1.4

X

1

2

1

2

A

A

rate

rate

X

10.0

20.0

M/s 0.085

M/s 0.34

M

M 4 = 2x

x = 2

Rate of reaction = k [CH3CHO]x

Rate = k [CH3CHO]2

(The power of the [ ] of a reactant)Choose any two experiments