Reaction Rates and Chemical Equilibrium

Activation Energy is being supplied

Activated Complex

Collision Theory Reactions can occur:

Very fast – such as a firecracker Very slow – such as the time it took for dead plants

to make coal Moderately – such as food spoilage

A “rate” is a measure of the speed of any change that occurs within an interval of time

In chemistry, reaction rate is expressed as the amount of reactant changing per unit time. Example: 3 moles/year, or 5 grams/second

I wonder what happens if I mix

these two solutions…

WOW, that was

really FAST

It was also really FUN

I wonder if I should be wearing my

goggles?

I. Reaction RatesA. Collision Model: molecules, ions or atoms

must touch (or collide) to react• However, only a small fraction of collisions produces

a reaction. Why?

1. Particles lacking the necessary KE to react will bounce apart unchanged

• Negative/positive ions repeling each other, e- clouds repel each other strongly at short

distances

• Ions/molecules with very high KE can overcome repulsive forces to get close enough to react

• particles must hit in the correct orientation of e- clouds

2. Collisions must have enough energy to produce the reaction: must equal or exceed the

“activation energy” - the minimum energy needed to react.

Will a AA battery start a car?• Think of clay clumps thrown together gently – they don’t

stick, but if thrown together forcefully, they stick tightly to each other.

3. An “activated complex” is an unstable arrangement of atoms that forms momentarily (typically about 10-13 seconds) at the peak of the activation-energy barrier This is sometimes called the transition state

a. Results in either a) forming products, or b) reformation of reactants

Both outcomes are equally likely

- Page 543

a. Reactants b. Absorbed c. No; it could also revert back to the reactants

The collision theory explains why some naturally occurring reactions are very slow Carbon and oxygen react when charcoal burns, but this

has a very high activation energy (C + OC + O2(g)2(g) →→ CO CO2(g)2(g) + 393.5 + 393.5

kJ)kJ) At room temperature, the collisions between carbon and

oxygen are not enough to cause a reaction

Mechanisms and Rates Most reactions are several elementary steps Reaction progress curve for complex

reactions is several peaks and valleys Peaks are specific step activation energy and

valleys are energy of intermediate products

There is an activation energy for each elementary step.

Slowest step (rate determining) must have the highest activation energy.

+

Ea

First step is fastFirst step is fast

Low activation energyLow activation energy

Second step is slowHigh activation energy

+

Ea

+

Ea

Third step is fastLow activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Activated Complexes or Transition StatesTransition States

Reaction between ethene, CH2=CH2, and ethene, CH2=CH2, and

hydrogen chloride, HClhydrogen chloride, HCl Produces chloroethaneProduces chloroethane

After two molecules collideAfter two molecules collide A double bond between the two A double bond between the two

C is converted into a single bond C is converted into a single bond A H atom gets attached to one A H atom gets attached to one

of the Cof the C and a Cl atom to the otherand a Cl atom to the other The The reaction can only happenreaction can only happen if the hydrogen end if the hydrogen end

of the H-Cl bond approaches the C-C double bondof the H-Cl bond approaches the C-C double bond

(a) Two BrNO molecules approach each other at high speeds.(b) The collision occurs.(c) The energy of the collision causes Br-N bonds to break and Br-Br bonds to form.(d) The products: one Br2 and two NO molecules.

What happens in the collision What happens in the collision process?process?

Reaction progress of two Reaction progress of two BrNO molecules – terms of BrNO molecules – terms of

energyenergy

B. Factors Affecting RateB. Factors Affecting Rate1. Temperature

• Higher temp. faster particlesHigher temp. faster particles• More and harder collisionsMore and harder collisions

2. 2. Surface Area Surface Area (Particle size)• Smaller particles bigger surface areaSmaller particles bigger surface area

3. Concentration3. Concentration• Increasing conc. USUALLY increases rate Increasing conc. USUALLY increases rate • More concentrated closer together the More concentrated closer together the

particlesparticles• Collide more often Collide more often Faster reaction Faster reaction

4. Dissolving speeds up reactions.• Getting two solids to react w/each other is

slow.

5. Catalysts- substances that speed up a reaction without being used up

(enzyme)• Speeds up rxn by giving the rxn a new path• new path has a lower activation energy• More molecules have this energy• The reaction goes faster only aids reactions

that are thermodynamically possible

a. Inhibitor- a substance that blocks a catalyst; rxns slows or even stops

Endothermic Reaction with a Catalyst

Exothermic Reaction with a Catalyst

C. C. Heterogeneous ReactionsHeterogeneous Reactions: : reactants in reactants in different phasesdifferent phases

1. Homogeneous Reactions – only one 1. Homogeneous Reactions – only one phase phase for reactants & productsfor reactants & products

Heterogeneous Heterogeneous CatalysisCatalysis Example: the reaction between ethene and

hydrogen in the presence of a nickel catalyst.

1. Adsorption and activation of the reactants.

2. Migration of the adsorbed reactants on the surface.

Heterogeneous Heterogeneous CatalysisCatalysis

Example: the reaction between ethene and hydrogen in the presence of a nickel catalyst.

3. Reaction of the adsorbed substances. 4. Desorption (break away) of the products

II. Chemical Equilibrium – a dynamic state where the concentrations of all reactants &

products remain constant Recall: Equilibrium state & vapor pressure,

rate of evaporation = rate of condensation

A. Rxn. rates – forward rate = reverse rate NOT a condition in which products & reactants are in equal

amts NOT a static condition

Reversible Reactions Some reactions do not go to completion as

we have assumed They may be reversible – a reaction in which the

conversion of reactants to products and the conversion of products to reactants occur simultaneously

Forward: 2SO2(g) + O2(g) → 2SO3(g)

Reverse: 2SO2(g) + O2(g) ← 2SO3(g)

Reversible Reactions The two equations can be combined into one,

by using a double arrow, which tells us that it is a reversible reaction:

2SO2(g) + O2(g) ↔ 2SO3(g)

A chemical equilibrium occurs, and no net change occurs in the actual amounts of the components of the system.

Reversible Reactions Even though the rates of the forward and

reverse are equal, the concentrations of components on both sides may not be equal An equlibrium position may be shown:

A B or A B 1% 99% 99% 1%

Note the emphasis of the arrows direction It depends on which side is favored; almost all

reactions are reversible to some extent

B. Equilibrium Constant: Keq• Chemists generally express the position of equilibrium in

terms of numerical values, not just percent

1. Relate values to the amounts (Molarity) of reactants & products at equilibrium

Equilibrium Constants• Consider this reaction:

• The capital letters are the chemical, and the lower case letters are the balancing coefficient:

aA + bB cC + dD The equilibrium constant (Keq) is the ratio of

product concentration to the reactant concentration at equilibrium, with each concentration raised to a power (which is the balancing coefficient).

aA + bB cC + dD

2. The “equilibrium constant expression” has this general form:

[C]c x [D]d

[A]a x [B]b

(brackets: [ ] = molarity concentration)

KKeqeq ==

Note that Keq has no units on the answer; it is only a number because it is a ratio

3. Equilibrium constants tells us whether products or reactants are favored:

if Keq > 1, products favored at

equilibrium

if Keq < 1, reactants favored

Write the equilibrium constant for the Write the equilibrium constant for the following problems:following problems:

1. CH3OH (g) CH2O (g) + H2 (g)

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

ANSWERS:ANSWERS:

1. [CH2O] x [H2 ]

[CH3OH]

2. [HI]2

[H2] x [I2] 

Keq =

Keq =

- Page 557

Does this favor the reactants or products?

Calculating Equilibrium Calculate the equilibrium constant for the

following reaction.

3H2(g) + N2(g) 2NH3(g) if at 25ºC there 0.15 mol of N2, 0.25 mol of NH3, and 0.10 mol of H2 in a 1.0 L container.

Keq = [NH3]2

[H2]3 [N2]

[0.25 mol/L NH3]2

[0.10 mol/L H2]3 [0.15 mol/L N2]

== = = 416.67 L2/mol2

= 420 = 420 LL22/mol/mol22

For the following reaction, calculate the For the following reaction, calculate the value of the equilibrium constant:value of the equilibrium constant:

2NB3 (g) N2 (g) + 3Br2 (g)

[NBr3] = 2.07 x 10-3 M [N2] = 4.11 x 10-2 M

[Br2] = 1.06 x 10-3 M

ANSWER:ANSWER:

[N2] x [Br2]3

[NBr3]2

Keq = = 1.14 x 10-5

III. Applications

A. Le Chatelier’s Principle – if stress is applied to a system in equilibrium, the system changes in a way that relieves the stress (p. 598) The French chemist Henri Le Chatelier (1850-1936)

studied how the equilibrium position shifts as a result of changing conditions

What items did he consider to be stress on the equilibrium?

1) Concentration

2) Temperature

3) Pressure

Each of these will be discussed in detail

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 what replace what you’ve takenyou’ve taken awayaway..

Le Chatelier Le Chatelier Translated:Translated:

• When you add something to a system When you add something to a system at equilibrium, the system at equilibrium, the system shiftsshifts in in such a way as tosuch a way as to use up what you’ve use up what you’ve addedadded..

1. Concentration – adding more reactant produces more product, &

removing the product as it forms will produce more product

Changing Changing ConcentrationConcentration

If you add reactants (or increase their concentration), the forward reaction will speed up.

More product will form. Equilibrium “Shifts to the right”

If you add products (or increase their concentration), the reverse reaction will speed up.

More reactant will form. Equilibrium “Shifts to the left”

a. When a reactant or product is added to a system at equilibrium, the system shifts away from the added component

Changing Changing ConcentrationConcentration

If you remove reactants (or decrease their concentration), the reverse reaction will speed up

More reactant will form. Equilibrium “Shifts to the left”.

If you remove products (or decrease their concentration), the forward reaction will speed up

More product will form. Equilibrium “Shifts to the right”

b. When a reactant or product is removed from a system at equilibrium, the system shifts towards the removed component

LeChatelier Example #1LeChatelier Example #1A closed container of NA closed container of N22OO44 and NO and NO22 at at equilibrium. equilibrium. NONO22 is added is added to the to the container.container.

NN22OO4 4 (g)(g) + Energy + Energy 2 NO 2 NO22 (g)(g)

The equilibrium of the system The equilibrium of the system shifts to the _______ to use up the shifts to the _______ to use up the added NOadded NO22..

leftleft

2. Temperature – increasing the temp. causes the equilibrium position to shift in the

direction that absorbs heat• If heat is one of the products (just like a chemical), it is

part of the equilibrium• so cooling an exothermic reaction will produce more

product, and heating it would shift the reaction to the reactant side of the equilibrium:

C + OC + O2(g)2(g) →→ CO CO2(g)2(g) + 393.5 kJ + 393.5 kJ

Changing Temperature – Page 600

As you raise the temperature the reaction proceeds in the endothermic direction. (Adding energy shifts the equilibrium so energy is absorbed.)

Input heat, the reaction needs to cool itself down again. Too cool down it needs to absorb the extra heat; the back

reaction (endothermic) absorbs heat! The position of equilibrium moves to the left and the new

equilibrium mixture contains more A and B (less C and D)

a. Use heat as reactant or product to interpret shift of position

i. Increase temp. shifts away from energy

ii. Decrease temp. shifts towards energy

A + 2B A + 2B C + D C + D + Energy + Energy

LeChatelier Example #2LeChatelier Example #2A closed container of ice and water A closed container of ice and water at equilibrium. The at equilibrium. The temperaturetemperature is is raised.raised.

Ice + Energy Ice + Energy Water Water

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

rightright

LeChatelier Example LeChatelier Example #3#3

A closed container of water and its vapor closed container of water and its vapor at equilibrium. at equilibrium. Vapor is removedVapor is removed from from the system.the system.

water + Energy water + Energy vapor vapor

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

rightright

3. Pressure – changes in P will only effect gaseous equilibria

a. Increasing the pressure will usually favor the direction that has fewer molecules

N2(g) + 3H2(g) ↔ 2NH3(g)

• For every two molecules of ammonia made, four molecules of reactant are used up – this equilibrium shifts to the right with an increase in pressure

(a) A mixture of NH3(g), N2(g), and H2(g) at equilibrium. (b) The volume is suddenly decreased. (c) The new equilibrium position.

Changes in PressureChanges in Pressure As the pressure increases the reaction

will shift in the direction of the least moles gases.

At high pressure2H2(g) + O2(g) 2 H2O(g)

At low pressure 2H2(g) + O2(g) 2 H2O(g)

LeChatelier Example #4LeChatelier Example #4A closed container of NA closed container of N22OO44 and NO and NO22 at at equilibrium. The equilibrium. The pressurepressure is is increasedincreased..

NN22OO4 4 (g) + Energy (g) + Energy 2 NO 2 NO22 (g)(g)

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

leftleft

B. Solubility Equilibria Ionic compounds (aka salts) differ in their solubilities Most “insoluble” salts will actually dissolve to some extent

in water (slightly soluble)

The equilibria that we have considered thus far have been homogeneous (all the species are in the same phase, i.e. gas)

1. dissolution or precipitation of ionic compounds

2. heterogeneous 

Solubility Product Constant

Consider: AgCl(s) Ag+(aq) + Cl-(aq)

The “equilibrium expression” is:

[ Ag+ ] x [ Cl- ]

[ AgCl ] Keq =

What was the physical state of the AgCl?

H2O

AgCl existed as a solid material, and is not in a solution = a constant concentration!

the [ AgCl ] is constant as long as some undissolved solid is present (same with any pure liquid- do not change their conc.)

3. solubility product constant = Ksp

4. pure solid or pure liquid is not included in the equilibrium expression

From our example:

[Ag1+] x [Cl1-]Ksp =

Values of solubility product constants are given for some common slightly soluble salts

Ksp = [Ag1+] x [Cl1-] Ksp = 1.8 x 10-10

5. The smaller the numerical value of Ksp, the lower the solubility of the compound

AgCl is usually considered insoluble because of its low value

To solve problems:

a) write the balanced equation, which splits the chemical into its ions

b) write the “equilibrium expression”, and

c) fill in the values known; calculate answer

Do not ever include pure liquids nor solids in the expression, since their concentrations cannot change (they are constant) – just leave them out!

Do not include the following in an equilibrium expression: any substance with a (l) after it such as:

Br2(l), Hg(l), H2O(l), or CH3OH(l)

any substance which is a solid (s) such as: Zn(s), CaCO3(s), or H2O(s)

ALWAYS include those substances which can CHANGE concentrations, which are gases and solutions:

O2(g) and NaCl(aq)

A saturated solution of a slightly soluble salt in contact with undissolved  salt involves an equilibrium like the one below.

CaF2 (s)   <==>    Ca2+ (aq)  +   2F- (aq)

Write the solubility product constant for this heterogeneous equilibria.

Ksp = [Ca2+][F-]2 

Solid copper(II) chromate dissolves in water to give a solution that contains 1.1 x 10-5 g copper(II) chromate per liter at 23°C. Calculate Ksp for solid copper(II) chromate at the temperature.

CuCrO4 (s)   <==>    Cu2+ (aq)  +   CrO42-

(aq)

Ksp = [Cu2+][CrO42-] = (7.0 X 10-8)(7.0 X 10-8)

= 4.9 10-15

1.1 x 10-5 g CuCrO4

156.07 g CuCrO4

1 mole CuCrO4 = 7.0 x 10-8 mol CuCrO4

C. Common Ion Effect

A “common ion” is an ion that is found in both salts in a solution

example: You have a solution of lead (II) chromate. You now add some lead (II) nitrate to the solution.

The lead is a common ion This causes a shift in equilibrium (due to Le Chatelier’s

principle regarding concentration), and is called the common ion effect

1. solubility product constant (Ksp) can also be used to predict whether a precipitate will form or not:

a. if the calculated ion-product conc. is greater than the accepted

value for Ksp, then a precipitate will form