CH111 SOS Study Session

CH111 SOS SessionBy: Matt Nichols + Mike Golian

Goals of chapter 12

• Describe and predict changes in concentration with time

• Describe and predict how the rate of the reaction is affected by concentration.

• Use reaction rates to figure out mechanism: which bonds break and form in what order in a reaction

• Understand the temperature dependence of reactions.

• Understand how catalysts speed reactions.

Reactions

• Molecules only undergo a reaction if:– They have enough energy– The geometry of the collision is conducive to the reaction

• All reactions undergo elementary reactions (the most basic reactions)– Usually involve only one or two molecules

• Reaction mechanisms are multi-step reactions– Many elementary reactions simplified into one chemical

equation– The slowest elementary reaction in the reaction

mechanism is called the rate determining step

Reactions• Atoms, ions, and molecules recombine during chemical reactions• Reactions do not occur all at once, but are made up of sequential

molecular events making a reaction mechanism• A reaction mechanism is a step by step process, where each step is

referred to as an elementary reaction• Elementary reactions makes up each step in the sequential molecular

events ,and when put together make up the reaction mechanism• Elementary reactions are generally comprised 1,2 or 3 reactants• Overall reaction describes the starting materials and final products of a

reaction• Overall reactions are not generally elementary as they do not show the

individual steps of the reaction

Elementary reactions

• A bimolecular reaction is the most common elementary reaction

• Occurs with the collision of 2 atoms/molecules/ions• Characteristics feature is the collision of 2 species creating a

collision complex that results in rearrangement of chemical bonds

Elementary reactions continued

• A unimolecular reactions occurs when a single molecule fragments into 2 species or rearranges to form a new isomer

Elementary reactions cont’d

• A termolecular reaction occurs when 3 or more species collide and react

• Less common as the probability of 3 species colliding at the same time with the correct reaction geometry is small

• 3 species reaction commonly occur in 2 steps– with the collision of 2 species forming a complex in the first step– And the collision of the 3rd species with the complex from the first

step– This means that it can be written as a product of bimolecular and

unimolecular elementary reaction

Reaction intermediates

• A species that is produced in the first step and then consumed in a later step

• Will not appear in the overall reaction• Intermediate represented by the collision

Rate Law Relates the rate of reaction with the concentration of reactants through a

rate constant (k) experimentally determinedRate law can contain concentrations of chemical species that do not appear in

the overall chemical reaction - occurs when an intermediate is a rate determining step

aA + bB + cC... productsRate = k[A]m[B]n[C]p...

• Exponents in rate law do not depend on stoichiometric coefficients• Exponent is the order of reaction with respect to that species• Exponent dependant on the reaction mechanism• Sum of the exponents is the overall reaction order

– M+N+P =overall order of reaction

Rate constants

• Every reaction has characteristic rate constant• Has units of concentration/time• Remains constant over the reaction even with change

concentrations of reactants/products• Make sure units of K give a rate unit that makes sense

• A rate law has the form R= k[A]^2[B], when [A]=1.12M and [B]=0.87M and the rate of reaction is 4.78x10^-2 M s-1 what is the value of K including units?

First Order Reaction

The general rate law is Rate = k[A]

• Integrated rate law

• Plotting on a graph – Plot as vs. time

Gives rise to a linear graph

First order example

• Decomposition of hydrogen peroxide is a first order reaction

• 2H202 2H20 + 02

• Initial [H202] = 2.32 M• K= 7.30 x 10^-4 S^-1• What is the rate law?• What will the concentration of hydrogen peroxide be after

1200 s• For appearance of product

Half life of first order reaction

• The time it takes to reduce initial product concentration to 50% of original value

• t1/2 is The point where [A] = 0.5[Ao]• Half life is only dependant upon the rate constant• What is the half life of hydrogen peroxide from the previous

slide (K= 7.30 x 10^-4 S^-1)

Example – first order

• At what time will the 30.6% of the initial hydrogen peroxide have decomposed?

• K= 7.30 x 10^-4 S^-1 • Do not need initial concentration as they cancel out!!!• Assuming 100% initially

Zero Order Reaction

Rate = k[A]^0• Rate constant is independent of the initial reactant

concentrations• Will take form of y=mx+b, where [At]= -kt + [Ao]• Rate of reaction remains constant throughout the reaction

– Don’t confuse with 1st order decomposition reaction

• Has K units of concentration over time ( ie. M s^-1) as there are no units to cancel

• Plotting on a graph– Plot as [A] vs. time– Will be a straight line with slope equal to the negative

value of K

Second Order Reaction

Rate = k[A]2

• Integrated rate law

• Plotting on a graph – Plot vs. time

• Half-life

Second order example

• The data in the table below is from the decomposition of A products

• At this point it is very useful to graph your data• Data analyzed by each method to save time• What is the value of k with units?• Derive the half life equation!! ( simple substitution)• What is the half life time?

time, min [A], M ln[A] 1/[A]0 1 0 15 0.63 -0.46 1.6

10 0.46 -0.78 2.215 0.36 -1.02 2.825 0.25 -1.39 4

Experimental determination of reaction rate

• A + B 2C• What is the order of the reaction• What is the K value with units?

experiment [A], M [B], M rate = -d[A]/dt, M h-1

1 0.50 0.50 1.2

2 1.0 0.50 4.8

3 2.0 1.0 38.4

Pseudo first order reactions

• Can be possible to simplify some second order reactions where one reagent is in great excess of the other to a pseudo first order reaction

• Suppose we follow a 1L hydrolysis reaction of 0.01M ethyl acetate to completion ( creates ethanol and acetic acid)

• Initially there is ~ 1000g water 55.5 M, when completed there is 55.5-0.01= ~55.5M

• As Water concentration essentially remains constant, reaction rate can be treated as independent of water concentration ( exponent 0), which simplifies this to a first order reaction

• Can apply first order reaction kinetics

Summary of rate parameters

Rate Law

[4]

Integrated Rate Law

[4]

Units of Rate Constant (k)

Linear Plot to determine k

Half-life

[4]

What if the first step isn’t rate determining??

• All examples this far have had the first step being the rate determining step, and as no step can go faster than the slowest elementary reaction, all steps after this have no effect on the reaction rate

• For many reactions the rate determining step occurs in the 2nd or 3rd elementary reaction, thus the overall reaction rate is dependant upon this step and must be related to the initial reactants

• Further, some 2nd and 3rd elementary reactions are comprised of intermediates that do not appear in the overall reaction

• Doesn’t make sense to have a rate law with an intermediate…must relate to the initial reactants

Rate determining step not first problem

• H2 + Br2 2 HBr• Appears as simple 1:1 stoichiometry, however if 1st step were

to be rate determining a ½ Br2 would be necessary, which doesn’t exist

• Reaction mechanism:Br2 2Br (fast, reversible)

Br + H2 HBr + H (slow rate determining)H + Br2 HBr (fast)

Br2 2Br (K1)

2Br Br2 ( K -1)

Br + H2 HBr + H ( K2)***

H + Br2 HBr (K3)

---------------------------------------------------Net reaction H2 + Br2 2 HBr

rate = k2[Br][H2]

although this rate law accurately describes the reaction it cannot be used as it is very hard to measure the concentration of the intermediates in reactions as they are very short lived and hard to detect

Relate [Br] to [Br2]

• Assume forward and reverse rates are equal• Br2 2Br• Relate equations• K1 [Br2]=K-1 [Br]2 …………..rearrange• [Br]2= (K1/K-1) [Br2] ……square root both sides• [Br] = (K1/K-1) ^(1/2) [Br2] ^(1/2)…substitute back for [Br]

• Rate = k[H2] [Br2] ^(1/2)

Activation Energy• The minimum amount of energy needed above the average kinetic energy

for a molecule to undergo a chemical reaction• The activation energy directly affects the rate of a reaction• Activation arises from the energy requirement to distort and/or break

chemical bonds of the reactants• Activated complex is the highest point on an activation energy diagram,

and is not considered an intermediate as these are stable for a short period

Temperature and Reaction rates and the Temperature and Reaction rates and the Boltzman distributionBoltzman distribution

• At any T, a larger fraction of molecules can react if Ea is small compared to a larger Ea

• For any reaction a larger fraction of molecules can react if T is high rather than low

• Vertical line represent the activation energy requirement of a reaction• The greater the number of molecules to the right of the vertical line the

faster the reaction will proceed

Arrhenius Equation

-Rate constant is dependant upon activation energy, temperature of the system, and collision frequency, and collision orientation

-Arrhenius equation relates these variables– Energy, orientation and collisions taken into

account in the equation

• The Arrhenius equation can be rewritten to remove the exponent:

• When there are two different temperatures and the activation energy is being calculated, the Arrhenius equation can be modified into:

Example 1

• If T2 has a value k = 9.63x10^-5 s^-1, and• T1 = 298K and has a value K1 = 3.46x10^-5 s^-1• Ea = 106 kj/mol• What is the value of T2?

Example 2

• In a reaction that was done at 0 °C, the reaction coefficient was found to be 4.92x10-12 sec-1. In a second reaction at room temperature, the reaction coefficient was found to be 4.15x10-12 sec-1. What is the activation energy of the reaction?

Catalysts

• Molecules that are involved in a reaction but are regenerated at the end

• Help to speed up a reaction by providing a different reaction mechanism– Reaction mechanism has a lower activation energy

• Substrate – the species that is catalysed• Two types of catalysts:

– Homogenous– Heterogenous

Heterogenous Catalysis

• A catalyst that is not in the same phase as the reaction – Example is a Catalytic converted in car

• Four steps are involved in the catalysis of a substrate– Adsorption of substrate to catalyst– Migration to active sites– Reaction– Desorption of products

Homogenous catalysts

• A catalyst that is in the same phase as the reaction

• Example. Chlorine gas produced from CFC’s that react with ozone

• CF2Cl2 CF2Cl + Cl• Cl + O3 ClO + O2

• O3 + O 2O2

• Chlorine acts as the catalyst in ozone decomposition

Biological Catalysts

• Organisms have developed enzymes to catalyze reactions• Enzymes are highly substrate-specific• Follow the same mechanism as heterogenous catalysis• Derive the rate law for an enzyme

K1

K-1

K2

Dynamic Equilibrium

• State of a reaction in which there is no net change of the amount of products and reactants– Formation of products = formation of reactants

• Reaction is assumed to be reversible• Equilibrium is shown as:

aA + bB ↔ cC + dD

Equilibrium Constant

• Describes the reversibility of a reaction• Equation denoted as:

• aA + bB ↔ cC + dD• Many different forms are present:

– Kp rate of production (forward)

– Kd rate of decomposition ( reverse)

• Reaction has to be at equilibrium to use the equilibrium constant!

• Eq’m constant ( Keq) = (Kp/ Kd )

Properties of Equilibrium Constants

• Keq is related to the stoichiometry of the balanced net reaction

• Keq applies only under equilibrium conditions• Keq is independent of the initial conditions

– Is constant for any particular reaction at a given temperature, initial conditions can be any mixture of reactants and products and will reach a state of equilibrium determined by value of Keq

Homogenous Rules

3H

1N

2NH

P )(P)(P

)(PK

22

3

N2 (g) + H2 (g) ↔ NH3 (g)1 3 2

All components in the same phase (usually aqueous or gas)All components are used in the equilibrium equation

Heterogeneous equilibrium

• Concentration of a pure solid or liquid does not vary significantly

• The amount of a solid liquid can change but he number of moles per unit volume essentially remains fixed

• Means that their concentrations are always equal to their standard concentration

• If the concentrations are the same when they are entered into the equilibrium equation will results in a value of 1

• Allows us to omit pure solids and liquids from the equilibrium constant

Heterogeneous

• CaCO3 (s) ↔ CaO (s) + CO2 (g)

• By definition the equilibrium constant should be

• Pure liquids and solids are excluded from the equilibrium equation, results in…

][CaCO

)(P [CaO] K

3

COeq

2

2COp P K

Water in aqueous equilibrium

• Water is often a reagent in aqueous equilibrium• H2O (l) + CO2 (g) H2CO3 (aq)

• The concentration of carbon dioxide is in bars and concentration of carbonic acid is in moles

• What are the appropriate units for water as it is neither a pure liquid or a solute?

• Is expressed in its mole fraction Xsolvent = nsolvent / ntotal

]CO[H

)(P O][H K

32

CO2eq

2

Water in aqueous equilibrium

• The mole fraction of water varies only slightly as solutes are added

• 1L of pure water contains 55.5 moles, if a solute is dissolved to 0.5M the mole fraction of water is 0.99 ( very close to 1)

• Equilibrium calculations are rarely more accurate than 5%, thus water can be treated as a pure liquid as its concentration is essentially constant

Relationships involving chemical equilibrium

• Forward and reverse reactions are balanced exactly, the equilibrium constant can be expressed in the reverse order

• Reverse equilibrium constant in the reciprocal of the forward eq’m constant

• Multiplying reaction stoichiometry by a constant rasies Keq to that power

• If 2 reactions are combined their Keq are multiplied

Example problem

• N2(g) + O2(g) 2 NO(g) Kc1 = 2.3 x 1019

• 2 NO(g) + O2(g) 2 NO2(g) Kc2 = 3 x 106

• What is the net reaction?• If the reactions are combined to show the relationship

between N2(g), O2(g) and NO2(g) what is the equilibrium constant?

Magnitudes of chemical equilibria

• Equilibrium constants vary tremendously in value• If Keq is large (eg. 10^15) majority will be products

• If Keq is small (eg. 10^-15) majority will be reactants

• If Keq is neither small or large ( eg. 3) there will be a mix of both reactants and products

N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)

Reaction Quotient

• Denoted as Q– Uses the same formula as K

• Used to determine whether an equilibrium reaction is at equilibrium or not

• If not, can also be used to determine which way the reaction will proceed to get to equilibrium

• If Q < Keq, reaction proceeds toward reactants

• If Q = Keq, reaction is at equilibrium

• If Q > Keq ,reaction proceeds towards products

Gibbs’ Free Energy and Equilibrium

• Can elucidate the spontaneity of a reaction• Negative value indicates a reaction will proceed

forward in the direction written• ΔG = Δ Go + RT ln Q• Equation is used to relate ΔG with the concentration

of reactants through the reaction quotient Q• ΔG = 0 when Q= Keq , when the system in in

equilibrium

Example problem

• N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)

• The production of ammonia gas has a Δ Go formation= -16.4 kJ/mol

• Δ Go formation = 0 for N2 (g) and H2 (g)

• What is Δ Goreaction ?

• What is the Keq ?

Temperature and Equilibrium

• Temperature affects equilibrium just like it affects the rate of a reaction– Will cause the Keq to change

• Exothermic reaction has a negative Δ Ho

Which makes the first term positive, increasing temperature will lower the value of the first term, which lowers Keq

Example problem

• CH4 (g) + H2O (g) CO (g) + 3H2 (g)

• Hydrogen gas is produced by the above reaction• This is an endothermic reaction and for it to proceed forward

reaction temperatures are often carried out above 1000K• Δ Ho = +206 kJ/mol, Keq =1.3x10^-25 at 298 k

• What is Keq at 1500 K? what does value imply about hydrogen gas production compared to 298K?

CH4 H20 CO H2Δ H kJ/mol -74.6 -241.85 -110.5 0Δ S 186.3 188.835 197.7 130.68

Le Chatalier’s Principle

• When a change is imposed on a system, the system will react in the direction that reduces the amount of change

• What kind of change can remove equilibrium conditions from a system?– Change in concentration of a reactant/product– Change in temperature– Change in pressure

Le Chatalier’s Principle Cont’d

• Can use the reaction quotient to qantitatively evaluate LCP

• If any change causes the value of Q to increase, products will be consumed and reactants produced

• If any change that causes the value of Q to decrease, reactants will be consumer and products produced

• If the change has no effect on Q then the change doesn’t affect equilibrium

Le Châtelier’s Principle

• Any change to a chemical system will prompt the system to counteract the change

• Used to predict which way an equilibrium shifts if there is a change to it

Catalysts and Equilibrium• Don’t cause any change in Keq• Acts in a manner to reduce the net activation energy• Does not alter the thermodynamic changes that accompany

the given reaction– Total amount of energy gained lost in the overall conversion of

reactants to products remains the same

Temperature and Keq

• Temperature is the only variable that can change the value of Keq

• The change is dependant upon the change in enthalpy (Δ Ho )

• Increase in temperature shifts the equilibrium towards the endothermic reaction

• Decrease in temperature shifts the equilibrium towards the exothermic reaction

How does change affect a NaCl solution

• NaCl (s) Na + (aq) + Cl -

(aq)

• what happens If:– More NaCl (s) is added?

– More CaCL 2(s)

– The solution is heated– More Ca(N03)2 is added?

– More water is added?

Working with equilibria

• There are 7 steps that should be considered when working with equilibria problems, they are;– Determine what is being asked for– Identify the major chemical species– Determin what chemical equilibria exist– Write the Keq expressions

– Organize the data and unknowns– Perform the calculation– Does the result make sense?

Example 1

• You have decided to make strong homemade vinegar by dissolving acetic acid in water at 1M. At eq’m conditions; [HH33OO+ +

= [CC22HH33OO22 --]=4.2x10^-3M

• HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --

• What is the equilibrium constant?

Effect of changes in pressure or volume

• What happens when;• adding or removing a gaseous reactant/product ?

– use reaction quotient Q to evaluate which direction the system will move

• Addition of an inert gas to constant volume mixture?– Increase total pressure of system, however partial

pressures remain unchanged as does Keq

• Change in pressure by changing the volume of the system– Increase in volume decreases pressure and decrease in volume

increase pressure (ideal gas law) Keq unchanged

Example of pressure change in equilibrium

• The given reaction below once at equilibrium has its volume reduced by a factor of 10 by applied external pressure and constant temperature

• Initially at eq’m in 10L have: 0.68 mol SO3 , 0.32 mol SO2, 0.16 mol O2, where Keq = 2.8x10^2 at 1000K

• What are the values in moles of the components at the new volume (1L)? T=1000K

Working with equilibria and ICE tables

• Write the equation and get the stoichiometry right• Write the Keq expression

• Write the initial conditions• Determine the direction the reaction will proceed

( use Q)• Determine the change in concentration ( don’t forget

about the coefficients)• Determine concentrations at equilibrium• solve

Working with small equilibrium constantsWorking with small equilibrium constants

• When Keq is small the changes are likely to be small in comparison with initial conditions

• Eg. You have a 1M solution of acetic acid, Keq = 1.8x10^-5

• HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --

HCHC22HH33OO22 HH33OO++ CC22HH33OO22 --

I 1 0 0C 1-x x xE 1-x x x

Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00Can neglect X as we can assume its very small as K value is small

Working with large equilibria

• When Keq is very large the reation will basically go to completion

• Is easier to work back from total completion to determine equilibrium concentrations

• Can eliminate the x term much in the same manner as we can in working with small equilibria

• That is because the amount of initial reactants compared to the products is very small

example

• The oxidation of NO to NO2 is an essential step in the production of nitric acid, if 10 bar of each NO and O2 are present initially, how much of each will remain after eq’m

• 2NO + O2 2NO2 Keq = 4.2x10^12

• Answer: x=2.2x10^-6

NONO OO22 NONO22

I 10 10 0C 10-x 10-x xE 10-x 10-x x

NONO OO22 NONO22

C 0 5 10C x 0.5x 10-xE x 5.0+0.5x 10-x

Types of aqueous equilibria

• In general there are 3 broad categories:– 1 proton transfer (acid/base) equilibria

• Acid donates a proton to water (Ka )

• base receives proton from water (Kb )

– Solubility• Salt can dissolve in water (Ksp )

• cation and anion can precipitate out (together)

– Complexation• Cation associates with electron donating species to form

complex(Kf )

• Complex can dissociate

Solubility and Ksp

• Solubility refers to the maximim amount of substance that is dissolved in a given volume of solvent– Is expressed in molarity ( Mol/L)

• It can also be expressed in a weight/volume ratio– Such as 1g/ml

Spectator Ions

• Ions that are present during a reaction but are not involved in it

• Eg. 1L of 0.1 M solutions of equal volume of AgNO3(aq) and NaBr (aq) are mixed

• What will be the precipitate?• What ions will be considered spectator ions?

Solubility cont’d

• Lead iodide PdI2 is “insoluble” solid used in bronzing

• PdI2 Pb+ + 2I-

• calculate the molar solubility of lead iodide Ksp = 7.1x10^-9

Common ion effect in solubility equilibria

• Suppose we have a saturated solution of PdI2 and to that solution we add another salt KI

• The addition of KI introduces the common ion I- • Referring back to LeChataliers principle the system will

respond by shifting the equilibrium to reduce the newly added reactant, in the case of this example…..to reduce I-

• PdI2 Pb+ + 2I-

• eq’m shifts to consume I-

• The solubility of a slightly soluble compound is also reduced in the presence of a second solute that furnishes a common ion– PdI2 solubility is reduced when KI is added

Solubility of PdI2 in the presence of KI

• What is the molar solubility of PdI2 in 0.1M KI?

• Ksp = [pb+][I-]2 = (s)(0.10+2s)2 = 7.1x10^-9

• As lead iodide is known to be “insoluble” can simplify equation by assuming 0.1M>>s

• this allows the removal of s from the equation• S = 7.1X10^-9/(0.10)2 • S= 7.1X10^-7

example

• Should precipitation of MgF (s) occur if a 22.5 mg sample of MgCl2-6H20 is added to 325mL of 0.035 M KF?