Exam 1 T-scoreABCDF 50118.7%34.2%30.1%15.5%1.4% 50616.9%39.0%29.9%10.6%3.5%...

Exam 1T-score A B C D F

501 18.7% 34.2% 30.1% 15.5% 1.4%506 16.9% 39.0% 29.9% 10.6% 3.5%All 17.8% 36.8% 30.0% 12.9% 2.5%

16.0% 35.0% 35.0% 14.0% 2.0%

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• Write the thermochemical equation for the reaction for CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

50.0mL of 0.400 M CuSO4 at 23.35 oC Tfinal 25.23oC

50.0mL of 0.600 M NaOH at 23.35 oCDensity final solution = 1.02 g/mL CH2O = 4.184 J/goC

Thermochemical EquationsThermochemical Equations

Chemical ThermodynamicsChemical Thermodynamics• Chemical reactions obey two fundamental

laws:

1. The law of conservation of mass States that matter can be neither created nor destroyed Explains why equations must balance and is the basis for

stoichiometry and equilibrium calculations Stoichiometry that allows us to compare apples and

oranges Equilibrium predictions of reversible reactions which

leads to Kinetics allowing us to determine how fast the

reaction will occur

2. The law of conservation of energy States that energy can be neither created nor destroyed Energy takes various forms that can be converted from one

to the other

Some Thermodynamic Some Thermodynamic TermsTerms

ThermodynamicsThermodynamics - The study of the relationship between heat, work, and other forms of energy of a system at equilibrium. Predicts whether a particular reaction is energetically possible in the direction as written and the composition of the reaction system at equilibrium. Thermodynamics does not say whether an energetically feasible reaction will actually occur as written. Thermodynamics tells nothing about the rate of the reaction or the pathway by which it will occur.

ThermochemistryThermochemistry - A branch of thermodynamics which focuses on the study of heat given off or absorbed in a chemical reaction.

TemperatureTemperature - An intensive property of matter; a quantitative measurement of the degree to which an object is either "hot" or "cold". There are 3 scales:

•FahrenheitFahrenheit - relative • 32 ◦F is the normal freezing point

temperature of water; 212 ◦ F is the normal boiling point temperature of water.

•CelsiusCelsius (centigrade) - relative • 0 ◦ C is the normal freezing point

temperature of water; 100 ◦ C is the normal boiling point temperature of water.

•KelvinKelvin - absolute • 0 K is the temperature at which the volume

and pressure of an ideal gas extrapolate to zero.

Standard States and Standard States and Standard Enthalpy Standard Enthalpy

ChangesChanges1.1. Thermochemical standard state conditionsThermochemical standard state conditions• The thermochemical standard T = 298.15 K.• The thermochemical standard P = 1.0000 atm.

– Be careful not to confuse these values with STP.

2.2. Thermochemical standard states of matterThermochemical standard states of matter• For pure substances in their liquid or solid phase the

standard state is the pure liquid or solid.• For gases the standard state is the gas at 1.00 atm of

pressure.• For gaseous mixtures the partial pressure must be 1.00

atm.• For aqueous solutions the standard state is 1.00 M

concentration.

Thermodynamics and WorkThermodynamics and Work• A system is that part of the universe in which we are interested (in

chemistry this is the reactant side of the chemical equation); the surroundings are everything else—the rest of the universe.

• System + surroundings = universe.• A closed systemclosed system can exchange energy but not matter with its

surroundings; an open systemopen system exchanges both, and an isolated isolated system system exchanges neither.

• State function State function — the property of a system that depends only on the present state of the system and not on its history.

1. State Functions are independent of pathway:– T (temperature), P (pressure), V (volume), E (change in

energy), H (change in enthalpy – the transfer of heat), and S (entropy)

2. Examples of non-state functions are:– n (moles), q (heat), w (work)

• A change in state function depends only on the difference between the initial and final states, not on the pathway used to go from one to the other.

• Thermodynamics is concerned with state functions and does not deal with how the change between the initial and final state occurs.

Thermochemical Thermochemical EquationsEquations

• Thermochemical equationsThermochemical equations are a balanced chemical reaction plus the H value for the reaction. – For example, this is an exothermic thermochemical

equation.

• The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles.

• 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions.

moles 6 moles 5 moles 8 mole 1

kJ 3523 OH 6 CO 5O 8 HC )(22(g)2(g))12(5

H2O(S) + 6.02 kJ →H2O(l)

this is an endothermic thermochemical equation.

Caloric Theory of HeatCaloric Theory of Heat•Served as the basis of thermodynamics. •Is now known to be obsolete •Based on the following assumptions

•Heat is a fluid that flows from hot to cold substances. •Heat has a strong attraction to matter which can hold a lot of heat. •Heat is conserved. •Sensible heat causes an increase in the temperature of an object when it flows into the object. •Latent heat combines with particles in matter (causing substances to melt or boil) •Heat is weightless.

The only valid part of the caloric theory is that heat is weightless.

Heat is NOT a fluid, and it is NOT conserved.

Some Thermodynamic Some Thermodynamic TheoriesTheories

1. Divides the universe into two parts:

A.A. SystemSystem. - The substances involved in the chemical and physical changes under investigation: In chemistry lab, the system is the REACTANTSREACTANTS inside the beaker.

B. SurroundingsSurroundings - Everything not included in the system, i.e. the rest of the universe.

2.A BOUNDARYBOUNDARY separates the system and the surroundings from each other and can be tangible or imaginary.

A. Heat is something that is transferred back and forth across boundary between a system and its surroundings

B. Heat is NOT conserved.

Some Thermodynamic Some Thermodynamic TheoriesTheoriesKinetic Theory of HeatKinetic Theory of Heat

• The set of conditions that specify all of the properties of the system is called the thermodynamic state of a systemthermodynamic state of a system.

• For example the thermodynamic state could include:– The number of moles and identity of each substance.– The physical states of each substance.– The temperature of the system.– The pressure of the system.

Some Thermodynamic Some Thermodynamic TheoriesTheories The kinetic theory of heat is based upon the last postulate in

the kinetic molecular theory which states that the average kinetic energy of a collection of gas particles is dependent only upon the temperature of the gas.

where R is the ideal gas constant (0.08206 L-atm/mol-K) and T is temperature (Kelvin) The kinetic theory of heat can be summarized as follows:

When heat enters a system, it causes an increase in the speed at which the particles in the system move.

There are two basic ideas of importance for thermodynamic systems.

1. Chemical systems tend toward a state of minimum potential minimum potential energy.energy.

2. Chemical systems tend toward a state of maximum disordermaximum disorder.

The Three Laws of The Three Laws of ThermodynamicsThermodynamics

When:S > 0 disorder increases disorder increases (which favors

spontaneity).S < 0 disorder decreases disorder decreases (does not

favor spontaneity).

When:H > 0 disorder increases disorder increases (which favors

spontaneity).H < 0 disorder decreases disorder decreases (does not favor

spontaneity).

The First Law of The First Law of ThermodynamicsThermodynamics

• The first law is also known as the Law of Conservation of Law of Conservation of

Energy.Energy.

Energy is neither created nor destroyed in chemical reactions and physical changes.

•The energy of the universe does not change. •The energy in a system may change, but it must be complemented by a change in the energy of its surroundings to balance the change in energy.

The term internal energy is often used synonymously with the energy of a system. It is the sum of the kinetic and potential energies of the particles that form the system. The change in energy of the system is identical in magnitude but opposite in sign to the change in energy of the surroundings.

If a system is more complex than an ideal gas, then the internal energy must be measured indirectly by observing any changes in the temperature of the system. The change in the internal energy of a system is equal to the difference between the final and initial energies of the system:

The equation for the first law of thermodynamics can be rearranged to show the energy of a system in terms of the energy of its surroundings. This equation indicates that the energy lost by one must equal the energy gained by the other:


Esys = KEsys + PEsys 1. KE – kinetic energy:

translational, rotational, vibrational

2. PE – energy stored in bonds (Bond energy)

The energy of a system can change by the transfer of work and or heat between the system and its surroundings. Any heat that is taken, given off, or lost must be complemented by an input of work to make up for the loss of heat. Conversely, a system can be used to do any amount of work as long as there is an input of heat to make up for the work done.

This equation can be used to explain the two types of heat that can be added to a system:1. Heat can increase the temperature of a system. This is sensible heat. 2. Heat that does ONLY WORK on a system is latent heat.


• Any machine that converts energy to work is designed to want to maximize the amount of work obtained and to minimize the amount of energy released to the environment as heat

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

C3H8(g) + 5 O2(g) ) 3 CO2(g) + 4 H2O(g)


Most chemical reactions occur at Most chemical reactions occur at constant P, soconstant P, so

and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small)∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E∆∆H = change in H = change in heat content heat content of the systemof the system∆∆H = HH = Hfinalfinal - H - Hinitialinitial

and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small)∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E∆∆H = change in H = change in heat content heat content of the systemof the system∆∆H = HH = Hfinalfinal - H - Hinitialinitial

Heat transferred at constant P = qHeat transferred at constant P = qpp

qqpp = = ∆H ∆H where where H = enthalpyH = enthalpyHeat transferred at constant P = qHeat transferred at constant P = qpp

qqpp = = ∆H ∆H where where H = enthalpyH = enthalpy

-103.8 kJmol-1 0 kJmol-1 -393.5 kJmol-1 -241.8 kJmol-1

1. Exchange of heat (q) Endothermic and exothermic2. Work is performed (w)

E = q + w

Solids, Liquids, SolutionsChanges in volume are negligibleTherefore w is effectively zero

E = q + 0 = H

H is change in enthalpy which is the transfer of heat and is measuredexperimentally by determining changesin temperature.

GasesWhy only gases?

Because changes in volume results in work

w = FdF = Pressure x Area d = h

W = P (A h) = V


heat transfer outheat transfer out(exothermic), -q(exothermic), -q

heat transfer inheat transfer in(endothermic), +q(endothermic), +q

SYSTEMSYSTEM

∆E = q + w

w transfer inw transfer in(+w)(+w)Compression of systemCompression of system

w transfer outw transfer out(-w)(-w)Expansion of systemExpansion of system

By convention except for some engineers whose frame of reference is the work done on the surroundings.hi hf

A(hf-hi)<0 V

hihf

A(hf-hi)>0 V

w = -PVE =H + w = H – PV = H – PV)


E = E = H – H – PV) PV)

Constant Volumew = -PVV = 0

E = q + 0 = HCheck the temperature change

Constant PressureApply some stoichiometry

And the Ideal Gas LawPV=nRT

(PV)=(nRT)Hold Temperature constant k1

(PV)=(nRk1)Combine constants and multiply through by -1

(PV) = -R1nw = -PV = -R1n

E = H + w = H - R1n

E H n

E = H exothermic No change

E = H endothermic

No change

E > H exothermic increase

E > H endothermic

decrease

E < H exothermic decrease

E < H endothermic

increase

Thermochemical Thermochemical EquationsEquations

Write the thermochemical equation for CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

50.0mL of 0.400 M CuSO4 at 23.35 oC 50.0mL of 0.600 M NaOH at 23.35 oC Tfinal 25.23oC

CH2O = 4.184 J/goC

Density final solution = 1.02 g/mL

The Second Law of The Second Law of ThermodynamicsThermodynamics

• The second law of thermodynamics states, “In In spontaneous changes the universe tends spontaneous changes the universe tends towards a state of greater disordertowards a state of greater disorder..”

• Spontaneous processes have two requirements:1. The free energy change of the system must be

negative.2. The entropy of universe must increase.• Fundamentally, the system must be capable

of doing useful work on surroundings for a spontaneous process to occur.

Changes in S are usually quite small compared to E and H. Notice that S has units of only a fraction of a kJ while E and H values are much larger numbers of kJ.

Enthalpy changes are not the only factors that determine whether a process is spontaneous. Most spontaneous reactions are exothermic, but there are many that are not exothermic, however, reactions can be both spontaneous and highly endothermic.

Entropy (S) – is the measure of the disorder in a system. Entropy of the universe is unchanged in reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process. Entropy is a state function described by the equation:

where k is a proportionality constant equal to the ideal gas constant (R) divided by Avogadro's number (6.022 x 10-23) and lnW is the natural log of W, the number of equivalent ways of describing the state of a system.

In this reaction, the number of ways of describing a system is directly proportional to the entropy of the system.


Hand W ln WRoyal flush (AKQJ10 in one suit) 4 1.39Straight flush (five cards in sequence in one suit)

36 3.58

Four of a kind 624 6.44Full house (three of a kind plus a pair)

3,744 8.23

Flush (five cards in the same suit) 5,108 8.54Straight (five cards in sequence) 10,200 9.23Three of a kind 54,912 10.91Two pairs 123,552 11.72One pair 1,098,240 13.91No pairs 1,302,540 14.08

Total 2,598,960

Number of Equivalent Combinations for Various Types of Poker Hands


Entropy of Reaction (S) •The difference between the sum of the entropies of the products and the sum of the entropies of the reactants:

In the above reaction, n and m are the coefficients of the products and the reactants in the balanced equation.

As with H, entropies have been measured and tabulated.

When:S > 0 disorder increases (which favors spontaneity).S < 0 disorder decreases (does not favor spontaneity).


Natural processes that occur in an isolated system are spontaneous when they lead to an increase in the disorder, or entropy, of the system.

Isolated system - System in which neither heat nor work can be transferred between it and its surroundings. This makes it possible to ignore whether a reaction is exothermic or endothermic. If Ssys > 0, the system becomes more disordered through the course of the reaction If Ssys < 0, the system becomes less disordered (or more ordered) through the course of the reaction.


There are a few basic principles that should be remembered to help determine whether a system is increasing or decreasing in entropy.

• Liquids are more disordered than solids. •WHY? - Solids have a more regular structure than liquids.

• Gases are more disordered than their respective liquids.

•WHY? - Gases particles are in a state of constant random motion.

• Any process in which the number of particles in the system increases consequently results in an increase in disorder.

• In general for substances with the same molar mass and number of atoms in its three states of matter, Sº values fall in the :

SSgas gas > S> Sliquidliquid > S > Ssolidsolid

Does the entropy increase or decrease for the following reactions?

• CaCO3(s) → CaO(s) + CO2(g)

• N2(g) + 3H2(g) 2NH3(g)

• NH4NO3(s) → NH4+

(aq) +

NO3-(aq)

• H2O(g) H2O(l)


→→

→→

Entropy, SEntropy, S• The Third Law of Thermodynamics states, “The entropy of

a pure, perfect, crystalline solid at 0 K is zero.”• This law permits us to measure the absolute values of the

entropy for substances.– To get the actual value of S, cool a substance to 0 K,

or as close as possible, then measure the entropy increase as the substance heats from 0 to higher temperatures.The coldest place in nature is the depths of outer space. There it is 3 degrees above Absolute Zero.

– Notice that Appendix L has values of S not S.Bose-Einstein Condensation in a gas: a new form of matter at the coldest temperatures in the universe...

A. Einstein S. Bose

Predicted 1924... ...Created 1995

Cornell and Wieman cooled a small sample of atoms down to only a few billionths (0.000,000,001) of a degree above Absolute Zero

Entropy, SEntropy, S

BEC

When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder; the greater the number of atoms or molecules in the gas, the greater the disorder.

The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it; the greater the number of microstates, the greater the entropy.

Entropy and Entropy and TemperatureTemperature

S increases slightly with T

S increases a large amount with phase changes

Calculating Calculating SS from Standard Molar from Standard Molar Entropy ValuesEntropy Values• Similar molecular structures have similar Sº values

1. Those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds

2. Those with higher entropies are soft crystalline substances that contain larger atoms and increased molecular motion and disorder

• Absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity

• Substances with strong hydrogen bonds have lower values of Sº, reflecting a more ordered structure

• To calculate Sº for a chemical reaction from standard molar entropies, the “products minus reactants” rule is used; here the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation

Entropy, SEntropy, S• Entropy changes for reactions can

be determined similarly to H for reactions.This is only true, i.e. conserved, for the system. This is not included for the surroundings.

oreactants

n

oproducts

n

o298 SnSnS

Entropy, SEntropy, S• Calculate the entropy change for the following

reactions at 25oC. 240 Jmol-1K-1 304.2 Jmol-1K-1

210.6 Jmol-1K-1 219.7 Jmol-1K-1 240 Jmol-1K-1

C3H8(g)+ 5O2(g) 3CO2(g) + 4H2O(g)

270.2 Jmol-1K-1 0 Jmol-1K-1 197.6 Jmol-1K-1 188.7 Jmol-1K-1

Absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is an ideal that cannot be achieved, however, the combination of these two ideals constitutes the basis for the third law of thermodynamics.

• The third law of thermodynamics has two important consequences:

1. It defines as positive the sign of the entropy of any substance at temperatures above absolute zero

2. It provides a fixed reference point that allows the measurement of the absolute entropy of any substance at any temperature

The Third Law of The Third Law of ThermodynamicsThermodynamicsThe entropy of any perfectly The entropy of any perfectly

ordered, crystalline substance at ordered, crystalline substance at absolute zero is zeroabsolute zero is zero..

G and SpontaneityG and Spontaneity• In the mid 1800’s J. Willard Gibbs determined the

relationship of enthalpy, H, and entropy, S, and temperature, T, that best describes the maximum useful energy obtainable in the form of work from a process at constant temperature and pressure. G is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner– The relationship also describes the spontaneity of

a system.• The relationship is a new state function, G, the

Gibbs Free Energy.

G = H-TS at constant T and P

H = q whether a process is reversible or irreversibleTS = qrev (Suniv = 0)G = q – qrev

Free Energy Change, Free Energy Change, G, G, and Spontaneityand Spontaneity

• The change in the Gibbs Free Energy, G, is a reliable indicator of spontaneity of a physical process or chemical reaction.– G does not tell us how quickly the process

occurs.• Chemical kinetics, the subject of Chapter

16, indicates the rate of a reaction.• Sign conventions for G.

– G > 0 reaction is nonspontaneous– G = 0 system is at equilibrium– G < 0 reaction is spontaneous

The Temperature The Temperature Dependence of Dependence of

SpontaneitySpontaneity• Free energy has the relationship

G = H -TS.• Because 0 ≤ H ≥ 0 and 0 ≤ S

≥ 0, there are four possibilities for G.

Forward reaction

H S G spontaneity< 0 > 0 < 0 at all T’s.< 0 < 0 T dependent at low T’s.> 0 > 0 T dependent at high T’s.> 0 < 0 > 0

Nonspontaneous at all T’s.

G H S

Low T

High T

G = 0 Equilibrium

G < 0 Spontaneous

G > 0 Non Spontaneous

). .

). .

). .). .

). .

). . ). . ). .

). .

). .

Spontaneity is favored when

H < 0 and/or S > 0G = H -TS

). .

). .

). .

. .

G and SpontaneityG and Spontaneity• Changes in free energy obey the same type

of relationship we have described for enthalpy and entropy changes.

Calculate Go298 for the reaction in

-23.56 kJmol-1 -394.4 kJmol-1 -237.2 kJmol-10 kJmol-1

The Temperature The Temperature Dependence of Dependence of

SpontaneitySpontaneity• Determine the temperature at which the following

system is at equilibrium, spontaneous, non-spontaneous.

C3H8(g) + 5 O2(g) ) 3 CO2(g) + 4 H2O(g)

1. We know that So298= -1077.4 kJmol-1K-1,

2. We know that Ho298= -2219.9 kJmol-1,

3. and that Go298= -2108.5 kJmol-1.

Chemical Chemical KineticsKinetics– Study of reaction rates, or

the changes in the concentrations of reactants and products with time

– By studying kinetics, insights are gained into how to control reaction conditions to achieve a desired outcome, its mechanism

– Chemical kinetics of a reaction depend on various factors

1.1.Physical states and surface Physical states and surface areas of reactantsareas of reactants

2.2.Reactant concentrationsReactant concentrations3.3.TemperatureTemperature4.4.Solvent and catalyst propertiesSolvent and catalyst properties

– Reaction rateReaction rate = change in concentration of a reactant or product with time. Three “types” of rates

1.1.initial rateinitial rate2.2.average rateaverage rate3.3.instantaneous rateinstantaneous rate

The Rate of ReactionThe Rate of Reaction• Consider the hypothetical reaction,

A(g) B(g)

• equimolar amounts of reactant A will be consumed while product B will be formed as indicated in this graph:

The Rate of ReactionThe Rate of Reaction• Mathematically, the rate of a reaction can be

written as:aA(g) + bB(g) cC(g) + dD(g)

1. Differential rate law – Expresses the rate of a

reaction in terms of changes in the concentration of one or more reactants, [R], over a specific time interval, t

– Describes what is occurring on a molecular level during a reaction

2. Integrated rate law – Describes the rate of a

reaction in terms of the initial concentration, [R]0, and the measured concentration of one or more reactants, [R], after a given amount of time, t

– Used for determining the reaction order and the value of the rate constant from experimental measurements

In general, for a A + b B → x X with a catalyst C

Rate = k [A]m[B]n[C]p

The exponents m, n, and p • are the order of reactant• The overall order of reaction is the sum of the order of reactants• can be 0, 1, 2 or fractions• must be determined by experimentRate law must provide a rate with the units M/sThe proportionality constant, k, is called the rate constant. 1. Value is characteristic of the reaction and reaction conditions 2. A given reaction has a particular value of the rate

constant under a given set of conditions, such as temperature, pressure, and solvent

The Rate of ReactionThe Rate of Reaction

The Rate of ReactionThe Rate of Reaction• The rate of a simple one-step reaction is

directly proportional to the concentration of the reacting substance.

• [A] is the concentration of A in molarity or moles/L.

• k is the specific rate constant.– k is an important quantity in this chapter.

Ak = Rateor ARate

C + BA (g)(g)(g)

– Reaction whose rate is independent of concentration

– Its differential rate law is rate = k

– One can write their rate in a form such that the exponent of the reactant in the rate law is 0

rate = – [A] = k[reactant]0 = k(1) = k t

– Since rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of –k (concentration decreases with time); a graph of the concentration of any product as a function of time is a straight line with a slope of +k

Zeroth-Order ReactionsZeroth-Order Reactions

– Reaction rate is directly proportional to the concentration of one of the

reactants– Have the general form A products– Differential rate for a first-order reaction is

rate = – [A] = k[A] t– If the concentration of A is doubled, the rate of the

reaction doubles; – If the concentration of A is increased by a factor of 10,

the rate increases by a factor of 10– Units of a first-order rate constant are inverse seconds, s–1

– First-order reactions are very commonThe order of a reaction can be expressed in terms of either

each reactant in the reaction or the overall reaction.• For example:

First-Order ReactionsFirst-Order Reactions

• Two kinds of second-order reactions

1. The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of the reactant and has the form 2A products.

– Differential rate law is rate = – [A]

2t

– Doubling the concentration of A quadruples the rate of the reaction

– If the [A] is halved the rate of the reaction will decrease by a factor of 4. (½)2 = ¼

– Units of rate constant is M–1s–1 or L/mols – Concentration of the reactant at a given time is described by the

following integrated rate law:

Second-Order ReactionsSecond-Order Reactions

2. The second kind has a rate that is proportional to the product of the concentrations of two reactants and has the form A + B products.

– Reaction is first order in A and first order in B– Differential rate law for the reaction is rate = – [A] = – [B] = k[A] [B] t t– Reaction is first order both in A and in B and has an overall reaction order of 2

Factors That Affect Factors That Affect Reaction RatesReaction Rates

• There are several factors that can influence the rate of a reaction:

1. The nature of the reactants.2. The concentration of the reactants.3. The temperature of the reaction.4. The presence of a catalyst.• We will look at each factor

individually.

Phase and Surface Area Effects

• If reactants are uniformly dispersed in a single homogeneous solution, the number of collisions per unit time depends on concentration and temperature.

• If the reaction is heterogeneous, the reactants are in two different phases, and collisions between the reactants canoccur only at interfaces between phases; therefore, the number of collisions between the reactants per unit time is reduced, as is the reaction rate. The rate of a heterogeneous reaction depends on the surface area of the more condensed phase.

Nature of ReactantsNature of Reactants• This is a very broad category that encompasses the

different reacting properties of substances. • For example sodium reacts with water explosively at room

temperature to liberate hydrogen and form sodium hydroxide.

• Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

• The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

Nature of ReactantsNature of Reactants

• However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide.

• The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.

Dr Bunhead and some Pub tricks 1Dr, Bunhead and some more Pub tricks

Orchestra on heliumHuman Beatbox on Helium

Fun with your Wii

Just blowing things upA tentA trailerA grand pianomelons

Some things You might want to consider when fueling your car

Flour

34

Rb85

Nature of ReactantsNature of Reactants

Concentrations of Concentrations of ReactantsReactants

– The increase in the molecule numbers is indicative of an increase in concentration. A(g) + B (g) Products

A B

A B

A B BA B

A BA BA B

4 different possible A-B collisions



• Two substances cannot react with each other unless their constituent particles come into contact; if there is no contact, the rate of reaction will be zero.

• The more reactant particles that collide per unit time, the more often a reaction between them can occur.

• The rate of reaction usually increases as the concentration of the reactants increases.

Concentrations of Concentrations of Reactants: Reactants:

The Rate-Law ExpressionThe Rate-Law Expression• Example 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A(g) + B(g) 3 C(g)

Experiment

Number

Initial [A]

(M)

Initial [B]

(M)

Initial rate of formation of C

(M/s)

1 0.10 0.10 2.0 x 10-4

2 0.20 0.30 4.0 x 10-4

3 0.10 0.20 2.0 x 10-4


The Rate-Law ExpressionThe Rate-Law Expression• The following data were obtained for the following

reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction?

2 A(g) + B(g) + 2 C(g) 3 D(g) + 2 E(g)

Experiment

Initial [A]

(M)

Initial [B]

(M)

Initial [C]

(M)

Initial rate of formation of D

(M/s)

1 0.20 0.10 0.10 2.0 x 10-4

2 0.20 0.30 0.20 6.0 x 10-4

3 0.20 0.10 0.30 2.0 x 10-4

4 0.60 0.30 0.40 1.8 x 10-3


The Rate-Law ExpressionThe Rate-Law Expression• Consider a chemical reaction between

compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks.

Experiment

Initial Rate

(M/s)

Initial [A]

(M)

Initial [B]

(M)

1 4.0 x 10-3 0.20 0.050

2 1.6 x 10-2 ? 0.050

3 3.2 x 10-2 0.40 ?

• The integrated rate equation relates time and concentration for chemical and nuclear reactions.– From the integrated rate equation we can predict the

amount of product that is produced in a given amount of time.

– Integrated rate law for a zeroth-order reaction produces a straight line and has the general formula

[A] = [A]0 – akt, where [A]0 is the initial concentration of reactant A; the rate

constant must have the same units as the rate of the reaction, M/s, in a zeroth-order reaction

– Equation has the form of the equation for a straight line (y = mx + b); y = [A], mx = – akt, and b = [A]0

– Occur most often when the reaction rate is determined by available surface area

Zeroth-Order ReactionsZeroth-Order Reactions

Concentration vs. Time: Concentration vs. Time: The Integrated Rate The Integrated Rate

EquationEquation


EquationEquation• These reactions are 1st order in the reactant and 1st order

overall.

– Integrated rate law for a first-order reaction can be written in two different ways, one using logarithms and one using exponentials

Exponential form, [A] = [A]0e–kt, where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant, and e is the base of the natural logarithms, which has the value 2.718. Concentration of A will decrease in a smooth exponential curve over time

First-Order ReactionsFirst-Order Reactions

Rearranging the logarithmic expression : ln[A] = ln[A]0 – kt; the equation has the form of the equation for a straight line; y = ln[A] and b = ln[A]0; and a plot of ln[A] vs. t for a first-order reaction gives a straight line with a slope of –ak and an intercept of ln[A]0

oror

• An example of a reaction that is 1st order in the reactant and 1st order overall is:

a A productsThis is a common reaction type for many chemical reactions and all simple radioactive decays.

• Two examples of this type are:2 N2O5(g) 2 N2O4(g) + O2(g)

238U 234Th + 4He


EquationEquationFirst-Order ReactionsFirst-Order Reactions

• Solve the first order integrated rate equation for t.

• Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0.


EquationEquation

14.4 Using Graphs to Determine Rate Laws, Rate Constants, and Reaction Orders

For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of – k.For a first-order reaction, a plot of the logarithm of the concentration of a reactant versus time is a straight line with a slope of – k.For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.Properties of reactions that obey zeroth-, first-, and second-order rate laws are summarized in the following table.

• Cyclopropane, an anesthetic, decomposes to propene according to the following equation.

The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C.


EquationEquation

• Refer to Previous Example: How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds?– The integrated rate laws can be used for any unit that

represents moles or concentration.– In this example we will use grams rather than mol/L.


EquationEquation

• The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.

CS2(g) CS(g) + S(g)


EquationEquation

• For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is:

• Where: [A]0= mol/L of A at time t=0; [A] = mol/L of A at time t; k = specific rate constant; t = time elapsed since beginning of reaction; and a = stoichiometric coefficient of A in balanced overall equation.


EquationEquationSecond-Order ReactionsSecond-Order Reactions

Rearranging the logarithmic expression : 1/[A] = 1/[A]0 + akt; the equation has the form of the equation for a straight line; y = 1/[A] and b = 1/[A]0; and a plot of 1/[A] vs. t for a first-order reaction gives a straight line with a slope of ak and an intercept of 1/[A]0

Temperature EffectsTemperature Effects• Increasing the temperature of a system increases

the average kinetic energy of its constituent particles.

• As the average kinetic energy increases, the particles move faster, so they collide more frequently per unit time and possess greater energy when they collide, causing increases in the rate of the reaction.

• Rate of all reactions increases with increasing temperature and decreases with decreasing temperature.

Temperature: Temperature: The Arrhenius Equation The Arrhenius Equation

• Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (Ea), and (3) the specific rate constant (k).

• If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 >T1.

Temperature: Temperature: The Arrhenius Equation The Arrhenius Equation • Consider the rate of a reaction for which Ea=50 kJ/mol, at

20oC (293 K) and at 30oC (303 K). How much do the two rates differ?

• For reactions that have an Ea50 kJ/mol, the rate approximately doubles for a 100C rise in temperature, near room temperature.

• Consider:2 ICl2 ICl(g)(g) + H + H2(g)2(g) I I2(g)2(g) + 2 HCl + 2 HCl(g)(g)

• The rate-law expression is known to be R=k[ICl][H2].

Solvent EffectsSolvent Effects• The nature of the solvent can affect the reaction rates of

solute particles.• Solvent viscosity is also important in determining reaction

rates. 1. In highly viscous solvents, dissolved particles diffuse

much more slowly than in less viscous solvents and collide less frequently per unit time.

2. Rates of most reactions decrease rapidly with increasing solvent viscosity.

Catalyst Catalyst EffectsEffects• Catalyst is a substance

that participates in a chemical reaction and increases the rate of the reaction without undergoing a net chemical change itself.

• Catalysts are highly selective and often determine the product of a reaction by accelerating only one of several possible reactions that could occur.

CatalystsCatalysts• Catalysts change reaction rates by providing an alternative

reaction pathway with a different activation energy.

• Homogeneous catalysts exist in same phase as the reactants. At least one of the reactants interacts with the solid surface (in a physical process called adsorption) in such a way that a chemical bond in the reactant becomes weak and then breaks.

• Heterogeneous catalysts exist in different phases than the reactants. the number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture– Catalysts are often solids.

EnzymesEnzymes• Enzymes are catalysts that occur naturally in living

organisms and are almost all protein molecules with typical molecular masses of 20,000–100,000 amu.

• Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism.

• Some are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings.

• A reactant in an enzyme-catalyzed reaction is called a substrate.

• Enzymes can increase reaction rates by enormous factors and tend to be very specific, typically producing only a single product in quantitative yield.

• Enzymes are expensive, and often cease functioning at temperatures higher than 37ºC, and have limited stability in solution.

• Enzyme inhibitors cause a decrease in the rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring.

• Thermodynamics

– Deals with state functions and can be used to describe the overall properties, behavior, and equilibrium composition of a system

– Provides a significant constraint on what can occur during a reaction process

• Kinetics– Concerned with the particular pathway by which

physical or chemical changes occur, so it can address the rate at which a particular process will occur

– Describes the detailed steps of what actually occurs on an atomic or molecular level

Comparing Comparing Thermodynamics and Thermodynamics and

KineticsKinetics


KineticsKinetics• The following table gives the numerical values of the equilibrium constant K that correspond to various values of Gº

– If Gº + 10 kJ/mol or Gº –10 kJ/mol, an equilibrium is ensured to lie all the way to the left or to the right, respectively

– If Gº is quite small (10 kJ/mol), significant amounts of both products and reactants are present at equilibrium

• Most reactions have equilibrium constants greater than 1, with the equilibrium strongly favoring either products or reactants

• In many cases, reactions that are strongly favored by thermodynamics do not occur at a measurable rate, and reactions that are not thermodynamically favored do occur under certain nonstandard conditions

• A reaction that is not thermodynamically spontaneous under standard conditions can be made to occur spontaneously by varying reaction conditions, by using a different reaction to obtain the same product, by supplying external energy, or by coupling the unfavorable reaction to another reaction for which Gº<< 0


KineticsKinetics

Rescuers Search for Six Missing From Georgia Sugar Refinery Blast AP Friday, February 08, 2008

PORT WENTWORTH, Ga. — Six people remained missing early Friday after an explosion and fire at a sugar refinery that left dozens injured.Officials had not determined what caused the explosion but said they suspect sugar dust, which can be volatile.

Feb. 7: Smoke billows from behind the main plant of the Imperial Sugar Company during a fire at the plant in Port Wentworth, Ga.

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