Chemical EquationChemical Equation

A condensed shorthand A condensed shorthand statement which expresses a statement which expresses a chemical reaction in terms of chemical reaction in terms of formulas and symbols.formulas and symbols.

Indicates the reactants and the Indicates the reactants and the product/s of a chemical reactionproduct/s of a chemical reaction

A balanced equation gives a A balanced equation gives a quantitative relationship of the quantitative relationship of the reactants and the products.reactants and the products.

Chemical ReactionsChemical Reactions

CombinationCombinationDecompositionDecompositionDisplacementDisplacementReplacementReplacement

CombinationCombination

simplest conceivable reactionsimplest conceivable reactiona reaction in which 2 or more a reaction in which 2 or more

substances unite to form a single substances unite to form a single compoundcompound

ex.ex. Fe + S Fe + S FeS FeS

HH22O + SOO + SO33 H H22SOSO44

HH22O + NaO + Na22O O 2NaOH 2NaOH

CaO + SiOCaO + SiO22 CaSiO CaSiO33

2FeCl2FeCl22 + Cl + Cl22 2FeCl 2FeCl33

DecompositionDecomposition

reverse of combination reactionreverse of combination reactiona reaction in which a substances a reaction in which a substances

yields two or more products.yields two or more products.breaking up of binary compounds breaking up of binary compounds

into its elementsinto its elementsex.ex. 2HgO 2HgO 2Hg + O 2Hg + O22

2KClO2KClO33 2 KCl + 3O 2 KCl + 3O22

CaCOCaCO33 CaO + CO CaO + CO22

DisplacementDisplacement

a reaction in which compounds a reaction in which compounds combine freeing one of its combine freeing one of its constituent element constituent element

ex. Zn + 2HCl ex. Zn + 2HCl ZnCl ZnCl22 + H + H22

Fe + CuSOFe + CuSO44 FeSO FeSO44 + Cu + Cu

ClCl22 + NaBr + NaBr 2NaCl + Br 2NaCl + Br22

CaCOCaCO33 + SiO + SiO22 CaSiO CaSiO33 + CO + CO22

ReplacementReplacement

a reaction in which there is an a reaction in which there is an interchange of elements (or interchange of elements (or radicals) between two compoundradicals) between two compound

ex. AgF + NaCl ex. AgF + NaCl AgCl + NaF AgCl + NaF

AgNOAgNO33 + NH + NH44Cl Cl AgCl + NH AgCl + NH44NONO33

3Ca(NO3Ca(NO33))22 + 2Na + 2Na33POPO44

CaCa33(PO(PO44))22 + 6NaNO + 6NaNO33

BaClBaCl22 + NaSO + NaSO44 BaSO BaSO44 + 2NaCl + 2NaCl

Other Types of ReactionsOther Types of Reactions

CombustionCombustion NeutralizationNeutralization Endothermic/ Endothermic/

ExothermicExothermic Oxidation/ Oxidation/

ReductionReduction PrecipitationPrecipitation

CombustionCombustion

Combustion is a vigorous Combustion is a vigorous and exothermic reaction and exothermic reaction that takes place between that takes place between certain substances certain substances (particularly organic (particularly organic compounds) with oxygencompounds) with oxygen

ValenceValence

property of an elementproperty of an elementcombining capacities of combining capacities of

many elementsmany elementsassociated with the number associated with the number

of electrons in the outermost of electrons in the outermost principal quantum level of an principal quantum level of an atom atom

RadicalsRadicals

Groups of atoms that Groups of atoms that remains intact throughout remains intact throughout many chemical reactions many chemical reactions and which exhibit constant and which exhibit constant valence in all compounds valence in all compounds in which it is present. in which it is present.

StoichiometryStoichiometry

Greek for “Greek for “measuring measuring elementselements””

The calculations of quantities The calculations of quantities in chemical reactions based in chemical reactions based on a balanced equation.on a balanced equation.

We can interpret balanced We can interpret balanced chemical equations several chemical equations several ways.ways.

In terms of In terms of ParticlesParticles

Element - atomsElement - atomsMolecular compound Molecular compound

(non-metals) - molecule(non-metals) - moleculeIonic Compounds (metal Ionic Compounds (metal

and non-metal) - formula and non-metal) - formula unit unit

2H2H2 2 + O+ O22 2H2H22O O

Two molecules of hydrogen and one molecule Two molecules of hydrogen and one molecule of oxygen form two molecules of water.of oxygen form two molecules of water.

2 Al2 Al22OO33 AlAl ++ 3O3O22

2formula unitsAl2O3 form 4 atoms Al

and3moleculesO2

2Na + 2H2O 2NaOH + H2

Look at it differentlyLook at it differently

2H2H2 2 + O+ O22 2H2H22OO2 dozen molecules of hydrogen and 1 2 dozen molecules of hydrogen and 1

dozen molecules of oxygen form 2 dozen dozen molecules of oxygen form 2 dozen molecules of water.molecules of water.

2 x (6.02 x 102 x (6.02 x 102323) molecules of hydrogen ) molecules of hydrogen and 1 x (6.02 x 10and 1 x (6.02 x 102323) molecules of ) molecules of oxygen form 2 x (6.02 x 10oxygen form 2 x (6.02 x 102323) molecules ) molecules of water.of water.

2 moles of hydrogen and 1 mole of 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.oxygen form 2 moles of water.

In terms of In terms of MolesMoles

2 Al2 Al22OO33 AlAl ++ 3O3O22

2Na + 2H2Na + 2H22O O 2NaOH + H 2NaOH + H22

The coefficients tell us how The coefficients tell us how many moles of each kindmany moles of each kind

36.04 g H2 + O236.04 g H2O

In terms of massIn terms of mass

The law of conservation of mass appliesThe law of conservation of mass appliesWe can check using molesWe can check using moles

2H2H2 2 + O+ O22 2H2H22OO

2 moles H2

2.02 g H2

1 moles H2

= 4.04 g H2

1 moles O2

32.00 g O2

1 moles O2

= 32.00 g O2

In terms of massIn terms of mass

2H2H2 2 + O+ O22 2H2H22OO

2 moles H2O18.02 g H2O

1 mole H2O=36.04 g H2O

2H2 + O2 2H2O

36.04 g (2H2 + O2)= 36.04 g H2O

Your turnYour turn

Show that the following Show that the following equation follows the Law equation follows the Law of conservation of mass.of conservation of mass.

2 Al2 Al22OO33 AlAl ++ 3O3O22

Mole to Mole conversionsMole to Mole conversions

How many moles of OHow many moles of O22 are produced are produced when 3.34 moles of Alwhen 3.34 moles of Al22OO33 decompose? decompose?

2 Al2 Al22OO33 AlAl ++ 3O3O22

3.34 moles Al2O3 2 moles Al2O3

3 mole O2

= 5.01 moles O2

Your turnYour turn 2C2C22HH22 + 5 O + 5 O22 4CO 4CO22 + 2 H + 2 H22OO

If 3.84 moles of CIf 3.84 moles of C22HH2 2 are burned, how are burned, how many moles of Omany moles of O22 are needed? are needed?

If 2.47 moles of CIf 2.47 moles of C22HH2 2 are burned, how are burned, how many moles of COmany moles of CO22 are formed? are formed?

How many moles of CHow many moles of C22HH2 2 are needed to are needed to produce 8.95 mole of Hproduce 8.95 mole of H22O?O?

Mass in Chemical Mass in Chemical Reactions Reactions

For example...For example... If 10.1 g of Fe are added to a solution of If 10.1 g of Fe are added to a solution of

Copper (II) Sulfate, how much solid Copper (II) Sulfate, how much solid copper would form?copper would form?

Fe + CuSOFe + CuSO44 Fe Fe22(SO(SO44))33 + Cu + Cu

2Fe + 3CuSO2Fe + 3CuSO44 Fe Fe22(SO(SO44))33 + 3Cu + 3Cu

10.1 g Fe55.85 g Fe

1 mol Fe= 0.181 mol Fe

2Fe + 3CuSO2Fe + 3CuSO44 Fe Fe22(SO(SO44))33 + 3Cu + 3Cu

0.181 mol Fe2 mol Fe

3 mol Cu= 0.2715 mol Cu

0.2715 mol Cu1 mol Cu63.55 g Cu

= 17.25 g Cu

Another way of doing it ...Another way of doing it ...

10.1 g Fe55.85 g Fe1 mol Fe

2 mol Fe3 mol Cu

1 mol Cu63.55 g Cu

= 17.24 g Cu

ExamplesExamples

To make silicon for computer chips To make silicon for computer chips they use this reactionthey use this reaction

SiClSiCl44 + 2Mg + 2Mg 2MgCl 2MgCl22 + Si + Si

How many grams of Mg are needed How many grams of Mg are needed to make 9.3 g of Si?to make 9.3 g of Si?

How many grams of SiClHow many grams of SiCl44 are are needed to make 9.3 g of Si?needed to make 9.3 g of Si?

How many grams of MgClHow many grams of MgCl2 2 are are produced along with 9.3 g of silicon?produced along with 9.3 g of silicon?

The U. S. Space Shuttle boosters The U. S. Space Shuttle boosters use this reactionuse this reaction

3Al3Al(s)(s) + 3NH + 3NH44ClOClO44 Al Al22OO33 + AlCl + AlCl33 + 3NO + 6H + 3NO + 6H22OO

How much Al must be used to react How much Al must be used to react with 652 g of NHwith 652 g of NH44ClOClO44 ? ?

How much water is produced?How much water is produced?How much AlClHow much AlCl3 3 is produced?is produced?

More ExamplesMore Examples

If 6.45 moles of water are If 6.45 moles of water are decomposed, how many decomposed, how many liters of oxygen gas will be liters of oxygen gas will be produced at STP?produced at STP?

And still another And still another exampleexample

To compute it...To compute it...

If 6.45 grams of water are decomposed, If 6.45 grams of water are decomposed, how many liters of oxygen will be how many liters of oxygen will be produced at STP?produced at STP?

2H2H22O O 2H 2H22 + O + O22

6.45 g H2O 18.02 g H2O

1 mol H2O2 mol H2O

1 mol O2

1 mol O2

22.42 L O2

4.01 L O2

Your TurnYour Turn

How many liters of COHow many liters of CO22 at STP will at STP will be produced from the complete be produced from the complete combustion of 23.2 g Ccombustion of 23.2 g C44HH10 10 ? ?

What volume of oxygen will be What volume of oxygen will be required?required?

Did you get itDid you get it

2C2C44HH1010 + 13O + 13O22 8CO 8CO22 + 10H + 10H22OO

23.2 g C4H10 58.1 g C4H10

1 mol C4H10

2 mol C4H10

8 mol CO2

1 mol CO2

22.42 L CO2

35.81 L CO2

Did they got itDid they got it

2C2C44HH1010 + 13O + 13O22 8CO 8CO22 + 10H + 10H22OO

23.2 g C4O10 58.1 g C4H10

1 mol C4H10

2 mol C4H10

13 mol O2

1 mol O2

22.42 L O2

58.19 L O2

Gases and ReactionsGases and Reactions

QuizQuiz

How many liters of CHHow many liters of CH4 4 at STP are required at STP are required to completely react with 17.5 L of Oto completely react with 17.5 L of O2 2 ??

CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO

17.5 L O2 22.4 L O2 1 mol O2

2 mol O2 1 mol CH4

1 mol CH4 22.4 L CH4

= 8.75 L CH4

22.4 L O2 1 mol O2

1 mol CH4 22.4 L CH4

Avogadro told usAvogadro told us

Equal volumes of gas, at the same Equal volumes of gas, at the same temperature and pressure contain temperature and pressure contain the same number of particles.the same number of particles.

Moles are numbers of particlesMoles are numbers of particlesYou can treat reactions as if they You can treat reactions as if they

happen liters at a time, as long as happen liters at a time, as long as you keep the temperature and you keep the temperature and pressure the same. pressure the same.

ExampleExample

How many liters of COHow many liters of CO2 2 at STP are at STP are produced by completely burning 17.5 produced by completely burning 17.5 L of CHL of CH4 4 ??

CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO

17.5 L CH4 1 L CH4 1 L CO2 = 17.5 L CO2

Limiting ReagentLimiting Reagent

If you are given one dozen loaves of If you are given one dozen loaves of bread, a gallon of mustard and three bread, a gallon of mustard and three pieces of salami, how many salami pieces of salami, how many salami sandwiches can you make.sandwiches can you make.

The limiting reagent is the reactant The limiting reagent is the reactant you run out of first.you run out of first.

The excess reagent is the one you The excess reagent is the one you have left over.have left over.

The limiting reagent determines how The limiting reagent determines how much product you can makemuch product you can make

How do you find out?How do you find out?

Do two stoichiometry problems.Do two stoichiometry problems.The one that makes the least The one that makes the least

product is the limiting reagent.product is the limiting reagent.For exampleFor exampleCopper reacts with sulfur to form Copper reacts with sulfur to form

copper ( I ) sulfide. If 10.6 g of copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how copper reacts with 3.83 g S how much product will be formed? much product will be formed?

If 10.6 g of copper reacts with 3.83 g If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be S. How many grams of product will be formed?formed?

2Cu + S 2Cu + S Cu Cu22SS

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting Reagent

Your turnYour turnIf 10.1 g of magnesium and 2.87 L of If 10.1 g of magnesium and 2.87 L of

HCl gas are reacted at STP, how HCl gas are reacted at STP, how many liters of gas will be produced?many liters of gas will be produced? .128.128

Identify the limiting reagent.Identify the limiting reagent.How many grams of solid is How many grams of solid is

produced?produced?6.16.1

How much excess reagent How much excess reagent remains?remains?8.5448.544

Your Turn AgainYour Turn Again

If 10.3 g of aluminum are If 10.3 g of aluminum are reacted with 51.7 g of CuSOreacted with 51.7 g of CuSO44 how much copper will be how much copper will be produced?produced?

How much excess reagent How much excess reagent will remain?will remain?

Yield Yield

The amount of product made in a The amount of product made in a chemical reaction.chemical reaction.

There are three typesThere are three typesActual yieldActual yield- what you get in the lab - what you get in the lab

when the chemicals are mixedwhen the chemicals are mixedTheoretical yieldTheoretical yield- what the balanced - what the balanced

equation tells you you should make.equation tells you you should make.Percent yieldPercent yield = Actual x 100 % = Actual x 100 %

Theoretical Theoretical

ExampleExample

6.78 g of copper is produced 6.78 g of copper is produced when 3.92 g of Al are reacted when 3.92 g of Al are reacted with excess copper (II) sulfate.with excess copper (II) sulfate.

2Al + 3 CuSO2Al + 3 CuSO4 4 Al Al22(SO(SO44))33 + + 3Cu3Cu

What is the actual yield?What is the actual yield?What is the theoretical yield?What is the theoretical yield?13.8513.85

What is the percent yield?What is the percent yield?4949

Energy in Chemical Energy in Chemical ReactionsReactionsHow Much?How Much?

In or Out?In or Out?

Chemical Chemical ThermodynamicsThermodynamics

EnergyEnergy

Energy is measured in Joules or Energy is measured in Joules or caloriescalories

Every reaction has an energy Every reaction has an energy change associated with itchange associated with it

ExothermicExothermic reactions release reactions release energy, usually in the form of heat.energy, usually in the form of heat.

EndothermicEndothermic reactions absorb reactions absorb energyenergy

Energy is stored in bonds between Energy is stored in bonds between atoms atoms

C + OC + O22 CO CO22En

erg

y

Reactants Products

C + O2

C + O2

395kJ

+ 395 kJ

In terms of bondsIn terms of bonds

COO C

O

O

Breaking this bond will require energy

CO

OOO C

Making these bonds gives you energyIn this case making the bonds gives you more energy than breaking them

ExothermicExothermic

The products are lower in The products are lower in energy than the reactantsenergy than the reactants

Releases energyReleases energy

CaCO3 CaO + CO2En

erg

y

Reactants Products

CaCO3

CaO + CO2

176 kJ

CaCO3 + 176 kJ CaO + CO2

EndothermicEndothermic

The products are higher in The products are higher in energy than the reactantsenergy than the reactants

Absorbs energyAbsorbs energy

Chemistry Happens inChemistry Happens in

MOLESMOLESAn equation that includes energy is An equation that includes energy is

called a called a thermochemical equationthermochemical equationCHCH44 + 2 O + 2 O22 CO CO22 + 2 H + 2 H22O + 802.2 kJO + 802.2 kJ

1 mole of CH1 mole of CH44 makes 802.2 kJ of makes 802.2 kJ of energy.energy.

When you make 802.2 kJ you make When you make 802.2 kJ you make 2 moles of water2 moles of water

CHCH44 + 2 O + 2 O22 CO CO22 + 2 H + 2 H22O + 802.2 kJO + 802.2 kJ

If 10. 3 grams of CHIf 10. 3 grams of CH44 are burned are burned completely, how much heat will be completely, how much heat will be produced?produced?

10. 3 g CH4

16.05 g CH4

1 mol CH4

1 mol CH4

802.2 kJ

=514 kJ

CHCH44 + 2 O + 2 O22 CO CO22 + 2 H + 2 H22O + 802.2 kJO + 802.2 kJ

How many liters of OHow many liters of O22 at at STP would be required to STP would be required to produce 23 kJ of heat?produce 23 kJ of heat?

How many grams of How many grams of water would be produced water would be produced with 506 kJ of heat?with 506 kJ of heat?

EnthalpyEnthalpy

The heat content a substance has at The heat content a substance has at a given temperature and pressurea given temperature and pressure

Can’t be measured directly because Can’t be measured directly because there is no set starting pointthere is no set starting point

The reactants start with a heat The reactants start with a heat contentcontent

The products end up with a heat The products end up with a heat contentcontent

So we can measure how much So we can measure how much enthalpy changesenthalpy changes

EnthalpyEnthalpy

Symbol is HSymbol is HChange in enthalpy is Change in enthalpy is HHdelta Hdelta HIf heat is released the heat content of If heat is released the heat content of

the products is lowerthe products is lowerH is negative (exothermic)H is negative (exothermic)If heat is absorbed the heat content of If heat is absorbed the heat content of

the products is higherthe products is higherH is positive (endothermic)H is positive (endothermic)

En

erg

y

Reactants Products

Change is down

H is <0

En

erg

y

Reactants Products

Change is upH is > 0

Heat of ReactionHeat of Reaction

The heat that is released or absorbed in The heat that is released or absorbed in a chemical reactiona chemical reaction

Equivalent to Equivalent to HHC + OC + O22(g)(g) CO CO22(g)(g) +393.5 kJ +393.5 kJ

C + OC + O22(g)(g) CO CO22(g)(g) H = -393.5 kJH = -393.5 kJ

In thermochemical equation it is In thermochemical equation it is important to say what stateimportant to say what state

HH2(g)2(g) + 1/2O + 1/2O2 (g)2 (g) H H22OO(g)(g) H = -241.8 kJH = -241.8 kJ

HH2(g)2(g) + 1/2O + 1/2O2 (g)2 (g) H H22OO(l)(l) H = -285.8 kJH = -285.8 kJ

Heat of CombustionHeat of Combustion

The heat from the reaction that The heat from the reaction that completely burns 1 mole of a completely burns 1 mole of a substancesubstance

CC22HH4 4 ++ 33 OO2 2 2 CO 2 CO22 + 2 H + 2 H22O O

CC22HH6 6 ++ OO2 2 CO CO22 + H + H22OO

2 C2 C22HH6 6 ++ 5 O 5 O2 2 2 CO 2 CO22 + 6 H + 6 H22OO

CC22HH6 6 ++ (5/2) O (5/2) O2 2 CO CO22 + 3 H + 3 H22OO

Standard Heat of FormationStandard Heat of Formation

The The H for a reaction that produces H for a reaction that produces 1 mol of a compound from its 1 mol of a compound from its elements at standard conditionselements at standard conditions

Standard conditions: 25°C and 1 atm.Standard conditions: 25°C and 1 atm.Symbol isSymbol is H f

0

The standard heat of formation of an element is 0

This includes the diatomics

What good are they?What good are they?

There are tables of heats of There are tables of heats of formationsformations

The heat of a reaction can be The heat of a reaction can be calculated by subtracting the heats calculated by subtracting the heats of formation of the reactants from of formation of the reactants from the productsthe products

H = H (products ) - H (reactants)f0

f0

ExamplesExamples

• CHCH4(g)4(g) + + 22 OO22(g) (g) CO CO22(g)(g) + 2 H + 2 H22OO(g)(g)

H f0

CH4 (g) = -74.86 kJH f

0O2(g) = 0 kJ

H f0

CO2(g) = -393.5 kJ

H f0

H2O(g) = -241.8 kJ H= [-393.5 + 2(-241.8)]-[-74.68 +2 (0)] H= 802.4 kJ

Why Does It Work?Why Does It Work?

If HIf H22(g)(g) + 1/2 O+ 1/2 O22(g)(g) H H22O(g) O(g) H=-285.5 kJH=-285.5 kJthen: then:

HH22O(g) O(g) HH22(g)(g) + 1/2 O+ 1/2 O22(g) (g) H =+285.5 kJH =+285.5 kJ

If you turn an equation around, you If you turn an equation around, you change the signchange the sign

2 H2 H22O(g) O(g) HH22(g)(g) + O+ O22(g) (g) H =+571.0 kJH =+571.0 kJ

If you multiply the equation by a If you multiply the equation by a number, you multiply the heat by that number, you multiply the heat by that number.number.

Why does it work?Why does it work?

You make the products, so you need You make the products, so you need their heats of formationtheir heats of formation

You “unmake” the products so you have You “unmake” the products so you have to subtract their heats. to subtract their heats.

Practice ProblemDiborane, B2H6 is a highly reactive boron hydride. Calculate the H for the synthesis of diborane from its elements according to the equation

2 B (s) + 3 H2 (g) B2H6 (g) and the following data:

2B(s) + 3/2O2(g) B2O3 -1273 kJB2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) -2035 kJH2(g) + 1/2O2(g) H2O(l) -286 kJH2O(l) H2O(g) 44 kJ

2B(s) + 3/2O2(g) B2O3 -1273 kJ

B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g)

-2035 kJ

H2(g) + 1/2O2(g) H2O(l) -286 kJ

H2O(l) H2O(g) 44 kJ

2 B (s) + 3 H2 (g) B2H6 (g)

Hess’s Law

In going from a particular set of reactants to a particular set of products, the enthalpy change is the same whether the reaction takes place in one step or in a series of steps

QuizQuiz

Find the enthalpy change for the Find the enthalpy change for the chemical reaction given below:chemical reaction given below:

2 SO2 SO3(g) 3(g) 2 SO 2 SO2(g)2(g) + O + O2(g)2(g)

SO3 (g)

SO2 (g)

-396 kJ/mol

-297 kJ/mol

H f0

2(-297) - [ 2(-396)] = 198

SO3 (g)

SO2 (g)

-396 kJ/mol

-297 kJ/mol

H f0

198 kJ/mol

1st Law of Thermodynamics

Energy can neither be created nor destroyed

the capacity to do work or cause heat flow

Types of Energy

Potential Energy - energy due Potential Energy - energy due to position or compositionto position or composition

Kinetic Energy - energy due to Kinetic Energy - energy due to motion and depends on mass motion and depends on mass and velocity of the matter.and velocity of the matter.

2nd Law of Thermodynamics

In any spontaneous process there is always an increase in the entropy of the universethe entropy of the universe is

increasing

Entropy

A thermodynamic functionA thermodynamic functionSymbol is S Symbol is S It is a measure of It is a measure of

randomness or disorderrandomness or disorderA driving force for A driving force for

spontaneous processspontaneous process

Entropy (S)Entropy (S)

Change in entropy is Change in entropy is S (delta S)S (delta S)If the reaction is accompanied by If the reaction is accompanied by

production of heat production of heat S is positiveS is positiveIf the reaction involves heat If the reaction involves heat

absorption absorption S is negativeS is negativeIf the reaction neither produces If the reaction neither produces

nor absorbs heat nor absorbs heat S is 0S is 0

Change in Entropy (Change in Entropy (S)S)

• S > 0 S > 0 spontaneous reactionspontaneous reaction• S < 0 S < 0 not spontaneous reactionnot spontaneous reaction• S = 0 S = 0 system at equilibriumsystem at equilibrium

Factors Affecting Factors Affecting SS

Heat flowHeat flow TemperatureTemperature

SSunivuniv = = SSsyssys + + SSsurrsurr

Consider liquid water, H2O (l)

H2O (l) H2O (g) 44 kJ

SSsyssys SSsurrsurr ssunivuniv spontaneous?spontaneous?

+ + + Yes

– – – No

+ – Ssys>Ssurr Yes

+ – Ssys<Ssurr No

– + Ssys>Ssurr No

– + Ssys<Ssurr Yes

What have you What have you learned?learned?

Ssurr is temperature dependent

Ssurr =HT

Practice ProblemPractice Problem

Calculate the Calculate the SSsurrsurr for the for the

evaporation of water at 25 °C evaporation of water at 25 °C and at 100 °C. and at 100 °C. .15.12.15.12

H2O (l) H2O (g) 44 kJ

Gibbs Free EnergyGibbs Free Energy

Free EnergyFree Energy

Symbol is G (in honor of Josiah Symbol is G (in honor of Josiah Willard Gibbs)Willard Gibbs)

Another thermodynamic functionAnother thermodynamic functionrelated to the spontaneity of a related to the spontaneity of a

reaction or processreaction or processuseful in dealing with temperature useful in dealing with temperature

dependence of spontaneitydependence of spontaneitydefined as G = H - TSdefined as G = H - TS

Change in Free Energy (Change in Free Energy (G)G)

Change in free energy is Change in free energy is G (delta G)G (delta G)

G = G = H - TH - TSS

G° = G° = H° - TH° - TS°S°

Suniv =GT

Note: A process at constant T and P is spontaneous in the direction in which the free energy decreases

Case Result

SS ppoossiittiivvee,, HH nneeggaaiittiivvee

Spontaneous at all temperature

SS ppoossiittiivvee,, HH ppoossiittiivvee

Spontaneous at high temperatures

SS nneeggaattiivvee,, HH nneeggaattiivvee

Spontaneous at low temperature

SS nneeggaattiivvee,, HH ppoossiittiivvee

NOT spontaneous at any temperature

BrBr2 (l)2 (l) Br Br2 (g)2 (g)

At what temperature is the process spontaneous at 1 atm?H° = 31.0 kJ/mol S° = 93.0 J/K•mol

G° = H° - TS° =

T = H°/S°

= Ans. = T > 333 °K or T > 60 °K

HH22OO (s) (s) H H22OO (l) (l)

At what temperature is the process spontaneous at 1 atm?H° = 6033 J/mol S° = 22.1 J/K•mol

G° = H° - TS° =

T = H°/S°

= Ans. = C

HH22OO (l) (l) H H22OO (g) (g)

At what temperature is the process spontaneous at 1 atm?H° = 44.4 kJ/mol S° = 119 J/K•mol

G° = H° - TS° =

T = H°/S°

= Ans. = C

Chemical Chemical KineticsKinetics

Chemical Chemical ReactionReaction

Can be classified into several Can be classified into several types like neutralization, types like neutralization, precipitation, redox, etc.precipitation, redox, etc.

Affected by several factors like Affected by several factors like nature of reactants, pressure, nature of reactants, pressure, temperature, etc.temperature, etc.

It is defined or represented by It is defined or represented by its reactants and productsits reactants and products

Chemical Chemical KineticsKinetics

The area of chemistry that is involved in rates of chemical reactions

Reaction Rates?Reaction Rates?

defined as the change in concentration of reactant or product per unit time

Rate =t

Concentration (mol/L)Time

( 1 s) NO2 NO O2

0 0.0100 0.0000 0.000050 0.0079 0.0021 0.0011

100 0.0065 0.0035 0.0018

150 0.0055 0.0045 0.0023

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

(g)(g)

Calculate the average rate at which the concentration of NO2 changes over the first 50 seconds.Rate =

t

t

=

]t = s - ]t = s

50 s - 0 s=

= - 4.2 x 10-5 M/s

Collision modelCollision model A model based on the A model based on the

idea that atoms/molecules idea that atoms/molecules must collide to reactmust collide to react

The model used to The model used to account for the observed account for the observed behavior and behavior and characteristics of reaction characteristics of reaction rates.rates.

Activation Activation EnergyEnergy

The threshold energy that The threshold energy that must be overcome to must be overcome to produce a chemical produce a chemical reactionreaction

2 BrNO2 BrNO (g) (g) 2 NO 2 NO (g)(g) + Br + Br2 (g)2 (g)

2 Br-N bonds must be broken2 Br-N bonds must be broken 1 Br-Br bond must be formed1 Br-Br bond must be formed the kinetic energy possessed the kinetic energy possessed

by the reacting molecules, by the reacting molecules, BrNO + BrNO is required to BrNO + BrNO is required to break the bond and rearrange break the bond and rearrange the atoms to form the productsthe atoms to form the products

Pote

nti

al

Energ

y

Reaction Progress

}E

Ea

2 BrNO

2NO + Br2

ON----Br

ON----Br

2 BrNO

2NO + Br2

bond breaking

bond forming

Activated Activated Complex or Complex or

Transition StateTransition State

The arrangement of The arrangement of atoms found at the top atoms found at the top of the potential energy of the potential energy barrier as reaction barrier as reaction proceeds to completionproceeds to completion