2 Chemical Reactions involve: reactants turned into productsan energy change in form of heatEnergy changes accompany all chemical reactions and are due to rearranging of chemical bonding.
3 The addition of energy is always a requirement for breaking of bonds but the breaking of bonds in and of itself does not release energy.Energy release occurs whenNEW BONDS are formed.
4 If more energy is released when new bonds form than was required to break existing bonds, then thedifference will result in an overall release of energy….EXOTHERMICIf more energy is required to break existing bonds thenis released when new bonds form,then energy is being absorbed…..ENDOTHERMIC
5 the study of energy transformations THERMODYNAMICS:the study of energy transformationsTo study thermodynamics (thermochemistry)situations need to be viewed a little differently……..System : All substances under going a physical orchemical changeSurroundings : everything else that is not a partof the system
6 System: all of the reactants and products Concentrated Sulfuric acid is added todistilled water in a beaker.H2SO4(l) + H2O(l) HSO4-(aq) + H3O+ (aq)System: all of the reactants and productsSurroundings: water molecules NOT being used,beaker, air…… etc.( this is an exothermic process, heat is given off….what does that mean? )
7 Energy is a property that is determined by specifying the condition or “state” ( temperature, pressure, etc)of a system or substance.(is NOT really solid, liquid or gas….)That’s why its called a STATE FUNCTION(where its value is determined only by itspresent condition and not on its history)Ohio
8 EXOTHERMIC REACTION Energy is – Given Off! Activation Energy Energy Reaction CoordinateActivation EnergyH < 0Energy is –Given Off!
9 Energy is + Absorbed! ENDOTHERMIC H > 0 Activation Energy Products Reaction CoordinateActivation EnergyH > 0ReactantsProductsENDOTHERMICEnergy is +Absorbed!
10 Internal Energy of a System Internal energy can be increased by:an increase in temperaturea phase changethe initiation of a chemical reactionInternal energy can be Decreased by:a decrease in temperature
11 ∆E = Efinal - Einitial Total Energy of a System cannot be determined, but the CHANGE may be measured.∆E = Efinal - Einitial
12 The Laws of Thermodynamics First Energy cannot be created or destroyed but only transferred from one body to another or changed from one form to another. Second Every spontaneous change increases the entropy of the universe/system. Third The entropy of a perfect crystalline substance (no disorder) is zero at 0 K.
13 ∆Esystem = q + w In a chemical system, the energy exchanged between a system and its surroundings can be accounted for byheat (q) and work (w).∆Esystem = q + w∆E < 0 (negative) means system gave away energy to surroundings∆E > 0 (positive) means system absorbed energy from surroundingsIf q < 0 then system gave away heat to surroundings (-)If w < 0 then system did work on surroundings (-)ALL ENERGY MEASURED IN JOULES!
14 2NH3(g) N2(g) + 3H2(g)More moles of gas are produced than reacted. Thissystem has done work on its surroundings. The systemis ‘pushing back’ to make room for the expanding gas.
15 VERY IMPORTANT AND USEFUL: Only pressure/volume work (expansion or contractionof a gas) is of importance in a chemical reactionsystem, and only when there is an increase or decreasein the amount of of gas present!
16 For q: + means the system gained heat - means the system lost heat w=∆E-Depends on values of q and w-For q: + means the system gained heat- means the system lost heatFor w : + means work is done ON system- means system does work ON surroundings
17 Useful when volumes of gases are shown in a More on Work; w ………The expansion/contraction of a gas against constantexternal pressure can be expressed:w = -P∆VUseful when volumes of gases are shown in achemical reaction.conversion needed: J/liter atmosphere**** If there is no change in the total volume of gasbefore and after a reaction, then there is no workdone by OR on the system!****
18 ∆E = q Remember! To DO work, something MUST MOVE! If a reaction occurs in a rigid container, only the pressure,NOT the volume may change. If no volume changeoccurs, nothing moved…… NO WORK is done on orby the system!That means that any ∆E change is due ONLY to heattransfers. So……∆E = q
19 B) AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) Example:For the following reactions performed atconstant external pressure, is work done on thesystem by the surroundings, by the systemon the surroundings or is the amount of work negligible?A) Sn(s) + 2F2(g) SnF4 (s)B) AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)C) C(s) + O2(g) CO2(g)D) SiI4(g) + heat Si(s) + 2I2(g)
21 ∆H = q (at constant pressure) Most times chemical reactions occur where the externalpressure is constant….most often at atmospheric pressure and in aqueous formsHere, a pressure change or volume change does nothappen. So the heat exchange is the importantthing to consider…….∆H = q (at constant pressure)∆H is ENTHALPY, the exchange of heat betweena system and its surroundings at constant pressureCommonly called: heat of reaction
22 The amount of ∆H in any given reaction is directly proportional to the quantity ofreactants AND depends on the conditionsthat the reaction takes place.So… ∆H is associated with a balancedchemical equation and the temperature andpressure are specified and units of enthalpychange is given inJoules/mole
23 A thermochemical equation: 2HgO(s) + heat 2Hg(l) + O2(g) ∆H = kJ(endo)K2O(s) + CO2(g) K2CO3(s) ∆H = kJ (exo)Also:HgO(s) + heat Hg(l) + ½ O2(g) ∆H = kJ2K2O(s) + 2CO2(g) 2K2CO3(s) ∆H = kJWhat’s going on here with these diff equations?
24 Standard temperature = 25oC One special set of conditions used withchemical equations are the Standard StateConditions:Standard temperature = 25oCStandard pressure = 1 atmStandard concentration = 1 molarWhen these conditions are met, then the ∆Hmeasure is calledStandard Enthalpy of Reaction: ∆Horxn(has an o by the letter)
25 Standard Enthalpy of Reaction: ∆Horxn The amount of energy associated withwith reactions at standard states has beencalculated for many reactions and can befound on lists of ∆Horxn(some equation data can be made availablefrom the q equation at standard conditions)If the sign of the kJ/mol is negative thenthe reaction is exothermic.
26 First, let’s look at heat used/given off If there is no list available to you, then thesenumbers can be calculated in a variety of ways.First, let’s look at heat used/given offby the formation of compounds fromtheir elements.
27 Standard Enthalpy of Formation (∆H°f ) First Way:Standard Enthalpy of Formation (∆H°f )change in enthalpy (heat) that accompanies the formation of 1 mole of that substance from its elements, with all substances in their standard states
28 The standard enthalpy of formation for ethanol (C2H5OH):2C(s) + 3H2(g) + 1/2O2(g) C2H5OH(l)Hof = kJ/molonly elements as reactantsonly 1 mole of product made
29 Remember: the standard enthalpy of the formation of element has gotto be zero.Why is that?
30 H = Hproducts - Hreactants Second Way:Another way of calculating H is to usecombinations of Hof of multiple compoundsH = Hproducts - HreactantsFe2O CO 2Fe + 3CO2Calculate the standard enthalpy of abovereaction using this Hof data:CO = kJ/mol CO2 = kJ/molFe2O3 = kJ/mol
31 Fe2O3 + 3 CO 2Fe + 3CO2 H = Hproducts - Hreactants Hrxn = [ 3(-393.7)] – [3(-110.4) + (-822.2)]No #’s needed for the 2Fe!
32 According to Hess’s law, we can do some Third Way:What if asked to find Hof for SrCO3?But there isn’t a listing for enthalpy offormation for SrCO3 from its elements?According to Hess’s law, we can do somecreative equation manipulation!
33 Add these equations together and get: Sr + 3/2 O2 + C SrCO3
34 Hess's LawThe enthalpy change for any reaction depends on the products andreactants and is independent of the pathway or the number of steps between the reactant and product.
35 Find other reactions that are easier to perform and when summed up, equal the original desired reaction.According to Hess's Law,the sum of the heat flows for these reactions is equal to the heat flow for the original reaction.
36 Find the standard enthalpy of formation of N2H4N2 + 2O2 2NO Horxn= +67.4N2H4 + 3O2 2NO2 + 2H2O Horxn=2H2 + O2 2H2O Horxn =We need:N2 + 2H2 N2H4(coefficient of 1 before N2H4)
38 Another kind of equation that can be used Fourth Way:Another kind of equation that can be usedis the heat of combustion: HocCombustion is the adding of oxygen toa compound or element. When using ahydrocarbon, usually CO2 and H2O are formed.Combustion of CO: CO + 1/2O2 CO2Combustion of Al : 2Al + 3/2O2 Al2O3You may have to create this equation if giventhe kJ/mol of heat.
39 Must write the equations for combustion for each of these items Calculate the standard enthalpy of formationof ethane, C2H6, given the followingcombustion data; C = -393, H2 = -286 andC2H6 = -1560kJ/molMust write the equations for combustionfor each of these items2. Arrange those elements to give you:2C + 3H2 C2H63. Add up the correct kJ’s
41 Hrxn = ΣHbonds broken - ΣHbonds formed Fifth Way:Still another way to describe energy changes in compound includes the energy changes from breaking and forming bondsHrxn = ΣHbonds broken - ΣHbonds formedadd up the bonds broken and formed ina given equation and that can equal theheat of reaction.(pgs in text)
42 Bond Energy (kJ/mol) C=O 743 H--H 436 H-C 412 C--C 348 H-N 391 N--N 170H-O463O--O145H-F565F--F154H-Cl427Cl-Cl239H-Br366Br-Br193H-I299I--I151C-CC=C612C--N305CºC837C--O360C--S272C--F485O=O498C--Cl339C--Br276C--I238NºN945C=O
45 Spontaneous Processes and Entropy Why does a chemical reaction go naturally inone particular direction?A spontaneous process is one that, once initiated,continues to proceed in the forward directionwith no further input of energy.Using certain characteristics of a reaction, we candescribe the spontaneity of the reaction.
47 What are the forces that could lend a reaction to becomespontaneous or nonspontaneous?
48 Entropymeasure of disorderMovement from order to disorder[Many times overcoming a high positiveactivation energy (high + H)]
49 Degree of disorder of a system = S Called EntropyIncrease in disorder (entropy) willresult in + quantity.Decrease will be – quantity.Spontaneous reactions usually have + valuefor Entropy
50 Entropy can take many forms: In terms of Solid, Liquid and Gas Phases, thephases that provide more random motion wouldhave greater entropy.Within a gas sample, the sample that has the highertemperature or volume would provide morerandom motion and have greater entropy.Substances that are made of larger particles comparedto substances of smaller particles, have greaterentropy.Mixtures of gases, solids and solutions are usuallyviewed has having greater entropy.
51 Entropy is a STATE function……. Means it depends on the quantity and conditionsof the substance.More substance reacting = more entropy
52 ∆Srxn = Sproducts – Sreactants Calculating EntropyThere are starting points of Entropy for substances. Measured at the standard state conditions. These are listed in appendix of the text. (So )H2O(l) H2O(g)Is liquid water more or less ordered than gaseous water?∆Srxn = Sproducts – Sreactants∆Srxn = ( ) J/mol = 119 J/mol+ Entropy means greater disorder!
53 Consider this spontaneous reaction: N H2 2NH3Why is this Spontaneous,Is the product more or less ordered?
54 N2 + 3H2 2NH3 Using : Srxn = Sproducts – Sreactants = J/molNegative sign says the product is more ordered.Entropy is not the reason this reaction occurs.so what DID cause this reaction to be spontaneous?
55 Suniv = Sreaction + Ssurr To determine the spontaneity of a process,more than a few things need consideration.Second (Law of Thermodynamics)Every spontaneous change increases the totalentropy of a system and its surroundings.For a chemical process to be spontaneous,it must increase the entropy of the universe!Suniv = Sreaction + Ssurr+ Suniv = Spontaneous Suniv = NOT Spontaneous(What is meant by the term….. Universe, here?)
56 -H Ssurr = T The entropy of the surroundings can be calculated: We know how to calculate the Entropy of the reaction….The entropy of the surroundingscan be calculated:-HSsurr =(Sometimes q)T
57 Suniv = Srxn + Ssurr Putting these two ideas together….. Suniv = [Sproducts -Sreactants] + -HT(let’s see this in action…….)
58 Consider this spontaneous situation: 2H2(g) + O2(g) 2H2O(l) kJ/mole∆So for the following are:H2(g) = 131 J/mol O2(g)= 205J/mol H2O(l) = 70 J/molS = Sproducts – Sreactants∆Sorxn = [2(70)] – [2(131)+ (205)]∆Sorxn = J/mole (lower entropy)(Seems correct, water is more ordered than the two gases)
59 Why is this reaction spontaneous? Disorder is not increased! What’s the force, here?-HrxnSsurr =-(-572 kJ/mol)298K=TThis equals 1919 Joules/mole (converted to Joules)Need to calculate the entropy for the universe……..(OH! A lot bigger than -327!)
60 Suniv = Srxn + Ssurr Suniv = -327 J/mole + 1919 J/mole Answer: J/mole+ answer means that entropy for the overallprocess, including system AND surroundingsDID, in fact, increase!Suniv > so IS spontaneous
61 ∆Gorxn ∆Gorxn = ∆Horxn – (T)∆Sorxn Further study of these calculations and processes leadto a culmination formula that connects ∆H, heat or enthalpyand ∆S, entropy or disorder:This is another way to calculate/provethat a reaction is spontaneous……Gibbs Free Energy: The amount of energy that isavailable to do work on the UNIVERSE in a chemicalreaction.∆Gorxn∆Gorxn = ∆Horxn – (T)∆Sorxn∆Gorxn < 0 FOR spontaneous reactions( WORK? What work?)
62 2H2(g) + O2(g) 2H2O(l) + 572 kJ/mole (where’s the work?)2H2(g) + O2(g) 2H2O(l) kJ/moleRemember negligible work?There is not much energy available to do work on thissystem; three moles of gas on one side, NONE on theother. Negligible to zero work done on the system.Most of the energy here is given off as enthalpy, not work.But according to the Second Law, for this reaction toproceed as spontaneous, enough energy as work must be givenaway to the surroundings so that the increase in entropy ofthe surroundings offsets the decrease in the entropy of thesystem.( how much energy is that?)
63 ∆Gorxn = ∆Horxn – (T)∆Sorxn 2H2(g) + O2(g) 2H2O(l) kJ/mole∆Hosurr > (0.327kJ/mol K given to surroundings) * (298K) =97kJ/mole available to do work on surroundings∆Gorxn = ∆Horxn – (T)∆Sorxn∆Gorxn = needs to be negative to have aspontaneous reaction∆Gorxn = kJ/mol – [ (298K)(-0.327kJ/mole)]∆Gorxn = kJ/molNumber is – so reaction is spontaneous
64 ∆Gorxn = ∆Horxn – (T)∆Sorxn For any reaction to be spontaneous the sole conditionis that Gibbs Free Energy needs to be negative!∆Gorxn = ∆Horxn – (T)∆Sorxn
65 Go is negative for any reaction for which Ho is Go is negative for any reaction for which Ho is negative and So is positive. Go is negative for any reaction where both the enthalpy and entropy terms are negative.
66 NO Yes Keys to determining why a reaction happens H Sreaction G + at high temp+ at low temp-- at low temp+ at high tempNOYes
67 Our description of why a reaction occurs sometimes needs to include HOWneed to describe a step-wise progressionon exactly how a compound is formed.needs to include all energies along the wayThe Born Haber Cycle(normally associated with an ionically bonded compound)
68 Include these energy changes as needed: Standard enthalpy of formationStandard enthalpy of vaporization (substance into gas)First Ionization energy ( gas losing an electron)Enthalpy of bond dissociation (molecules to atoms)First electron affinity (gas gaining an electron)Enthalpy of Lattice (energy associated with onemole of solid is formed from its gaseous ions)
69 Formation of LiF Li(s) + ½ F2(g) (pgs. 363 – 367 of text) Li+ (g) + F (g)Li+ (g) + ½ F2(g)77kJ-328 kJLi+ (g) + F- (g)520kJLi (g) + ½ F2(g)161kJLi(s) + ½ F2(g)-1047 kJOverall -617 kJLiF
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