Thermochemistry

Heat energy transferred between two objects as a result of the temperature difference between them.

Temperature A measure of kinetic energy ·                    if molecules move slowly (seldom hit thermometer) then “cold” ·                    if hit frequently then “hot”

1st Law of Thermodynamics The energy of the universe is constant. i.e. the energy of the universe is conserved Many kinds of energy They are interconvertible Chemical, light, heat, electrical, etc

E = Efinal  Einitial  E if energy leaves system + E if energy enters system Note the E of a system doesn’t depend on how system got there -- i.e. it is a state function When we look at energy we must define a system    System has energy (E)   if no energy could enter or leave a system then E = 0 (because energy is conserved) but this is not possible so we look at E of system where E = Efinal  Einitial giving a  E if energy leaves system a + E if energy enters system Note the E of a system doesn’t depend on how system got there -- i.e. it is a state function

State Function A function or property whose value depends only on the present state (condition) of the system, not on the path used to arrive at that condition. Examples of state functions distance between points dollars in checkbook altitude of airplane   Examples of non-state functions ·         Distance traveled ·         Work done to get dollars So now what is E (E) and how do we measure it?

E = q + w Heat gain or loss Work done = -PDV Matches our earlier convention that Ein is + and Eout is – Work   What happens when a chemical reaction occurs? either 1.     energy gained or 2.     energy lost Historically, energy gain and loss is measured by looking at temperature and volume change. (1st engineers were interested in the amount of work that could be done by a steam engine by the expansion of water into steam.) E = q + w   heat gain or loss work done =- PV Matches our earlier convention that Ein is + and Eout is --.   So given a reaction volume  so PV =  temperature  so q =  chemical energy has been converted into work and energy. We can quantify al follows w = (2 atm)(2L) = 4 Latm = 404J Conversion factor 1 atm = 101 x 105 kg/m sec2 1 Joule = 1 kg m2/sec2

Enthalpy H = qP = E + PV H = Hfinal  Hinitial = Hproducts  Hreactants Enthalpy   Back to E = q + w = q + PV or the energy of a reaction is heat output + some work This can be rearranged to q = E + PV (note that at constant V, qV = E) H = enthalpy of a system H = change in enthalpy or defining a new term H = qP = E + PV = Enthalpy We will be working with enthalpy as chemists. Why do we worry about all of these changing definitions?? To be exact -- E and H are not the same -- but most reactions are done at atmospheric pressure and the work done is to push back the atmosphere. This is small (typically, about 2 KJ/1000 KJ of energy generated) So E and H are very close in value but H is easier to measure and they both give us essentially the same information So lets study enthalpy --- H = Hfinal  Hinitial or = Hproducts  Hreactants What does H tell us? for a reaction -- H = + means H of reactants < H of products endothermic heat is brought in from outside H is  means H of reactants > H of products exothermic heat is lost

Tells how much heat is required to change the temp of a substance. Some specific heats are Al 0.902 J/g oK Cu 0.385 J/g oK H2O 4.184 J/g oK   Quantity of heat supplied Tells how much heat is required to change the temp of a substance. Temperature change (always Tf-Ti) The higher the specific heat of a substance, the more heat required to heat it or cool it.   Lakes and other large bodies of water must absorb (or lose) much heat to change their temperature -- this is why they are so important to the weather -- they can store much heat. Also here a swimming pool takes many warm days to become warm -- doesn’t change temperature as quickly as air -- higher heat capacity (more E per T) When we wrap food in foil -- foil is cool to the touch but open and food is very hot Al has low heat capacity, it loses heat quickly and without much heat transfer. McDonalds warns -- “caution: filling is hot” in apple pies C is higher for apples (lots of water) than for crust.

A 55.0 g piece of metal was heated in boiling water to a temperature of 99.8oC and dropped into an insulated beaker with 225 mL of water (d = 1.00 g/ml) at 21.0 oC. The final temperature of the metal and water is 23.1oC. Calculate the specific heat of the metal assuming that no heat was lost to the surroundings.

Octane, C8H18, a primary constituent of gasoline, burns in air. C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(l) Suppose that a 1.00 g sample of octane is burned in a calorimeter that contains 1.20 kg of water. The temperature of the water and the bomb rises from 25.00oC to 33.20oC. If the specific heat of the bomb, Cbomb, is known to be 837 J/oC, calculate the molar heat of reaction of C8H18.

A quantity of ice at 0oC is added to 90. 0 g of water at 80oC A quantity of ice at 0oC is added to 90.0 g of water at 80oC. After the ice melted, the temperature of the water was 25oC. How much ice was added? specific heat of ice 2.06 J/goC 0.91 kJ/moloC specific heat of water 4.184 J/goC 7.54 kJ/moloC specific heat of steam 2.0 J/goC 0.92 kJ/moloC heat of fusion 333 J/g 6.01 kJ/mol heat of vaporization 2226 J/g 40.67 kJ/mol

50. 0 g of ice at -20. 0 oC are added to 342. 0 g of water at 86. 0 oC 50.0 g of ice at -20.0 oC are added to 342.0 g of water at 86.0 oC. What will be the final temperature of the sample? specific heat of ice 2.06 J/goC 0.91 kJ/moloC specific heat of water 4.184 J/goC 7.54 kJ/moloC specific heat of steam 2.0 J/goC 0.92 kJ/moloC heat of fusion 333 J/g 6.01 kJ/mol heat of vaporization 2226 J/g 40.67 kJ/mol

A 33. 14 g sample of copper and aluminum was heated to 119 A 33.14 g sample of copper and aluminum was heated to 119.25oC and dropped into a calorimeter containing 250.0 g of water at 21.00oC. The temperature rose to 23.05oC. Assuming no heat was lost to the surroundings, what is the percent copper in the sample?

Enthalpy Enthalpy transferred out of reactants  exothermic  H =  Enthalpy transferred into products  endothermic  H = + Lets go back to our definition of Enthalpy   Hreaction = Hproducts  Hreactants Enthalpy is a state function -- it doesn’t matter how you go from one place to another -- enthalpy and enthalpy changes are the same!! So in the reaction A  B H = ## Joules The H value is the same no matter how you get from AB

Enthalpy Hforward = Hreverse (For reversible reactions) H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ H2(g) + 1/2 O2(g)  H2O(g) H = 241.8 kJ

Enthalpy The H is proportional to the amount of substance undergoing change. H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ 2 H2O(g)  2 H2(g) + 1 O2(g) H = +483.6 kJ

Enthalpy The physical state of reactants and products is important. H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ H2O(l)  H2(g) + 1/2 O2(g) H = +285.8 kJ Additional energy must be put in to vaporize H2O

Enthalpy Enthalpy is a state function -- it doesn’t matter how you go from one place to another -- enthalpy and enthalpy changes are the same!! The H value is the same no matter how you get from AB

Hess’s Law The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Valid because enthalpy is a state function.

Determine the H for the sublimation of ice to water vapor at 0oC. H2O(s)  H2O(l) H = 6.02 kJ/reaction H2O(l)  H2O(g) H = 40.7 kJ/reaction ----------------------------------------------------- H2O(s)  H2O(g) H = 46.7 kJ/reaction

Calculate the enthalpy change for the formation of methane, CH4, from solid carbon (as graphite) and hydrogen gas. C(s) + 2 H2(g)  CH4(g) The enthalpies for the combustion of graphite, hydrogen gas and methane are given. C(s) + O2(g)  CO2(g) 393.5 kJ H2(g) + ½ O2(g)  H2O(l) 285.8 kJ CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) 890.3 kJ

Calculate the enthalpy change for the reaction S(s) + O2(g)  SO2(g) given 2 SO2(g) + O2(g)  2 SO3(g) H = 196 kJ 2 S(s) + 3 O2(g)  2 SO3(g) H = 790 kJ

Standard Heat of Formation The enthalpy change, Hfo, for the formation of 1 mol of a substance in the standard state from the most stable forms of its constituent elements in their standard states.

Benzene, C6H6, is an important hydrocarbon Benzene, C6H6, is an important hydrocarbon. Calculate its enthalpy of combustion; that is, find the value of Ho for the following reaction. C6H6(l)+15/2 O2(g)  6 CO2(g)+3 H2O(l) Given Hfo [C6H6(l)] = +49.0 kJ/mol Hfo [CO2(g)] = 393.5 kJ/mol Hfo [H2O(l)] = 285.8 kJ/mol

Nitroglycerin is a powerful explosive, giving four different gases when detonated. 2 C3H5(NO3)3(l)  3 N2(g) + ½ O2(g) + 6 CO2(g) + 5 H2O(g) Given the enthalpy of formation of nitroglycerin, Hfo, is 364 kJ/mol, calculate the energy liberated when 10.0 g of nitroglycerin is detonated.

Enthalpies from Bond Energies Calculate the enthalpy of formation of water vapor from bond energies. 2 H2(g) + O2(g)  2 H2O(g) (The experimental value is 241.8kJ/mol)

Oxygen difluoride, OF2, is a colorless, very poisonous gas that reacts rapidly and exothermically with water vapor to produce O2 and HF. Calculate the DHof for OF2. OF2(g) + H2O(g)  2 HF(g) + O2(g) DHorxn = -318 kJ The heats of formation for H2O(g) and HF(g) are -241.8 kJ/mol and -271.1 kJ/mol respectively.

Stoichiometry using Enthalpy Consider the following reaction: 2 Na(s) + Cl2(g) 2 NaCl(s) H = 821.8 kJ Is the reaction exothermic or endothermic? Calculate the amount of heat transferred when 8.0 g of Na(s) reacts according to this reaction.

Ba(OH)28H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(g) + 10 H2O(l) We generally expect that reactions evolving heat should proceed spontaneously and those that absorb heat should require energy to occur. Mix barium hydroxide and ammonium chloride Ba(OH)28H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(g) + 10 H2O(l) Requires energy (freezes H2O to get it) so why does it happen? Answer H is not the only thing that drives a reaction!!

Determination of H using Hess’s Law Hrxn is well known for many reactions, but it is inconvenient to measure Hrxn for every reaction. However, we can estimate Hrxn for a reaction of interest by using Hrxn values that are published for other more common reactions.  The Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps are added to lead to reaction of interest (indirect method). Standard conditions (25°C and 1.00 atm pressure). (STP for gases T= 0°C) 

Hess’s Law “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” - 1840, Germain Henri Hess (1802–50), Swiss

Calculation of H by Hess’s Law 3 C(graphite) + 4 H2 (g)  C3H8 (g) H= -104 3 C(graphite) + 3 O2 (g)  3 CO2 (g) H=-1181 4 H2 (g) + 2 O2 (g)  4 H2O (l) H=-1143 C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Appropriate set of Equations with their H values are obtained (or given), which containing chemicals in common with equation whose H is desired. These Equations are all added to give you the desired equation. These Equations may be reversed to give you the desired results (changing the sign of H). You may have to multiply the equations by a factor that makes them balanced in relation to each other. Elimination of common terms that appear on both sides of the equation .

Calculation of H by Hess’s Law 3 C(graphite) + 3 O2 (g)  3 CO2 (g) H=-1181 4 H2 (g) + 2 O2 (g)  4 H2O (l) H=-1143 C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104 Hrxn = + 104 kJ 1181 kJ 1143 kJ 2220 kJ C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

Calculation of H by Hess’s Law Calculate heat of reaction W + C (graphite)  WC (s) ΔH = ? Given data: 2 W(s) + 3 O2 (g)  2 WO3 (s) ΔH = -1680.6 kJ C (graphite) + O2 (g)  CO2 (g) ΔH = -393.5 kJ 2 WC (s) + 5 O2 (g)  2 WO3 (s) + CO2 (g) ΔH = -2391.6 kJ ½(2 W(s) + 3 O2 (g)  2 WO3 (s) ) ½(ΔH = -1680.6 kJ) W(s) + 3/2 O2 (g)  WO3 (s) ) ΔH = -840.3 kJ C (graphite) + O2 (g)  CO2 (g) ΔH = -393.5 kJ ½(2 WO3 (s) + CO2 (g)  2 WC (s) + 5 O2 (g)) ½ (ΔH = +2391.6 kJ) WO3 (s) + CO2 (g)  WC (s) + 5/2 O2 (g) ΔH = + 1195.8 kJ) W + C (graphite)  WC (s) ΔH = - 38.0

Hess’s Law Problem: Chloroform, CHCl3, is formed by the following reaction: Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g) Determine the enthalpy change for this reaction (ΔH°rxn), using the following: 2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g) ΔH°f = – 103.1 kJ/mol CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g) ΔH°rxn = – 890.4 kJ/mol 2 HCl (g) → H2 (g) + Cl2 (g) ΔH°rxn = + 184.6 kJ/mol C (graphite) + O2 (g) → CO2(g) ΔH°rxn = – 393.5 kJ/mol H2 (g) + ½ O2 (g) → H2O (l) ΔH°rxn = – 285.8 kJ/mol answers: a) –103.1 kJ b) + 145.4 kJ c) – 145.4 kJ d) + 305.2 kJ e) – 305.2 kJ f) +103.1 kJ This is a hard question. To make is easer give: C (graphite) + ½ H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔH°f = – 103.1 kJ/mol

Methods of determining H Calorimetry (experimental) Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method). Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method) Bond Energies used with Hess’s Law  Experimental data combined with theoretical concepts 

(3) Determination of H using Standard Enthalpies of Formation (Hf )  Enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.  C + O2  CO2 ∆Hf = -393.5 kJ/   Standard Enthalpy of formation Hf are measured under standard conditions (25°C and 1.00 atm pressure).

Calculation of H CH4(g) + O2(g)  CO2(g) + H2O(g) - C + 2H2(g)  CH4(g) ΔHf = -74.8 kJ/ŋ C(g) + O2(g)  CO2(g) ΔHf = -393.5 kJ/ŋ 2H2(g) + O2(g)  2H2O(g) ΔHf = -241.8 kJ/ŋ We can use Hess’s law in this way: H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients.   - n CH4(g) + n O2(g) n CO2(g) + n H2O(g) H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)] = - 560.5 kJ

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) H = nHf(products) - mHf(reactants) H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ Table of Standard Enthalpy of formation, Hf

(4) Determination of H using Bond Energies Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. This is the bond enthalpy. The bond enthalpy for a Cl—Cl bond, D(Cl—Cl), is measured to be 242 kJ/mol.

Average Bond Enthalpies (H) Average bond enthalpies are positive, because bond breaking is an endothermic process. NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.

Enthalpies of Reaction (H ) Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. In other words, Hrxn = (bond enthalpies of bonds broken)  (bond enthalpies of bonds formed)

Hess’s Law: Hrxn = (bonds broken)  (bonds formed) CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl) = [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)] = (655 kJ)  (759 kJ) = 104 kJ

Bond Enthalpy and Bond Length We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases.

2003 B Q3

2005 B

2002

Entropy The amount of randomness, or molecular disorder, in a system. S = more positive to indicate greater disorder.

Predict which has greater entropy O2(g) at 5 atm of O2 at 0.5 atm Br2(l) or Br2(g) 1 mol N2 (g) in 22.4 L or 1 mol N2(g) in 2.24 L CO2(g) or CO2(aq)

Predict entropy changes for freezing of one mole of water evaporation of 1 mol of Br2 precipitation of BaSO4 upon mixing of aqueous solutions of Ba(NO3)2 and H2SO4 2 C(s) + O2(g)  2 CO(g) 2 K(s) + Br2(l)  2 KBr(s) 2 MnO2(s)  2 MnO(s) + O2(g) O(g) + O2(g)  O3(g)

+ S is entropy favored  S is entropy disfavored  H is enthalpy favored + H is enthalpy disfavored

Gibbs Free Energy, G Determines whether a reaction is spontaneous and at what temperature it becomes spontaneous. Spontaneous -- A process that proceeds on its own with out any continuous external influence.

Energy Units 1 calorie = 4.184 J 1 food calorie = 1 Cal = 1 kcal = 1000 cal Given the reaction below for the combustion of glucose to form carbon dioxide and water, calculate the Calories/g for carbohydrates. C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l) DHrxn = -2801.6 kJ

M & M candies consist of 70% carbohydrates, 21% fat, and 4 M & M candies consist of 70% carbohydrates, 21% fat, and 4.6% protein as well as other ingredients that do not have caloric value. What quantity of energy is generated if 47.9 g of M&Ms (1 small package) were burned in a bomb calorimeter? How long will a I need to walk to use up the value of the M&Ms if 1 hour of walking uses up 400 Cal? 4 Cal/g carbs 4 Cal/g protein 9 Cal/g fat

G = H  TS If H = + and S =  never spontaneous G = + If H =  and S = + always spontaneous G =  If H = + and S = + or if H =  and S =  temperature determines spontaneity At T where G =  reaction is spontaneous At T where G = + reaction is nonspontaneous

Ca(s) + Cl2(g)  CaCl2(s) H = 59.8 kJ S = 273J/K G = H  TS spontaneous at low T non-spontaneous at high T, entropy takes precedence Reaction becomes spontaneous at temperature where G becomes zero -- or when G = zero reaction is spontaneous in neither direction -- equilibrium! 0 = 59.8 kJ  T(0.273kJ/K) T = 219K or 53oC Reaction is spontaneous below 53oC

NH3(g) + 2 O2(g)  HNO3(aq) + H2O(l) H = 413 kJ S = 386J/K 0 = 413kJ  T(0.386J/K) T = 1069K = 796oC Reaction is spontaneous below 796oC

C6H12O6(s)  2 C2H5OH(l) + 2 CO2(g) H = 70 kJ S = +780J/K 0 = 70kJ  T(+0.78J/K)  T = 90K spontaneous at all temperatures -- would need an impossible temperature to become non-spontaneous!!

C6H12(l) + 6 O2(g)  3 CO2(g) + 6 H2O(g) H =  S = + 6 CO2(g)+6 H2O(g) C6H12O6(s)+9 O2(g) H = + S = 