Chapter 6: Thermochemistry

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1 Chapter 6: Thermochemistry 9/27 Energy is the capacity to do work. Work: force X distance Potential Energy: available by virtue of an object s position Kinetic Energy: energy of motion Radiant Energy: in electromagnetic waves, (light) (kinetic) Thermal Energy: random motion of atoms and molecules (kinetic) Chemical Energy: store within chemical bonds (potential) Nuclear Energy: stored with atomic nuclei (potential) Energy changes in chemical reactions: Heat: transfer of thermal energy between two bodies that are at different temperatures. Temperature: measure of the thermal energy Temperature IS NOT the same as thermal energy Bathtub=lower temperature, greater thermal energy Coffee=higher temperature, lower thermal energy Thermal energy in an object depends on Temperature Mass Material Temperature vs. heat Analogy: bank account

2 How much money is in my account? How many transfers are in my account? Doesn t make sense. Consider a solution What is the temperature? How much heat is in it? Doesn t make sense How much energy? Kinetic energy=temperature Thermochemistry: study of heat change in chemical reactions System: the part of the universe being studied Exchanged=mass and energy Exothermic process: gives off heat thermal energy goes from system to surroundings 2H2 (g) + O2 (g) 2H2O (l) + penergy H2O (g) H2O (l) + energy Endothermic process: heat has to be supplied thermal energy goes from surroundings to system energy + 2HgO (s) 2Hg (l) + O2 (g) energy + H2O (s) H2O (l) Exothermic or exothermic? Making ice cubes from water=exo Conversion of frost to water vapor=endo Baking bread=endo Forming a He+ from He (g)=endo Nuclear fission=exo Cl2 2Cl=endo 2Na + S Na2S=exo Units of energy: Joule (J) 1 kg x m^2 x s^ 2

3 ~1 human heartbeat Lift 1 kg by 10 cm calorie (cal) Raise temperature of 1g of H2O by 1 C =4.184 J Calorie=1000 calories = 1 kcal Label on food information 37.2 J to cal 37.2 J 1 cal x = 8.81cal J cal to kj cal J 1 kj x x = kj 1 cal 1000 J kj to kcal kj 1000J 1 cal 1kcal x x x = kcal 1kJ J 1000 cal 29.4 cal to J 29.4 cal J x =123 J

4 1 cal What is energy? The ability to do work or produce heat Heat is microscopic Work is macroscopic Analogy: coffee in a fridge vs. pushing a desk Heat and work: the only ways to transfer energy Thermodynamics: State functions: properties that are determined by the state of the system, regardless of how that condition was achieved. ΔU=Ufinal Uinitial ΔP=Pfinal Pinitial ΔV=Vfinal Vinitial ΔT=Tfinal Tinitial 1st law of thermodynamics: energy can t be created or destroyed (can change form). ΔUsystem+ ΔUsurrounding = 0 or ΔUsystem= ΔUsurrounding C3H3 + 5O2 3CO2 + 4H2O Exothermic chemical reactions! Chemical energy lost by combustion=energy gained by the surroundings 9/30 ΔU= q + w ΔU is the change in internal energy of a system

5 q is the heat exchange between the system and surroundings w is the work done (or by) the system w= PΔV when a gas expands against a constant external pressure *******Table 6.1 Work done by the system: PxV=F x d^3=fd=w d^2 w=f x Δd P=F/A F=P x A w= P x A x Δd A x Δd =ΔV w= PΔV ΔV>0 PΔV<0 Wsys<0 *Work is not state function ΔW=Wf Wi Unit canceling w= PΔV (x)j=l x atm 1000cm^3=1000mL=1L=10^ 3m^3 1 atm=101.3x10^3 Pa=101.3x10^3N/m^2 N=kg x M s^2

6 x= J=1L x atm A gas expands from 1.6 L to 5.4 L (constant T). What is the work (in joules) (a) against a vacuum (b) against a pressure of 3.7 atm? w= PΔV (a) ΔV=5.4L 1.6L=3.8L P=0atm w= 0atm x 3.8 L = 0Latm=0joules (b) ΔV=5.4L 1.6L=3.8L P=3.7atm w= 3.7atm x 3.8 L = 14.1 Latm w= 14.1Latm x J = 1430J Latm 50g of water is cooled from 30 C to 15 C, losing 3140 J of heat. What is the energy change? ΔU= q + w ΔU= 3140 J w= PΔV=0 A balloon is heated by adding 900 J of heat. Expansion does 422 J of work on the atm. What is the energy change? ΔU= q + w ΔU= 900J 422J=478J 42.6L of gas expands 48.2L following addition of 1060 J of heat at 1.0 atm. What is the energy change? ΔU= q +w ΔU=1060J

7 w= PΔV w= (1atm)(48.2L 42.6L) w= (1atm)(5.6L) w= 5.6Latm J x = 570J Latm ΔU=1060J+( 570J) ΔU=490J 30.2 L of gas expands to 84.2 L following addition of 512 J of heat at 1.2 atm. What is the energy change? ΔU=q + w ΔU=512J w= PΔV ΔV=84.2L 30.2L=54.0L w= (1.2atm)(54.0L) w= 64.8Latm J x = 6600J Latm ΔU=512J 6600J= 6100J or 6.1kJ 612 J of heat is removed from 37.2 L of gas at 0.80 atm and the volume becomes 20.0L. What is the energy change? ΔU=q + w ΔU= 612J w= PΔV ΔV=20.0L 37.2L= 17.2L w= (0.80atm)( 17.2L)

8 w=14latm J x = 1400J Latm ΔU= 612J+1400J=800J 10/2 Enthalpy (ΔH) and the first law of thermodynamics At constant pressure: q=δh and w= PΔV ΔU=ΔH PΔV or ΔH=ΔU+PΔV Enthalpy (H) Definition: Thermodynamic potential Like U Measure of a system s potential energy Mathematically: H=U + PV ΔH=ΔU + PΔV (at constant pressure) Units: J or cal Recall: ΔU=q + w = q PΔV ΔH=q PΔV + PΔV ΔH=q *But only at constant P! Differences between q and ΔH Imagine: P1V1 P2V2 T1 T2 Lots of q/w combinations Heat and work are NOT state functions ΔH: Same regardless of path Enthalpy IS a state function Defined in terms of U, P, and V

9 ΔH=Hf Hi ΔH is intrinsic Amount matters Enthalpy and the first law of thermodynamics ΔU= q + w At constant pressure: q=δh and w= PΔV ΔU=ΔH PΔV ΔH=ΔU+PΔV Constant P: often good enough for chemistry Gas expands, 18 KJ work (done on surroundings) 79 kj heat released ΔH? = 79kJ ΔU? = ΔH PΔV = 79kJ 18kJ ΔU = 97kJ 418 J heat added P=1.12atm V:5.261L 6.189L ΔH? = 418J ΔU? = ΔH PΔV = 418J (1.12atm)(0.928L) =418J 105J J x = 105J Latm

10 ΔU =313J Exothermic: heat given off by the system to the surroundings ΔH<0 products<reactants Endothermic: heat is absorbed by the system from the surroundings ΔH>0 products>reactants 6.01 kj are absorbed for every 1 mole of ice that melts at 0 C and 1 atm. Endothermic reactions, positive, system absorbs heat kj are released for every 1 mole of methane that is combusted at 25 C and 1 atm. Exothermic reaction, negative, system gives off heat. Thermochemical equations Coefficients: always of moles of a substance H2O (s) H2O (l) ΔH=6.01kJ Reverse the reactions, change the sign of ΔH H2O (l) H2O (s) ΔH 6.01kJ Multiply both sides of the equation: ΔH changes bi same factor 2H2O (s) 2H2O (l) ΔH=2 x 6.01= 12.0kJ Note: change in state change in enthalpy! Thermochemical equations The physical states of all reactants and products must be specified in the thermochemical equations Heat evolved during combustion of 266 g of P4? P4 (s) + 5O2 (g) P4O10 (s) ΔH= 3013kJ 266gP4 1mol P kj x x = 6470kJ 123.9g P4 1mol P4 C6H6 (l) 3C2H2 (g) ΔH= +630kJ

11 ΔH for 3C2H2 (g) for C6H6 (l)? = 630kJ ΔH for making 1 mole C2H2 (g)? = 210kJ Which is favored? Forward or reverse? If C6H6 (g) were used, effect on ΔH? more negative, less positive 10/4 C4H10 (g) + 13/2 O2 (g) 4CO2 (g) + 5H2O (g) ΔH= kJ ΔH when balanced with whole numbers? kJ ΔH for 4CO2(g)+5H2O(l) C4H10 (g)+13/2 O2(g)? ΔH=3877kJ Which is favored? Forward or reverse? If liquid H2O were made, effect on ΔH? more negative 2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) ΔH= 367.5kJ Form 1.00 mol 25 C, 1.00 atm ΔU? ΔU=ΔH PΔV V=nRT/P ΔV=(1.00mol)(0.0821Latm/Kmol)(278.15K)/(1atm) ΔV=24.5L PΔV=(24.5L)(1atm)X 101.3J = 2480J or 2.48kJ ΔU= 367.5kJ 2.48kJ= 370.0kJ Specific heat (s): heat (q) required to raise the temperature of a 1 gram of a substance by 1 C Heat capacity: (C): heat (q) required to raise the temperature of a given mass (m) of the substance by 1 C C=ms Heat (q) absorbed or released: q=msδt q=cδt ΔT=(Tfinal Tinitial) What is the heat capacity of 75.0g of graphite? C=ms=(75.0g)(0.720J/g C)=54.0J/ C

12 How much heat is absorbed if 75.0g of graphite is warmed from 25.0 C to 95 C? q=msδt q=(54.0j/ C)(75 C)=3780J How much heat is given off when a 634 g copper bar cools from 64 C to 2 C? scu=0.385j/g C ΔT= 62 C q=msδt q=(634g)(0.385j/g C)( 62 C)= 15,000J 15,000J is given off How much heat is absorbed when 28 moles of ethanol is warmed from 9 C to 63 C? Molar mass of C2H2O=1300g/mol ΔT=(63 C ( 9 C)=72 C q=msδt q= (1300g)(2.46J/g C)(72 C)=230,000J or 230kJ Calculate the final temperature when 100g of Al at 40.0 C is immersed in 60.0g of H2O at 10.0 C. Assume no loss of heat to the container or other surroundings. qwater= qal (60.0g)(4.184J/g C)(Tf 10 C)=(100g)(0.900J/g C)(Tf 40 C) 251J/ C(Tf 10 C)=90J/ C(Tf 40 C) 90J/ C 90J/ C 2.79(Tf 10 C)=Tf 40 C 2.79Tf+27.9 C=Tf 40 C 2.79Tf= 67.9 C Tf=17.9 C

13 Constant volume calorimetry qsys=qwater+qbomb+qrxn qsys=0 qrxn= (qwater+qbomb) qcal=ccalδt Reaction at constant volume ΔH=qrxn ΔH~qrxn If 4.00 CH6N2 is combusted at constant volume, the temperature goes from C to C. If Ccal is kj/ C, what is the heat of reaction for 1 mol CH6N2? 2CH6N2+ 5O2 2N2 + 2CO2 + 6H2O qrxn= Ccal X ΔT ΔT=14.5 C =6.794kJ/ C(14.5 C) qrxn= 113kJ g CH6N2 x = 1.30x10^3kJ/molCH6N2 4.00g CH6N2 1 mol CH6N2 ~ΔHrxn Constant pressure calorimetry qsys=qwater+qcal+qrxn qsys=0 qrxn= (qwater +qcal) qwater=msδt qcal=ccal ΔT Reaction at constant P ΔH=qrxn

14 10/ g of NaOH are dissolved in mL of water at constant pressure. The temperature goes from 23.6 C to 47.4 C. What is the ΔH for the process in kj/mol NaOH? ΔH=qrxn= (smδt) =(4.184J/g C)(109.55g)(2.38 C) = 10,900J 1kJ 40.00gNaOH x x = 436kJ/mol 9.55g NaOH 1000J 1 mol NAOH No way to measure absolute enthalpy (H) of a substance. Measure the enthalpy change for every reaction of interest? Arbitrary scale: reference point standard enthalpy of formation (ΔH f). (ΔH f): heat change when one mole of a compound is formed from its elements at a pressure of 1 atm. (ΔH f) of any element in its most stable form=zero (ΔH f)(o2)=0 (ΔH f) (03)=142kJ/mol (ΔH f) (C, graphite)= 0 (ΔH f) (C, diamond)=1,90kj/mol Standard enthalpy of reaction (ΔH rxn) enthalpy of a reaction carried out at 1 atm. aa + bb cc + dd ΔH rxn=σnδh f(products) ΣmΔH f(reactants) The direct method for calculating ΔH rxn Make the compound from its elements In nearly 100% yield Look it up on table 4P(white) + 5O2 P4O10

15 = kj/mol Indirect method: Synthesis from elements doesn t work Side reactions Too slow/fast/dangerous Hess s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps. (Enthalpy is a state function. It does not matter how you get there, only where you start and end.) Arrange the equations to cancel everything except the overall reaction. Steps: 1. Write the enthalpy of formation reaction 2. Add the given rxns so that the result is the desired rxn P4O6 (s) + 2O2(g) P4O10 (s) P4(s) + 3O2 (g) P4O6 (s) ΔH= kJ P4(s) + 5O2 (g) P4O10 (s) ΔH= kJ P4O6 (s) P4(s) + 3O2 (g) ΔH=1640.1kJ P4(s) + 5O2 (g) P4O10 (s) ΔH= kJ ΔH= kJ 3H2 (g) + O3(g) 3H2O (l) 2H2(g) + O2 (g) 2H2O (l) ΔH= 483.6kJ 3O3 (g) 2O3 (g) ΔH=284.6kJ 2H2(g) + O2 (g) 2H2O (l) ΔH= 483.6kJ X 1.5 2O3 (g) 3O2 (g) ΔH=284.6kJ X 0.5

16 ΔH= 867.7kJ 2SO2 (g) + O2 (g) 2SO3 (g) SO2= 296.9kJ SO3= 395.2kJ O2=0 ΔH rxn=σnδh f(products) ΣmΔH f(reactants) =2( 395.2) 2( 296.9) 0 = = 196.6kJ 10/9 CH4 (g) + 4Cl2 (g) CCl4 (l) + 4HCl (g) CH4: 74.8kJ Cl2: 0kJ CCl4: 139.3kJ HCl: 92.30kJ =[4( 92.3kJ)+( 139.3kJ)] ( 74.8kJ) = 433.7kJ Enthalpy of solution (ΔHsoln): heat generated/absorbed when a certain amount of solute dissolves in a certain amount of solvent. ΔHsoln=Hsoln Hcomponets ^ Measure at constant pressure The aqueous solution process for ionic compounds overall: NaCl (s) Na+ (aq) + Cl (aq) Ionic bonds are broken (endothermic) Ions are solvated (hydrated)(exothermic) We can imagine two steps (Hess s law): 1. Ionic lattice breaks down Lattice energy (U): energy needed to break the ionic solid into gaseous ions (endothermic) 2. Ions get surrounded by water molecules Heat of hydration (ΔHhydr): energy produced when ions get surrounded by water (exothermic)

17 U + NaCl (s) Na+ (g) + Cl (aq) Na+ (g) + Cl (g) Na+ (aq) + Cl (aq) + ΔHhydr NaCl (s) Na+ (aq) + Cl (aq) ΔHsoln Calculate ΔHsoln: If 1 mol of NaCl is dissolved in water NaCl (s) Na+ (aq) + Cl (aq) From ΔH Table: NaCl (s): kj/mol Na+ (aq): kJ/mol Cl (aq): 167.2kJ/mol =[( kJ/mol)+( 167.2kJ/mol)] ( 411.0kJ/mol) =4.1kJ/mol Calculate ΔHsoln: If 50.g of FeCl3 is dissolved in water FeCl3 (s) Fe3+ (aq) + 3Cl (aq) From ΔH Table: FeCl3 (s): kj/mol Fe3+ (aq): 47.7kJ/mol Cl (aq): 167.2kJ/mol =[( 47.7kJ/mol)+3( 167.2kJ/mol)] ( 400.0kJ/mol) = 149.3kJ 1 mol FeCl3 50.0g FeCl3 x x = 46.1kJ mol g FeCl3 1 Which substance could be used for melting ice? exo LiCl: 37.1 CaCl2: 82.8(best) Which substances could be used for a cold pack? endo NaCl: 4.0(maybe) KCl: 17.2 NH4Cl: 15.2 NH4NO3: 26.6(best)

18 Heat of dilution: heat change when a solution is diluted: (more solvent is added). If ΔHsoln is endothermic (positive), dilution absorbs heat If ΔHsoln is exothermic (negative), dilution produces heat H2SO4 (l) H+ (aq) + HSO4 (aq) ΔHrxn= ( 811.3)= (exothermic) Very conc (98%) dilute (10%) = lots of heat! Always Add Acid

Thermodynamics. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Thermodynamics. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.