Thermochemistry. Thermochemistry

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1 Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk prperties matter and deduces a ew general laws It des nt require any knwledge/assumptins mlecules CHEM 1000A 3.0 Thermchemistry 2

2 Deinitins System The part the universe we chse t study Surrundings The rest the universe (nrmally we nly wrry abut the immediate surrundings) Prcess A physical ccurrence (usually invlving energy lw) CHEM 1000A 3.0 Thermchemistry 3 Additinal deinitins Open system A system where energy and matter can be exchanged with the surrundings Clsed system A system where energy but nt matter can be exchanged with the surrundings Islated system A system where neither energy nr matter can be exchanged with the surrundings CHEM 1000A 3.0 Thermchemistry 4

3 CHEM 1000A 3.0 Thermchemistry 5 Systems and Energy All systems will cntain energy In thermdynamics we are interested in the lw energy, particularly in the rms heat and wrk. Nte that heat and wrk ccur when there is a prcess. They nly exist when smething happens. The system has energy, (ten described as the capacity t d wrk), it des nt have heat r wrk. CHEM 1000A 3.0 Thermchemistry 6

4 Heat Heat Energy that is transerred between a system and its surrundings as a result temperature dierences Heat transer can change the temperature smething but it des nt always d (nly) that Heat transer can melt r vaprize material CHEM 1000A 3.0 Thermchemistry 7 In the case a material that des nt change phase, the increase in temperature a system T due t the input a given amunt heat q is given by q = c T heat (J) heat capacity temperature (J K -1 ) change(k) CHEM 1000A 3.0 Thermchemistry 8

5 The heat capacity is a cnstant that depends n the system. S it s nt particularly useul. It is better t be able t deine heat capacity in terms a particular cmpund q = nc T q = mc T Number mlar heat mass speciic heat mles capacity (J K -1 ml -1 ) (kg) (J K -1 kg -1 ) CHEM 1000A 3.0 Thermchemistry 9 Units The SI unit heat is a Jule (since it is an energy) The lder unit heat is the calrie which is deined as the heat required t raise the temperature 1 g water 1 O C 1cal = J CHEM 1000A 3.0 Thermchemistry 10

6 Sign cnventin and cnservatin energy When heat lws between a system and its surrundings, we deine: q t be psitive i heat is supplied t the system q t be negative i heat is withdrawn rm the system I there are n phase changes, cnservatin energy requires that q system + q surrundings = 0 r q system = -q surrundings CHEM 1000A 3.0 Thermchemistry 11 Wrk Oten when a chemical reactin ccurs, wrk is dne. (this is the principle an engine) Since wrk and heat are bth rms energy we must cnsider bth. CHEM 1000A 3.0 Thermchemistry 12

7 Wrk Wrk is dne when a rce acts thrugh a distance. Fr example when a mass is mved. Deinitin w = Fxd Wrk (J) Frce (N) Distance (m) CHEM 1000A 3.0 Thermchemistry 13 The same deinitins apply r wrk as r heat: w is psitive i wrk is dne n the system w is negative i wrk is dne by the system CHEM 1000A 3.0 Thermchemistry 14

8 Pressure-Vlume wrk This is the mst cmmn type wrk. A sample gas is held by a pressure 2.4 atm. The pressure is then changed t 1.3 atm. The gas expands Hw much wrk is dne? (Nte that because the gas is expanding, the system is ding wrk n the surrundings s wrk must be negative) CHEM 1000A 3.0 Thermchemistry 15 Pressure-Vlume wrk w = -Fxd = -PxAxd = -P V w = -P V CHEM 1000A 3.0 Thermchemistry 16

9 First Law Thermdynamics We have deined a system, wrk and heat. A clsed system can exchange heat and wrk with its surrundings The system has sme energy we call the internal energy, U. CHEM 1000A 3.0 Thermchemistry 17 First Law Thermdynamics The irst law thermdynamics U = q + w CHEM 1000A 3.0 Thermchemistry 18

10 The internal energy is a state unctin. This means that its value depends n the state the system (its pressure, temperature etc) nt n hw it gt there Heat and wrk are nt state unctins. Their values depend n where the system has been, nt n where it is. CHEM 1000A 3.0 Thermchemistry 19 Where is the internal energy? It is nt necessary that we knw where the internal energy resides, But: It can be energy due t mtin in the mlecules e.g. translatinal energy (kinetic), r energy due t mlecular rtatinal and vibratin. Fr an ideal gas the internal energy is the kinetic energy 3RT E = 2 CHEM 1000A 3.0 Thermchemistry 20

11 Thermchemistry and the First Law We have U = q + w (Fr chemical reactins the mst cmmn rm heat is the heat reactin q rxn ) Fr a cnstant vlume prcess: U = q + w = q - P V But V= 0 therere U =q v The subscript indicates a cnstant vlume prcess CHEM 1000A 3.0 Thermchemistry 21 Mre reactins ccur at cnstant pressure cnditins than cnstant vlume. In this case the vlume the system can change and wrk can be dne. Suppse the system has t change rm ne state t anther. It can d it at cnstant vlume r cnstant pressure but the heat invlved is dierent CHEM 1000A 3.0 Thermchemistry 22

12 q v = q p + w But q v = U and w = -P V S U = q p -P V q p = U + P V U, P and V are state unctins s a cmbinatin them will als be a state unctin. Deine enthalpy as: H = U + PV H = U + PV At cnstant pressure H = U + P V s H = q p CHEM 1000A 3.0 Thermchemistry 23 2 CO(g) + O 2 (g) 6 2 CO 2 (g) At cnstant vlume U = q v = kj At cnstant pressure Wrk is dne n the system. Yu get the heat rm reactin and rm the wrk H = q p = kj CHEM 1000A 3.0 Thermchemistry 24

13 Enthalpy and change state There is usually a lw heat when there is a change state (slid t liquid, liquid t vapur) This is usually expressed as the enthalpy usin r enthalpy vaprizatin. E.g. H 2 O(s) 6 H 2 O(l) H=6.01 kj ml -1 H 2 O(l) 6 H 2 O(g) H=44.0 kj ml -1 This is ten just called the heat usin r heat vaprizatin. CHEM 1000A 3.0 Thermchemistry 25 Enthalpy/thermchemistry Thermchemistry is the branch thermdynamics that deals with energy changes in chemical reactins. Nrmally a chemical reactin is expressed with the assciated enthalpy stated H 2 (g) + ½O 2 (g) 6 H 2 O(l) H = kj exthermic Standard cnditins {P = 1 bar, T must be speciied (usually 298K)} CHEM 1000A 3.0 Thermchemistry 26

14 reactin reactin CHEM 1000A 3.0 Thermchemistry 27 Hess s Law Enthalpy is a state unctin. Therere we can can use sme bvius prperties t determine unknwn enthalpies 1. Enthalpy changes are prprtinal t the amunt material 2. The sign the enthalpy changes i a reactin is reversed 3. (Hess s Law) I a reactin can be thught as prceeding thrugh a number steps, then the enthalpy change n reactin must be the sum the enthalpy changes r each step CHEM 1000A 3.0 Thermchemistry 28

15 Fr ½N 2 (g) + O 2 (g) 6NO 2 (g) We can think the reactin prceeding as written Or in tw steps ½N 2 (g) + O 2 (g) 6NO(g) + ½O 2 NO(g) + ½O 2 6 NO 2 (g) CHEM 1000A 3.0 Thermchemistry 29 ½N 2 (g) + ½O 2 (g) 6NO(g) H = kj ml -1 NO(g) + ½O 2 6 NO 2 (g) H = kj ml -1 ½N 2 (g) + O 2 (g) 6NO 2 (g) H kj ml -1 CHEM 1000A 3.0 Thermchemistry 30

16 Standard enthalpies rmatin We have seen that the enthalpy a reactin is calculated as a dierence between the enthalpies the prducts and reactants. We d nt have an abslute scale r enthalpies (what is the enthalpy O 2? Fr example) Just as we measure heights abve sea-level, we measure enthalpies with respect t the standard state CHEM 1000A 3.0 Thermchemistry 31 Standard State By deinitin: the enthalpy an element in its mst stable rm at a pressure 1 bar r a speciied temperature is zer. (usually tabulated r 298K) Eg O 2 (g), Br 2 (l), C(graphite), Na(s) We calculate all enthalpies rm this standard state CHEM 1000A 3.0 Thermchemistry 32

17 We can nw deine a standard enthalpy rmatin as the enthalpy change that ccurs in the rmatin ne mle the substance rm its elements in their standard states. Usually given the symbl H where the is r rmatin and the indicates standard cnditins CHEM 1000A 3.0 Thermchemistry 33 Examples standard enthalpies rmatin C(gr)+ ½O 2 (g) 6 CO(g) H (CO(g))= kj ml -1 ½H 2 (g) + ½F 2 (g) 6 HF(g) H (HF(g))= kj ml -1 ½N 2 (g) + ½O 2 (g) 6 NO(g) H (NO(g))= kj ml -1 CHEM 1000A 3.0 Thermchemistry 34

18 Nw we have standard enthalpies rmatin we can determine the enthalpy a reactin. H reactin = H prducts H reactants ( prducts) ν H ( reactants) = ν H multiplied by their stichimetric ceicients ν. p where Σ is a sum ver the standard enthalpies rmatin H r CHEM 1000A 3.0 Thermchemistry 35 Where d H values cme rm? All enthalpies rmatin can be traced t calrimetry experiments. Mst widely used is the bmb calrimeter CHEM 1000A 3.0 Thermchemistry 36

19 The bmb calrimeter is a cnstant vlume reactr, usually used r cmbustin reactins. The reagents are the system and the calrimeter is the surrundings. By measuring the heat transerred int the surrundings the heat the reactin can be determined. CHEM 1000A 3.0 Thermchemistry 37 I the increase in temperature the surrundings is T then the heat ging int the surrundings is q calrimeter = c T Since the reactin takes place at cnstant vlume U =q rxn = - q calrimeter T get H rxn we use H rxn = U + PV And the ideal gas law H rxn = U + nrt Then we use H = ν H prducts ν H reactants rxn p (cnverted t per mle) ( ) ( ) CHEM 1000A 3.0 Thermchemistry 38 r

20 Example: Calculate the standard heat rmatin CH 4 i a bmb calrimeter experiment gives a measured heat gain in the calrimeter kj per mle CH 4 The cmbustin reactin is the ne that ully xidizes the reagents: CH 4 (g) + 2O 2 (g) 6CO 2 (g) + 2H 2 O(l) U =q rxn = - q calrimeter = kj ml -1 H rxn = U + nrt n= 1-3 = -2 H rxn = (-2)x8.314x298x 10-3 kj ml -1 = kj ml -1 H rxn = ν H p ( prducts) ν H ( reactants) CHEM 1000A 3.0 Thermchemistry 39 r H H rxn rxn H H = = H ν H p ( prducts) ν H ( reactants) ( CO ) + 2 H ( H O) H ( CH ) 2 2 = H ( CO2 ) + 2 H ( H2O) H ( CH 4 ) = ( ) H ( CH 4 ) -1 ( CH ) = 74.8 kj ml rxn CH 4 (g) + 2O 2 (g) 6CO 2 (g) + 2H 2 O(l) I we knw the heats rmatin CO 2 and H 2 O then: 4 r 4 CHEM 1000A 3.0 Thermchemistry 40

21 Bnd Energies The Bnd Dissciatin Energy is the energy t break a mle bnds. H 2 (g) 6 2H(g) H = D(H-H) = kj ml -1 Fr cases where there are mre than ne identical bnd, the average is taken H 2 O(g) 6 H(g) + OH(g) H = D(H-OH) = kj ml -1 OH(g) 6 H(g) + O(g) H = D(H-O) = kj ml -1 D(HO in H 2 O) = kj ml -1 CHEM 1000A 3.0 Thermchemistry 41 Bnd Energies CHEM 1000A 3.0 Thermchemistry 42

22 Bnd Energies Hess s law can be used t shw H rxn = H(bnd breakage) H(bnd rmatin) H rxn = Σ BDE(reactants) Σ BDE(prducts) This will never be as precise as using heats rmatin as BDEs are always averages CHEM 1000A 3.0 Thermchemistry 43

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