Lecture 5: Standard Enthalpies

Transcription

1 Lecture 5: Standard Enthalpies Reading: Zumdahl 9.6 Outline What is a standard enthalpy? Defining standard states. Using standard enthalpies t determine ΔH rxn Prblems Z9.54,Z9.55,Z9.58,Z9.63, Z9.76,Z9.79,Z9.85 1

2 Definitin f ΔH f We saw in the previus lecture the utility f having a series f reactins with knwn enthalpies. It wuld be nice t have a basic (systematic) set f reactins ne culd use t determine the heat f a reactin, rather than having t be clever t determine hw t add (arbitrary) whle reactins. What if ne culd deal with cmbinatins f cmpunds directly rather than dealing with whle reactins t determine DH rxn? Standard Enthalpy f Frmatin is The change in enthalpy that accmpanies the frmatin f 1 mle f a cmpund frm its elements with all substances at standard state. We envisin taking elements at their standard state, and cmbining them t make cmpunds, als at standard state. 2

3 What is Standard State? Standard State: A precisely defined reference state. It is a cmmn reference pint that ne can use t cmpare thermdynamic prperties. Definitins f Standard State: Fr a gas: P A = 1 atm. Fr slutins: [C A ] = 1 M (ml/l). Fr liquids and slids: pure liquid r slid Fr elements: The frm in which the element exists under cnditins f 1 atm and 298 K. (i.e. 25C) The text has a nice table summarzing this and a shrt discussin abut using the standard pressure as 1 bar rather than 1 Atm (which differ by 1%). We will nt cncern urselves with this distinctin. 3

4 Sample Standard States Standard elemental states (cnt.): Hydrgen: H 2 (g) (nt atmic H) Oxygen: O 2 (g) Carbn: C (gr). Graphite, slid, as ppsed t diamnd We will dente the standard state using the superscript. Example: ΔH rxn a reactin carried ut under standard state 4

5 Imprtance f Elements We will use the elemental frms as a primary reference when examining cmpunds. Pictrially, fr chemical reactins we envisin taking the reactants t the standard elemental frm, then refrming the prducts. 5

6 Example: Cmbustin f Methane CH ( g) + 2 O ( g) CO ( g) + 2 H O( l) Using Hess Law, and the elements as intermediates in reactins. Writing the abve (net) reactin as a sum f 4 separate reactins. 6

7 Elements and ΔH f Fr elements in their standard state: Δ = H f 0 elements With respect t the graph, we are envisining chemical reactins as prceeding thrugh elemental frms. In this way we are cmparing ΔH f fr reactants and prducts t a cmmn reference (zer) reactants ΔH rxn prducts q 7

8 Tabulated Values f ΔH f In Zumdahl, tables f ΔH f are prvided in Appendix 4 (pg A21). The tabulated values represent the ΔH f fr frming the cmpund frm elements at standard cnditins. C gr + 2H g CH g Δ H = 75 kj / ml ( ) ( ) ( ) 2 4 C gr + O g CO g Δ H = 394 kj / ml ( ) ( ) ( ) 2 2 H g + O g H O Δ H = 286 kj / ml ( ) ( ) ( ) f f f 8

9 Burn Methane (Hess Law) Visual f energy t burn methane ging thrugh the standard states. ( ) + 2 ( ) ( ) + 2 ( ) + 2 ( ) ( ) + 2 ( ) CH C gr H g C gr H g O g CO g H O X 9

10 Using ΔH f t determine ΔH rxn Since we have cnnected the ΔH f t a cmmn reference pint, we can nw determine ΔH rxn by lking at the difference between the ΔH f fr prducts versus reactants. Mathematically: ΔH rxn = ΣΔH f *c(prducts) - ΣΔH f *c(reactants) Example CH + 2O g 4 2 ( ) CO2( g) + 2H 2O ( ) H f ( H2O) + ΔH f ( CO2 ) 2 ΔH f ( O2 ) 1 ΔH f ( CH 4 ) ( ) 1 ( 394) 2 ( 0) 1 ( 75 ) kj / ml 2 ( 286) kJ / ml Δ H = 2 Δ 1 rxn Δ H = rxn rn x { } Δ H = + = 10

11 Hw t Calculate ΔH rxn When a reactin is reversed, ΔH rxn changes sign. If yu multiply a reactin by an integer, ΔH rxn is als multiplied (it is an variable) Elements in their standard states are nt included in calculatins because ΔH f = 0 fr elements. 11

12 Example: Burning Methanl Determine the ΔH rxn fr the cmbustin f methanl. CH 3 OH (l) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH f in Appendix 4: CH 3 OH(l) kj/ml CO 2 (g) kj/ml H 2 O(l) -286 kj/ml 12

13 Example -- Burning (cnt.) CH 3 OH (l) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH rxn = SDH f *c(prducts) - ΣΔH f *c(reactants) = ΔH f (CO 2 (g)) + 2ΔH f (H 2 O(l)) - ΔH f (CH 3 OH(l)) = (-393.5) +(2 )(-286) - (-238.6) kj/ml = kj/ml Exthermic! 13

14 Sample Prblem: Z9.63 Using the fllwing reactin: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g) ΔH rxn = kj Determine the ΔH f fr ClF 3 (g) ΔH f NH 3 (g) -46 kj/ml HF(g) -271 kj/ml 14

15 Anther Example (cnt.) Given the reactin f interest: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g) ΔH rxn = kj ΔH rxn = ΣΔH f *c(prducts) - ΣΔH f *c(reactants) ΔH rxn = 6ΔH f (HF(g)) - 2ΔH f (NH 3 (g)) - 2ΔH f (ClF 3 (g)) 15

16 Anther Example (end) ΔH rxn = 6ΔH f (HF(g)) - 2ΔH f (NH 3 (g)) - 2ΔH f (ClF 3 (g)) kj = (6ml)(-271 kj/ml) - (2)(-46 kj/ml) - (2)ΔH f (ClF 3 (g)) kj = kj - (2)ΔH f (ClF 3 (g)) 338 kj = - (2)ΔH f (ClF 3 (g)) ΔH f (ClF 3 (g)) = -169 kj/ml 16

17 A Reactin at nn-standard Temperature Prblem Z9.85 ΔH vap fr water (t steam) at the nrmal biling pint (373.2 K) is kj/mle. Given the fllwing heat capacity data, what is ΔH vap at K (rm temperature) fr 1 ml? C P (H 2 0(l)) = 75 J/ml.K C P (H 2 0(g)) = 36 J/ml.K Energies (such as enthalpies) are path independent, s we need a path fr the change frm water t steam at a lwer temperature than the nrmal biling pint. (Des water evaprate at rm temperature?) 17

18 Heating Water Prblem Z9.85 We knw hw much energy it takes t increase the temperature frm T t the nrmal biling temperature (100C), using the mlar heat capacity. Assume ne mle. q = C Δ T = ΔH P p p T = 100 C T = 20C b Δ T = T T = 80K warm b This applies t any substance, s it applies t steam (vapr r gas) as well as the liquid frm, water. This, f curse, wrks nly if the heat capacity is independent f temperature (which it is nt, exactly), but the heat capacities ver small temperature ranges d nt change much. 18

19 Diagram the different changes A (Hess s Law) diagram f the changes we knw and the ne we want. Can g directly frm water t steam at 67C r heat water t 100C let it g t gas (knw the heat required fr that) and then cl the steam back t 67C withut it cndensing t water. H 2 O(l) H 2 O(g) ( ) p ( ) warm ( ) ( ) Δ H = C ΔT ( ) / Δ H T = B TB = kj ml K Δ H g = C g ΔT p cl H 2 O(l) T = K Δ H( T) =? H 2 O(g) 19

20 Set Up Algebra and Slve The sum f the heats f the three steps must equal the heat fr the single step (which is what we are after) ( ) (, ) + ( ) + (, ) ( ) ( ) ( ) ( ) =Δ H ( ) { ( ) ( )} vap TB + Cp Cp g ΔTwarm + 3 = { 75 36} 80 J / mle Δ H T = ΔH warm ΔH T ΔH cl g vap vap B Δ H T = C Δ T +Δ H T + C g ΔT vap p warm vap B p cl = = 44. kj / mle This is nt a big crrectin. The prcess is endthermic at the biling pint and nly slightly mre endthermic at rm temperature. (Why desn t water bil at rm temperature, t?) Why des water bil at all if it is always s endthermic? See Prblem Z9.4. Use the heats f frmatin t get DH vap. What number d yu get? Why? 20

21 A reactin ccurs (a review frm 142) We have a chemical reactin, where a little bit f the material reacts. The balanced chemical reactin, written in terms f the species S i, and the stichimetric cefficients (C) is: react c i S i () () () () prd c i S i Because all species are attached by the balanced reactin, if sme f ne is used, thers are als used and thers are generated. n i = n i ± c i X ( ) 0 ( ) ( ) Δ n i = n i n i =± c i X () () 0 () () Here, need a minus sign with the cefficients fr reactants. X is the extent f advancement. If we knw X, which is the number f mles that change change, then we knw hw many mles f each species change. 21

22 Ammnia as an example f Stichimetry 3H + 1N 2NH Cnsider the synthesis f Ammnia: Given the balanced reactin, and stichimetric cefficients and initial cncentratins, set up ICE Table Initial Cnc Stic Change New Cnc Ceffs [H 2 ] =.1 3(-);react -3x [H 2 ]=.1-3x [N 2 ] =.2 1(-);react -x [F 2 ]=. 2-x [NH 3 ] =.3 2(+);prd +2x [HF]=.3+2x Prblem: Have a mixture f different initial mles f each the substances, and we lse 0.2 mles f ammnia, hw much f each species is present at the end? 22

23 Use the relatins t slve fr all species The ammnia change gives X, r delta X. Δ n( NH ) = c( NH ) x = 2 x 0.2 x = = 0.1mles 2 Nw we knw the advancement f the reactin. (Ntice it is negative, meaning the reactin is ging frm prducts t reactants, r it is running backward, as written). We can slve fr the cncentratins f each species. 0 n() i = ni () i + c( i) x fr i = 1, 2, 3 x =.1 n = ( 0.1) = 0.4 n n H N 2 2 NH = ( 0.1) =.3 3 = ( 0.1) =.1mles 23

24 Applicatin t Reactin Energy When a certain amunt f ne substance is used up (r generated) hw much heat (at cnstant pressure), q p, is released? (The reactin can be either endthermic r exthermic). Δn( i) qp = Δ Hrxn = x ΔHrxn c i () T crrectly determine the sign f the heat, ne must multiply the cefficient, c, fr reactants by minus ne; and use the cefficients fr prducts as is. Fr ammnia, the heat f frmatin is: -46 kj/mle. 0.2 Δ n( NH3) = c( NH3) x x = = 0.1mles 2 Δ H = 2Δ H = 2 46= 98 kj / mle rxn f ( 3) ( ) Δn NH qp = Δ Hrxn = x Δ Hrxn = 9.8kJ Endthermic in c NH3 this directin 24