Answer Key, Problem Set 8b (full)

Chemistry 11 Mines, Fall 017 Answer Key, Prblem Set 8b (full) 1. NT1;. NT; 3. 6.58 (with extra parts added); 4. NT3; 5. 6. & 6.3; 6. 6.93; 7. 6.99; 8. 6.75 (i.e., determine the H fr the thermchemical equatin); 9. 6.4; 10. NT4; 11. NT5; 1. 6.85 ------------------- General Early Ideas and Definitins (KE, PE, sys vs surrundings, heat, E, sign cnventins) 1. NT1. Define each f the fllwing and give the symbl assciated with the last tw: (a) kinetic energy (KE) energy f mtin (b) ptential energy (PE) energy f psitin (and ften cnsidered stred energy) (c) internal energy (f a system) the sum ttal f the kinetic energy and ptential energy f all the nanscpic entities in the system. Symbl is E. (d) heat energy that transfers (r flws ) frm a htter substance t a cler ne. Symbl is q. (Nte: KE, PE, and E are prperties f a substance r system, but heat is nt. Heat nly exists during the transfer; nce it gets there it is n lnger cnsidered t be heat it merely cntributes t the internal energy getting larger.). NT. (a) When yu change the temperature f a substance r system, what prperty always changes with it, n matter what? l ),average kinetic energy (per particle) in the system must change as T changes. The average kinetic energy depends nly n temperature (frm kinetic mlecular thery). (b) Fr a sample f a pure substance that is nt underging any physical change, what is the qualitative and quantitative relatinship between heat and temperature? Qualitative: As heat flws int a pure substance, its temperature must increase (if it is nt underging sme physical change). As heat flws ut f a pure substance, its temperature must g dwn. Quantitative: q = Cs x m x T This is cnsistent with the qualitative answer abve, because it shws that if q is psitive (heat transfers in), then T is psitive (T ges up), and if q is negative (heat transfers ut), then T is negative (T ges dwn). (c) Explain hw it can be that water is being heated and yet its temperature is nt increasing. What happens t the heat energy that flws int the water as the water is biling? Shrt answer. In this scenari, all the energy that transfers int the system as heat ends up as increased ptential energy (f the nw separated mlecules) rather than kinetic energy. Explanatin. Since average kinetic energy and T are prprtinal, if the average kinetic energy des nt g up, T will nt g up (r mre prperly wrded, if T stays the same, the average kinetic energy must stay the same). Since energy is cnserved, if energy transfers int a system (net), it must end up smewhere. Since the internal energy f the system is the sum f KE and PE, if the energy desn t end up increasing KE, it must increase PE (because thse are the nly tw chices!). It takes energy t separate things that are attracted t ne anther is the idea in PS8a that I referred yu t. This is basically just anther way f saying that ptential energy increases when tw things that attract ne anther are separated. In this case, the mlecules in the liquid are attracted t ne anther, s energy is required t actually d the vaprizatin prcess. During biling, this energy transfers int the system as heat during heating, and it ends up raising the ptential energy f the system withut raising the average kinetic energy f the particles. That is hw T can remain the same even thugh the system is still being heated. Once all the liquid is vaprized, if energy cntinues t enter the system, the temperature f the system (nw a gas) will g up, as average kinetic energy will nw increase again. Enthalpy Changes, General (meaning f, endthermic vs exthermic, sign f, predictin f end/ex) 3. 6.58. (plus added parts (d) and (e)). Is each prcess exthermic r endthermic? Als indicate the sign f H. Strategy: Assess whether (nly) separatin f attracted particles is ccurring, r whether (nly) cming tgether f attracted particles is ccurring. If it is (nly) separatin, which requires energy, then PS8b-1

Answer Key, Prblem Set 8b the prcess is endthermic, and thus H > 0. If it is (nly) cming tgether, which releases energy, the the prcess is exthermie, and thus H < 0. NOTE: If a prcess invlves sme separatin and sme cming tgether (bnd breaking and bnd making), then yu cannt make a predictin withut additinal infrmatin (see NT3). (a) dry ice evaprating endthermic, H > 0 CO(s) CO(g); Mlecules (which attract ne anther) are separating frm ne anther. This requires energy (raises PE) (b) a sparkler burning exthermic, H < 0 Light and heat are being emitted frm the system. Exthermic means heat transfers ut frm system. (c) the reactin that ccurs in a chemical cld pack used t ice athletic injuries endthermic, H > 0 System getting cld as a result f a prcess ccurring means energy is being absrbed (energy will then transfer in frm the surrundings because f the temperature difference.) (d) O 3 (g) O (g) + O(g) endthermic, H > 0 This particular chemical change invlves nly the breaking f a bnd (n new bnds are made). Pulling apart bnded atms must require energy. (Nte that ne mlecule splits int a smaller ne plus a free atm.) (e) NO (g) N O 4 (g) (Better viewed as: O N(g) + NO (g) O NNO (g)) exthermic, H < 0 This particular chemical change invlves nly the making f a bnd (an N-N bnd) (n bnds are brken). The cming tgether f tw atms that attract ne anther t frm a bnd must release energy. (Nte: Tw mlecules cmbine int ne larger ne.) 4. NT3. (fllw up t prir prblem) Describe specifically why yu cannt predict whether the fllwing prcess is endthermic r exthermic withut additinal infrmatin. H (g) + Br (g) HBr(g) Shrtest Answer: Because unlike the chemical changes shwn in parts (d) and (e) f the prir prblem, the prcess here invlves bth bnd breaking and bnd making. As such, yu can t knw withut further infrmatin whether the amunt f energy t break the bnds that are brken in the reactants will be mre r less than the amunt f energy released upn making the bnds in the prducts. Mre Detailed Answer: Unlike the prcesses in parts (d) and (e) f the prir prblem, the prcess here des nt invlve nly the separatin r nly the cming tgether f particles that attract ne anther. Visualize the reactin in terms f actual atms and mlecules rather than just seeing the symbls in the equatin. Think f H as being H-H, where tw H atms are bnded, Br as Br-Br where tw Br atms are bnded tgether, and HBr as H-Br, where an atm f H is bnded t an atm f Br. S imagine the smallest amunt f this chemical reactin that culd ccur ( ne equatin unit s wrth f reactin in my made-up jargn): ne mlecule f H reacts with ne mlecule f Br t frm tw mlecules f HBr. In rder t rearrange the atms in the reactants t make the prducts in this reactin, yu can imagine that yu d have t break ne H- H bnd and ne Br-Br bnd in the reactants, but when the prducts are frmed, yu d make tw H-Br bnds. The breaking f the H-H and Br-Br bnds will require energy, but the making f the H-Br bnds will release energy. S withut knwing which bnds are strnger (i.e., hw much energy it wuld require t break H-H bnds, Br-Br bnds, and H-Br bnds), yu cannt say whether the verall prcess will require energy (be endthermic) r release it (be exthermic). PS8b-

Answer Key, Prblem Set 8b 5. 6. & 6.3. 6.. Frm a mlecular viewpint, where des the energy emitted in an exthermic chemical reactin cme frm? In shrt, it cmes frm the ptential energy f the atms/ins in the system. In an exthermic prcess, the rearrangement f atms/ins (r even mlecules, in the case f a physical change) results in a lwering f ptential energy. That energy is cnverted int KE in the system, which ultimately transfers ut int the surrundings as heat. (Technically, the cnversin f PE int KE initially makes the system htter, and then the energy wuld be transferred frm the system t the surrundings as heat (heat transfers frm htter matter t clder matter). Why des the reactin mixture underg an increase in temperature even thugh energy is emitted? As nted abve, the PE lst is initially cnverted int KE f nanscpic particles (thermal energy) in the system, raising the temperature f the system. The energy nly transfers ut after this initial T increase. When yu feel a reactin vessel and its ht, yu are really sensing heat energy transferring frm the system int yur fingers (the surrundings). 6.3. Frm a mlecular viewpint, where des the energy absrbed in an endthermic chemical reactin g? It ges int raising the PE f the system. The atms/ins (r mlecules) in the prducts (after rearrangement) have a greater PE than they did when they were arranged hwever they were befre in the reactants. Why des the reactin mixture underg a decrease in temperature even thugh energy is absrbed? Initially, when the prcess ccurs (think f it as ccurring rapidly), the system cnverts its wn thermal energy (KE f particles) int ptential energy as the prcess ccurs, lwering the T f the system. Energy frm the surrundings then transfers int the system because f the (created) T difference. Meaning f a Thermchemical Equatin / Stichimetry with energy 6. 6.93. Determine the mass f CO prduced by burning enugh f each f the fllwing fuels t prduce 1.00 x 10 kj f heat. Which fuel cntributes least t glbal warming per kj f heat prduced? Nte: I am leaving ff state designatins fr all gases. Als, I will leave ff the rxn subscript frm Tr, since if the H cmes after a chemical equatin, yu shuld knw it refers t 1 mle s wrth f reactin. Lastly, I am leaving ff the degree superscript here because standard states are nt invlved in this prblem and s it is unnecessary t designate the H as such. Answers: (a) 5.49 g CO (fr CH4) ; (b) 6.46 g CO (fr C3H8); (c) 6.94 g CO (fr C8H18) Strategy: Thus, CH4 cntributes the least t glbal warming per kj f heat prduced, since it prduced the least amunt f CO per 100. kj heat prduced. 1) In each case, 1.00 x 10 kj f heat is t be generated by the chemical reactin represented by the thermchemical equatin shwn. As such, the H fr the actual reactin will be 1.00 x 10 kj. The questin is asking abut the amunt f CO that wuld need t be prduced, s interpret the enthalpy change fr the thermchemical equatin in terms f energy and amunt f CO. Namely, that (fr (a)) 80.3 kj f energy is released fr every ne mle f CO that is made. It is per ne mle nly because the cefficient f CO in this balanced equatin is 1. Nte, hwever, that fr (b), the cefficient is nt 1, but 3, s the interpretatin wuld be: 043 kj f energy is released fr every three mles f CO that is made. ) Create the crrespnding energy-mle cnversin factr (with Hrxn and CO): 80.3 kj prduced (fr (a)) OR 043 kj prduced (fr (b)) etc., and use it t calculate mles f CO 1ml CO prduced 3 ml CO prduced prduced. (i.e., d the stichimetry). Nte that since kj f energy is knwn, yu must use the PS8b-3

6.94 Answer Key, Prblem Set 8b cnversin factr in its reciprcal frm: (e.g., fr (b): 3 ml CO prduced ) (s really, yu culd just g 043 kj prduced directly t this frm the balanced equatin I just directly translated the H frm the equatin int kj/ml f CO because I m used t thinking abut it this way.) 3) Calculate grams f CO frm mles using the mlar mass f CO : 1.01 + (16.00) = 44.01 g/ml Executin: (a) CH4 + O CO + HO; H = -80. kj 1.00 x 10 1ml CO prduced 44.01g kj prduced x x 80.3 kj prduced ml CO 5 5.4854...49 g CO (b) C3H8 + 5 O 3 CO + 4 HO; H = -043 kj 1.00 x 10 3 ml CO prduced 44.01 g kj prduced x x 043 kj prduced ml CO 6.46 5.. 6.46 g CO (c) C8H18 + 5 O 8 CO + 9 HO; H = -5074.1 kj 1.00 x 10 8 ml CO prduced 44.01g kj prduced x x 5074.1kJ prduced ml CO 6.938.. g CO Calrimetry When a Prcess Des Occur in the System ( bringing multiple ideas tgether ) 7. 6.99. H O(l) H O (g) ; H = +44.01 kj Estimate the mass f water that must evaprate frm the skin t cl the bdy by 0.50 C. Assume a bdy mass f 95 kg and a specific heat fr the bdy f 4.0 J/g C. Answer: 78 g f HO Strategy: (Nte: I have frmulated this part f the key in reference t the prir prblem (#6). That said, yu may als wish t read steps 1 thrugh 4 f the strategy fr the next prblem in this key (Prblem #8 [6.75]) since that strategy applies equally well t this prblem. It just appraches it frm a mre general perspective.) Yu are given a thermchemical equatin and are asked abut an amunt f a reactant (r prduct) that must be used (r prduced). [The fact that the prcess in this prblem is technically physical change rather than chemical change desn t change the interpretatin f the thermchemical equatin.] Nte that this is the same kind f questin as in Prblem 6 abve. Hwever, what is different here is that yu are nt given the amunt f heat (t absrb) yu must calculate it based n the descriptin f the surrundings temperature change (here, the bdy is cnsidered the surrundings). S yu must d this kind f prblem in tw steps. Withut a frmal equatin, what yu d first is use the specific heat capacity, mass, and the magnitude f the temperature change fr the bdy t calculate qbdy (magnitude). Then yu use this as the energy needed t be used up by the water evapratin reactin. Mre specifically: 1) In equatin terms, ne can represent the first step I described abve as a calculatin f the H f the actual reactin (Hsys) frm the q f the surrundings (qbdy = qsurr) using: Hsys = -qsurr = -qbdy = -(Cs, bdy x mbdy x Tbdy) Nte that an ppsite sign that appears here frmally since the q bdy described abve was cnsidered as a psitive quantity (magnitude nly) when technically it is a negative quantity (heat transfers ut f the bdy and s q bdy is negative). PS8b-4

Answer Key, Prblem Set 8b ) Nw yu can d with this Hsys what yu did in Prblem #6. That is, use it, alng with an energymles cnversin factr frm the thermchemical equatin t figure 44.01kJ 1ml HO that evaprates ut the amunt f reactant that reacts. **Be careful, hwever, t make sure that yu cnvert ne f the energy units t be cnsistent with the ther! I will cnvert J t kj rather than kj t J, althugh either ne is bviusly crrect.** 3) Cnvert mles t grams using the mlar mass f HO t finish the prblem. Executin f Strategy: Hsys (4.0 J/g C x 95,000 g x (-0.50 C)) = +190,000 J (T is negative because yu are cling ) 1kJ +190,000 J x 190 kj 1000 J 1ml HO evaprates 18.0 g 190 kj x x 77.79.. 78 g HO evaprates 44.01kJ 1ml H O NOTE: This prblem is very similar t prblem #4 n PS8a! It might be wrth taking a lk at the prblem nw and cmparing the tw. Althugh the wrding is different (and the PS8a prblem asks fr mles instead f grams, and deals with an exthermic chemical reactin rather than an endthermic physical change), the thught prcess and sequence f steps is effectively the same. 8. 6.75. Zn(s) + HCl(aq) ZnCl (aq) + H (g) ; H =??? When 0.103 g f Zn is cmbined with enugh HCl t make 50.0 ml f slutin in a cffee-cup calrimeter, all f the zinc reacts, raising the temperature f the slutin frm.5 C t 3.7 C. Find H fr the thermchemical equatin. (Assume the slutin has d = 1.0 g/ml and C s 4.18 J/g C) Answer: H -160,000 J -160 kj -1.6 x 10 kj (per mle f reactin ) Strategy: (NOTE: This prblem is very similar t Parts B and C f Experiment 14! Perhaps take a lk at that nw while wrking this prblem!) 1) Recgnize this as a heat transfer kind f prblem. That means there will be a qsys = qsurr type f setup. ) Unlike sme f the earlier Mastering prblems n this set, this prblem des invlve a prcess (here, it is a chemical change) i.e., it des NOT invlve the flw f heat frm an initially htter bject r sample t a cler bject r sample. This means a cuple f things: First, we chse the atms that underg rearrangement t be the system. Secnd, it makes mre sense t view the pertinent equality frm (1) t be: Hsys = qsurr Writing H sys instead f q sys helps prevent yu frm (incrrectly) thinking that yu need a specific heat capacity fr the system, and trying t set the prblem up like the n prcess prblem f this set [which are in Mastering nly]! 3) Since n prcess ccurs in the surrundings (which is the slutin here), its T change depends nly n the heat transfer, Cs, and m accrding t: qsl n = Cs,sl n x msl n x Tsl n. Thus: Hsys = -(Cs,sl n x msl n x Tsl n) 4) Since all the variables n the right side are knwn *, yu can calculate Hsys crrespnding t the actual (amunt f) reactin that ccurred, which invlved 0.103 g f Zn. * 50.0 ml x 1.0 g/ml = 50. g slutin; C s,sl n is said t be 4.18 J/g C, T i and T f are given. 5) The H fr the thermchemical equatin must crrespnd t the reactin f 1 ml f Zn (since the cefficient f Zn is 1 in the balanced equatin. Thus, yu must either set up a prprtin, r simply divide kj by ml Zn t get kj/ml f Zn reacted. PS8b-5

Answer Key, Prblem Set 8b Executin f Strategy: Hsys (4.18 J/g C x 50. g x (3.7 C.5 C)) = -50.8 J (Nte: T = 1. C SF) ml Zn 0.103 g Zn 65.38 g/ml Zn J per ml f Zn reacted = 0.001575 ml Zn -50.8 J 0.001575 ml Zn -159197-160,000 J -160 kj Wrk vs Heat, Sign Cnventins, and Relatin t Change in Internal Energy (E, q, w relatinships) 9. 6.4. A system absrbs 196 kj f heat and the surrundings d 117 kj f wrk n the system. What is the change in internal energy f the system? Answer: 313 kj Reasning: In this prblem, yu need t realize that the internal energy f a system can nly be changed tw ways: via heating (r remving heat) r via wrk (being dne either by the system r n the system). Namely: Esys q + w (chemists cnventin) where q is heat and w is wrk. A psitive q means heat flws int the system (raising the system s energy) and a negative q means heat flws ut f the system. A psitive w (in a chemistry cntext) means wrk is dne n the system, raising its internal energy, and a negative w means wrk is dne by the system (n the surrundings), lwering its internal energy. Hess s Law Thus, fr this questin, q = +196 kj and w = +117, s Esys +196 + (+117) = +313 kj 10. NT4. Calculate the standard enthalpy change, H, fr the frmatin f 1 ml f strntium carbnate (the material that gives the red clr in firewrks) frm its elements. 3 Sr (s) C(graphite) O (g) SrCO (s) 3 The infrmatin available is (1) Sr (s) O (g) SrO (s) H -1184 kj () SrO (s) CO (g) SrCO (s) H -34 kj 3 (3) C(graphite) O (g) CO (g) H -394 kj Answer: -10 kj Strategy: Same as prir prblem. But nte that here it is easiest t skip the 3 rd substance (O) since it is present in mre than ne f the given equatins. Executin: 1) Sr(s) nly in (1), but there s a there and need a 1 take ½ x (1): H1 = ½ x (-1184 kj) = -59 kj ) C(graphite) nly in (3). Take equatin as is b/c crrect cefficient and side: H3 = -394 kj 3) **SKIP O BECAUSE IT IS IN MULTIPLE EQUATIONS** 4) SrCO3 nly in (). Take equatin as is. H = -34 kj Hverall -59 + (-394) + (-34) -10 kj PS8b-6

Answer Key, Prblem Set 8b Standard Enthalpies f Frmatin and Calculating Hrxn frm them 11. NT5. (i) What is the standard enthalpy f frmatin f a substance? (ii) D 6.8(b,d nly) in Tr and add parts (e) and (f): (e) F(g) ; (f) F(g) (iii) Why is the standard enthalpy f frmatin f any element (in its standard state) zer? (iv) Why is the answer in iii(f) abve psitive? Answers, Sme With Reasning Within: (i) The standard enthalpy f frmatin f a substance is the enthalpy change crrespnding t the frmatin f ne mle f a substance (in a specific state) frm its elements in their standard states. (ii) 6.8. Write an equatin fr the frmatin f [ne mle f] each cmpund frm its elements in their standard states, and find H f fr each frm Appendix IIB. Strategy: 1) Recgnizing that a standard enthalpy f frmatin is, by definitin, the change in enthalpy assciated with the frmatin f ne mle f a substance, write the frmula f the substance n the right side f an arrw (because it is being frmed, right?) with a cefficient f ne (and nly ne!). ) Lk at each element symbl in the prduct substance, and write the frmula fr the stable frm f each element, including the state designatin, n the left side f the equatin, withut a cefficient (fr nw). Nte that the mst stable frms f H, N, O, F, Cl, Br, and I are the diatmic mlecules, and the mst stable frm f C is graphite (written as C(s, graphite). Fr metals, the mst stable frms are generally the mnatmic atms (i.e, the same as their symbl, withut any subscripts). Yu will nt need t memrize any ther elements whse stable frm is nt their mnatmic atm. 3) Remember that the cefficient fr the substance n the right MUST REMAIN ONE. As such, add cefficients the left side, using fractins if needed, t balance the equatin fr each atm type. Answers: 3 (s) 3 (b) MgCO3(s): Mg C(s,graphite) O ( g ) MgCO ( s ) ; (MgCO (s)) = -1095.8 kj/ml Hf 3 NOTE: 3/ f a mlecule f O mathematically yields three O atms. Yu must keep the fractin here since the right side s cefficient must be ne. 1 C(s,graphite) H ( g ) O ( g ) CH3OH( l ; Hf (CH3OH( l )) = -38.6 kj/ml (d) CH3OH(l): ) NOTE: Physical states are critical in these prblems. The H f fr CH 3 OH(g) is greater than -38.6 kj (it is -01.0 kj) because the enthalpy (think PE here) f a substance in its gaseus state is always greater than in its liquid state it requires energy t turn a liquid int a gas! (Think abut that fr a mment and ask me if yu d nt see why this is s. (this reactin is n reactin nthing has Hf (F (g) = 0 kj/ml rearranged r changed) (e) F(g): F(g) F(g) ; ) When yu frm an element in its standard state frm its elements in their standard states, the equatin represents n change f any kind. The reactant is identical t the prduct. 1 (f) F(g): F ( g ) F ( g ) ; H f (F(g)) = +79.38 kj/ml (iii) Shrtest Answer: Frming a mle f an element frm a mle f its element(s) (bth in the same [standard] state) is ding nthing at all! See (ii)(e) abve! If there is n actual chemical r physical change, then there can be n enthalpy change! PS8b-7

Answer Key, Prblem Set 8b Mre Cmmentary, Mre Detailed Answer (Partial Rant ): I think it is abslutely silly fr Tr t write as a separate Rule n p. 75 (see 3. Standard Enthalpy f Frmatin ) that fr an element in its standard state, H f = 0. Yu shuld nt have t memrize this r think f the definitin fr H f f an element as being smehw different than the definitin fr the H f fr any substance. Just write ut the equatin fr the frmatin and if it is fr an element in its standard state, yu shuld see that nthing has happened and thus the H f must be zer! When yu frm a cmpund frm elements, yu necessarily must make and/r break sme bnds, but when yu fr an element frm an element (same frm and same state), there is n change physically and s there is n change in energy/enthalpy. This shuld make sense t yu. If nt, please reread and/r ask me. **NOTE: The chice f definitin fr H f essentially makes elements in their standard states ur reference state fr enthalpy. It arbitrarily makes elements the zers f ptential energy. (If there are alltrpes fr an element, ne f them is chsen fr the standard state f the element; yu dn t need t wrry abut this.) (iv) This H f is psitive because the prcess it represents invlves (nly) the breaking f a chemical bnd (cmpare this t Prblems 3(d) and 3(e) abve!), and we knw that it takes energy t break bnds (endthermic, psitive H). In fact, the H f fr F(g) is precisely half f the energy needed t break apart tw F atms that are bnded tgether. 1. 6.85 Hydrazine (N H 4 ) is a fuel used by sme spacecraft. It is nrmally xidized by N O 4 accrding t the equatin: N H 4 (l) + N O 4 (g) N O(g) + H O(g) Calculate H rxn fr this reactin (equatin) using standard enthalpies f frmatin. Answer: -380. kj General Cmment/Strategy: Finding the H fr a balanced chemical equatin can be thught f as finding the H (under standard cnditins) assciated with unfrming the number f mles f each reactant indicated by its cefficient and frming the number f mles f each prduct indicated by its cefficient. If we use H s f frmatin fr each substance, abbreviated as Hf s, alng with Hess s Law, this can be dne simply by summing the fllwing tw prcesses : 1) unfrming all the reactants [which is the ppsite f frming them], dented as Hunfrming_R s, & ) frming all the prducts, dented as Hfrming_P s Hverall Hunfrming_R s + Hfrming_P s -Hfrming_R s + Hfrming_P s Swapping the rder f the tw terms yields: Hfrming_P s Hfrming_R s n H n H p f (prducts) r f (reactants) which is just Equatin 6.15 in Tr! It s just prducts reactants using Hf s fr each reactant and prduct (times their respective mle amunts (cefficients). Yu just need t be VERY careful with signs!! Yu als need t make sure t get the prper values frm the table (with prper signs). And, f curse, make sure nt t frget t multiply each substance s Hf by its cefficient in the balanced equatin. Executin: PS8b-8

Answer Key, Prblem Set 8b Nte: I ve left ff units here fr cnvenience, but each cefficient s units are mles, and each standard enthalpy f frmatin s units are kj/ml. Thus the final units are just kj. [(+81.6) + (-41.8)] [1(+50.6) + 1(9.16)] -30.4 (59.76) kj -380.16-380. kj (values in parentheses cme frm Appendix) ========================= END OF OFFICIAL PROBLEM SET======================== Slutins t prblems n past years prblem sets (may be in Mastering nw): 6.11. If internal energy f the prducts f a reactin is higher than that f the reactants, what is the sign f E fr the reactin? In which directin des energy flw? Answer: E is psitive and energy flws frm the surrundings t the system. One culd draw an energy diagram t shw this as fllws: Energy Prducts Reactants E = Eprducts Ereactants > 0 Thermal Energy Transfer, (Specific) Heat Capacity, and N Prcess Calrimetry Situatins 6.48. Hw much heat is required t warm 1.50 kg f sand frm 5.0 C t 100.0 C? Answer: 95,000 J = 95 kj Reasning / Wrk: The specific heat capacity f a substance represents the amunt f energy needed t raise 1.0 g f it by 1 degree Celsius. Frm Table 6.4, the specific heat f sand is 0.84 J/(gC). 1.50 kg = 1500 g, s the energy needed wuld be 1500 times as great t raise the sample by 1 degree. Ging frm 5 t 100 degrees represents a change in T f +75 degrees, s multiply by 75 times mre. In equatin frm, we have that heat energy needed (q) equals: q = Cs x m x T qsand 0.84 J 1000 g C 95000 J 95 kj g x 1.50 kg x x (100.0-5.0) 94500 J r C kg 6.66. A 3.5-g irn rd, initially at.7 C is submerged int an unknwn mass f water at 63. C, in an insulated cntainer. The final temperature f the mixture upn reaching thermal equilibrium is 59.5 C. What is the mass f the water? Answer: 35 g Strategy: 1) Recgnize this as a heat transfer kind f prblem. Thus yu will have a qsys = -qsurr type f setup. ) Further recgnize that this prblem invlves NO CHEMICAL OR PHYSICAL CHANGES i.e., n prcess ther than heat flw frm a htter bject r sample t a cler bject r sample. This means that EACH sample (the system and the surrundings ) changes temperature nly as a result f a heat flw. Thus yu will have TWO specific heat capacities, TWO delta T s and TWO masses. In ther wrds qsys = -qsurr reduces t Cs,sys x msys x Tsys = (Cs,surr x msurr x Tsurr) PS8b-9

Answer Key, Prblem Set 8b 3) The abve reduces further in this prblem t: Cs,irn x mirn x Tirn = (Cs, water x mwater x Twater) 4) Nting that T Tf Ti, and Cs,water= 4.18 J/g C and Cs,irn = 0.449 J/g C (Table 6.4), all f the unknwns are given in the prblem except fr mwater, which can be determined algebraically. Executin f Strategy: 0.449 J/g C x 3.5 g x (59.5 C.7 C) = (4.18 J/g C x mwater x (59.5 C 63. C)) 537.0 J = -(-15.466) J/g x mwater mwater 537.0 J 15.466 J/g 34.7 g = 35 g Meaning f a Thermchemical Equatin / Stichimetry with Energy 6.60. What mass f natural gas (CH 4 ) must burn t emit 67 kj f heat? CH4(g) + O(g) CO(g) + HO(g) Hrxn = -80.3 kj NOTE: I d nt like the subscript rxn here, althugh that is the cnventinal way f representing the H fr a thermchemical equatin. I think it shuld be symblized Heqn because it nly reflects the H fr an amunt f reactin indicated by the cefficients f the balanced equatin, interpreted as mles. Since I like t distinguish any actual reactin (with actual amunts) frm the equatin, the subscript rxn can be cnfusing r misleading. Please make sure that yu realize that the H f reactin here des nt mean the H f the actual reactin (with actual amunts). Please g back t yur Exp 14 lab reprt frm and nte the difference between the Hsys (fr yur actual disslutin r reactin) and the H fr the thermchemical equatin (which Tr wuld call Hrxn) Answer: 5.34 g CH4 Strategy: 1) 67 kj f heat is t be generated by the chemical reactin represented by the thermchemical equatin shwn, which has a stated Hrxn = -80.3 kj. Since, 67 kj f heat is t be generated, H fr the actual reactin will be 67 kj. The questin is asking abut the amunt f CH4 that wuld need t be burned (excess O is assumed), s interpret the enthalpy change fr the thermchemical equatin in terms f energy and amunt f CH4. Namely, that 80.3 kj f energy is released fr every ne mle f CH4 that is burned. It is per ne mle nly because the cefficient f CH4 in this balanced equatin is 1. 80.3 kj prduced ) Create the crrespnding energy-mle cnversin factr (with Hrxn and CH4): 1ml CH4 reacted, and use it t calculate mles f CH4 reacted. (i.e., d the stichimetry). Nte that since kj f 1ml CH4 reacted energy is knwn, yu must use the cnversin factr in its reciprcal frm: (s 80.3 kj prduced really, yu culd just g directly t this frm the balanced equatin I just directly translated the H frm the equatin int kj/ml f CH4 because I m used t thinking abut it this way.) 3) Calculate grams f CH4 frm mles using the mlar mass f CH4: 1.01 + 4(1.008) = 16.04 g/ml Executin: PS8b-10 This is reasnable, because this is less than a mle f CH 4, and 67 kj is less than 80 kj. (Actually, yu culd g further and say that this is abut a third f a mle, which makes sense since 67 is abut a third f 80)

Answer Key, Prblem Set 8b 1ml CH4 reacted 16.04 g 67 kj prduced x x 5.338.. 5.34 g CH 4 80.3 kj prduced ml CH 4 Wrk vs Heat, Sign Cnventins, and Relatin t Change in Internal Energy (E, q, w relatinships) 6.39. Identify each energy exchange as primarily heat r wrk and determine whether the sign f E is psitive r negative fr the system. Answers: (a) heat, E is psitive; (b) wrk, E is negative; (c) heat, E is psitive General Reasning/Strategy: Heat r wrk? 1) If a htter sample f matter is put next t a cler ne, heat transfers frm the htter ne t the clder ne (this is the definitin f heat!). S if there is a scenari in which tw samples f matter at different temperatures are put next t each ther (and nthing else happens prcess-wise), the energy exchange is clearly heat. ) In rder fr there t be wrk, smething needs t mve n a macrscpic scale (wrk is frce acting ver a distance). S if there is a vlume change (in the absence f a physical r chemical change) r a change in psitin f a macrscpic bject, then the energy exchange is wrk rather than heat. 3) When neither f the abve is the case, mst likely there is a physical r chemical change ccurring (in what we wuld define as the system). In general, whenever there is a physical r chemical change ccurring in the system, the ptential energy changes assciated with the nanscpic rearrangement r repartnering lead t changes in KE energy, which ultimately leads t heat transfer either int r ut f the system [because the system wuld initially get htter r clder as PE was cnverted t KE r vice versa] (see Prblem #5 abve). S these situatins invlve (mstly) heat. (Even if there is sme wrk, it will generally be small cmpared t the value f the enthalpy change (heat at cnstant P) when there is a physical r chemical change happening). Psitive r negative? Heat: If heat transfers frm system t surrundings, heat is negative. If heat transfers frm surrundings t system, heat is psitive. Wrk: If wrk is dne by the system (n the surrundings), wrk is negative (chemists cnventin) and internal energy (E) decreases (if heat is zer). If wrk is dne n the system (by the surrundings), wrk is psitive and internal energy (E) increases (if heat is zer). NOTE: Expansin wrk dne by system (n surrundings) wsys is negative. Cmpressin wrk dne n system (by surrundings) wsys is psitive. Strategy Applied: (a) It is heat and nt wrk because the prcess ccurring in the system is a physical change (turning liquid water int a gas) and there is minimal macrcscpic mving f anything. It is psitive because evapratin requires energy (it invlves separating mlecules frm ne anther see Prblems #3 and #4 n this set (abve) r Prblem #5 n PS8a). S heat transfers int the sweat frm the bdy, qsys (and H) is psitive, and thus E is psitive (assume wrk is minimal). (NOTE: Althugh a bit f gas is prduced, and thus a bit f wrk is dne n the surrundings, that pales in cmparisn t the change in enthalpy assciated with the nanscpic rearrangements.) (b) It is wrk and nt heat because a gas expands against a pressure, where there is n physical r chemical change (the cntents f the balln remain a gas thrughut). It is negative because the cntents f the balln push the atmsphere away, thus ding wrk n it (and using its wn energy t d that wrk). (c) It is heat and nt wrk because a ht sample was put next t a clder ne. Heat transferred frm the flame (surrundings) t the slutin (system), s qsys is psitive. PS8b-11

Answer Key, Prblem Set 8b 6.43. The gas in a pistn (defined as the system) warms and absrbs 655 J f heat. The expansin perfrms 344 J f wrk n the surrundings. What is the change in internal energy fr the system? 6.78. Answer: 311 J Reasning: In this prblem, yu need t realize that the internal energy f a system can nly be changed tw ways: via heating (r remving heat) r via wrk (being dne either by the system r n the system). Namely: Esys q + w (chemists cnventin) where q is heat and w is wrk. A psitive q means heat flws int the system (raising the system s energy) and a negative q means heat flws ut f the system. A psitive w (in a chemistry cntext) means wrk is dne n the system, raising its internal energy, and a negative w means wrk is dne by the system (n the surrundings), lwering its internal energy. Thus, fr this questin, q = +655 J and w = -344 J, s Esys +655 + (-344) = +311 J NOTE: This is nt really a Hess s Law prblem as much as an Understanding what a thermchemical equatin means prblem. This was cvered in the PS8a handut and prblem number. This prblem was put here because yu need t use these ideas when yu manipulate the given equatins in a Hess s Law type prblem en rute t finding a set f equatins that will sum up t the target equatin (see my handut with a General Prcdure t Hess s Law type prblems). Answers: (a) 465 kj ( 3 x 155 kj). Prcess is exactly three times as much reactin as riginal. (b) -155 kj. Prcess is exact reverse f riginal s sign is ppsite. (c) -77.5 kj ( -½ x 155 kj). Prcess is exactly ne-half as much reactin as the reverse f the riginal. 6.8. Calculate H rxn (which means the H fr the target thermchemical equatin, r what I ften call H verall in a Hess s Law prblem such as this) fr the reactin (equatin): given: CH 4 + 4 Cl CCl 4 + 4 HCl (all gases) C(s) + H CH 4 ; H 1 = -74.6 kj C(s) + Cl CCl 4 ; H + Cl HCl; H = -95.7 kj H 3 = -9.3 kj Answer: Hverall -05.7 kj Explanatin: Fllwing the apprach described in my handut (and in that pain-stakingly made vide lecture that shuld be psted in DL nw), g substance by substance in the target equatin and check t make sure that a substance appears in nly ne place in the set f equatins given. Then make the psitin (reactant r prduct) and amunt needed (i.e., cefficient) match by either reversing the equatin r multiplying the equatin thrugh by a number (r bth). In this case, yu must be careful t skip Cl (the secnd reactant) during this prcess since it appears in mre than ne f the given equatins. Executin: 1) CH4 is nly in equatin 1, but it is a prduct there s the equatin must be flipped: H1 = +74.6 kj ) Cl is in equatins () and (3), s SKIP IT! PS8b-1

Answer Key, Prblem Set 8b 3) CCl4 is nly in equatin, and it is n the crrect side and with crrect cefficient, s it needs n mdificatin at all: H = -95.7 kj 4) HCl is nly in equatin 3, but it has a cefficient f there, and we need a 4. S multiply the equatin by tw: H3 x H3 x -9.3 kj -184.6 kj (Yu shuld really check t make sure these add up t the target, but I will mit that here fr nw.) Hverall +74.6 + (-95.7) + (-184.6) -05.7 kj 6.88(a,d)) Use standard enthalpies f frmatin t calculate H rxn fr each reactin (equatin). General Cmment/Strategy: Finding the H fr a balanced chemical equatin means finding the H (under standard cnditins) assciated with unfrming the number f mles f each reactant indicated by its cefficient and frming the number f mles f each prduct indicated by its cefficient. S we can use H s f frmatin f each substance, abbreviated as Hf s, alng with Hess s Law, t find the verall H fr any equatin! It is simply the sum f tw prcesses : 1) unfrming all the reactants [which is the ppsite f frming them], dented as Hunfrming_R s, & ) frming all the prducts, dented. Hfrming_P s Hverall = Hunfrming_R s + Hfrming_P s = -Hfrming_R s + Hfrming_P s Swapping the rder f the tw terms yields: = Hfrming_P s Hfrming_R s n H n H p f (prducts) r f (reactants) which is just 6.15 in Tr! It s just prducts reactants using Hf s fr each reactant and prduct (times their respective mle amunts (cefficients), and being VERY careful with signs!!) Yu als need t make sure t get the prper values frm the table (with prper signs). And, f curse, make sure nt t frget t multiply each substance s Hf by its cefficient in the balanced equatin. (a) H S(g) + 3 O (g) H O(l) + SO (g) (d) N O 4 (g) + 4 H N (g) + 4 H O(g) Answers: (a) -114.0 kj ( [(-85.8) + (-96.8)] [(+-0.6) + 3(0)] ) (d) -976.4 kj ( [1(0) + 4(-41.8)] [1(+9.16) + 4(0)] ) (values cme frm Appendix) (values cme frm Appendix) ========================================================== Mre Extra Prblems fr Practice (frm prir years answer keys) 6.63. The prpane fuel (C 3 H 8 ) used in gas barbeques burns accrding t the thermchemical equatin: C3H8 + 5 O 3 CO + 4 HO; H = -17 kj If a prk rast must absrb 1.6 x 10 3 kj t fully ck, and if nly 10% f the heat prduced by the barbeque is actually absrbed by the rast, what mass f CO is emitted int the atmsphere during the grilling f the prk rast? Answer: 950 g CO Reasning: 1) If 1.6 x 10 3 kj is needed, and nly 10% (0.10) f the heat prduced by the BBQ is actually absrbed, than yu need t generate 1/0.10 = 10 times as much heat as yu need, i.e., 1.6 x 10 4 kj. (Anther way t lk at this is t say that fr every 100 kj generated by the reactin, nly 10 kj (10% f 100) is absrbed. Thus again, 10x as much heat needs t be generated. Hence, 1.6 x 10 4 kj). PS8b-13

Answer Key, Prblem Set 8b ) Create an energy-mle cnversin factr (with H and CO) using the balanced equatin: 17 kj released, and use it t calculate mles f CO prduced. (i.e., d the stichimetry) 3 ml CO prduced 3) Calculate mles f CO t grams using mlar mass Executin: 3 ml CO prduced 44.01 g 17 kj prduced ml CO 950 g 4 1.6 x 10 kj x x 95.8.. 6.69. Tw substances, A and B, initially at different temperatures, cme int cntact and reach thermal equilibrium. The mass f substance A is 6.15 g and its initial temperature is 0.5 C. The mass f substance B is 5. g and its initial temperature is 5.7 C. The final temperature f bth substances at thermal equilibrium is 46.7 C. If the specific heat f substance B is 1.17 J/g C, what is the specific heat f substance A? Answer: 1.1 J/g C Strategy: 1) Like the last prblem, this ne nly invlves heat transfer (n prcess in system). Thus: ) Substitute in prperly and slve fr Cs,A Executin f Strategy: Cs,A x ma x TA (Cs,B x mb x TB) Cs,A x 6.15 g x (46.7 C 0.5 C) (1.17 J/g C x 5. g x (46.7 C 5.7 C)) Cs,A (-176.9 J) 1.097 J/g C 1.1 J/g C 161.13 g C 6.13. The internal energy f an ideal gas depends nly n its temperature. Which statement is true f an isthermal (cnstant-temperature) expansin f an ideal gas against a cnstant external pressure? Explain. (a) E is psitive (b) w is psitive. (c) q is psitive (d) E is negative Answers: (a) False. The prblem states that the internal energy f an ideal gas depends nly n its temperature. Thus, if the temperature des nt change ( isthermal ), Esys 0 (b) False. If the system expands, the system des wrk n the surrundings (pushes it away) w < 0 (cntributes t a decrease in E sys ) (c) True. Mathematical answer: If Esys 0 and Esys q + w and w < 0, then q > 0 Cnceptual answer: If the system des wrk n the surrundings, but its internal energy des nt change, the energy t d that wrk must be cming frm an input f heat frm the surrundings. (d) False. Esys 0 (See (a).) 6.136. Tw (samples f) substances A and B, initially at different temperatures, are thermally islated frm their surrundings and allwed t cme int thermal cntact. The mass f (the sample f) substance A is twice the mass f (the sample f) substance B, but the specific heat f substance B is fur times the specific heat f substance A. Which substance will underg a larger change in temperature? PS8b-14

Answer Key, Prblem Set 8b Answer: Sample A will change its temperature by a greater amunt (in abslute value). (Its change will, in fact, be twice the change f Sample B.) Explanatin: Since q = Cs x m x T, it shuld be clear that fr a given q, the greater the Cs, the smaller the change in T (a greater specific heat means harder t change its T. Als, fr a given q, the greater the m, the smaller the change in T (the greater the mass, the harder it is t change its T). In fact, T is inversely prprtinal t bth f these variables. S althugh sample A is x the mass f B (making the T half as great (x smaller) as B), B s 4x larger specific heat wuld make its T ne-quarter as great (4x smaller). S B s larger specific heat wins here, and its T will be half as great as A s. Ta C B mb 4 1 NOTE: Yu culd als shw mathematically that x x, meaning that the abslute Tb CA ma 1 1 value f A s temperature change is twice that f B s. PS8b-15