Answer Key, Problem Set 8b (full)

Size: px
Start display at page:

Download "Answer Key, Problem Set 8b (full)"

Transcription

1 Chemistry 11 Mines, Fall 017 Answer Key, Prblem Set 8b (full) 1. NT1;. NT; (with extra parts added); 4. NT3; & 6.3; ; ; (i.e., determine the H fr the thermchemical equatin); ; 10. NT4; 11. NT5; 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) (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

2 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-

3 Answer Key, Prblem Set 8b & 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) 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 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

4 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 : (16.00) = 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 g CO (b) C3H8 + 5 O 3 CO + 4 HO; H = -043 kj 1.00 x 10 3 ml CO prduced g kj prduced x x 043 kj prduced ml CO g CO (c) C8H O 8 CO + 9 HO; H = kj 1.00 x 10 8 ml CO prduced 44.01g kj prduced x x kJ prduced ml CO g CO Calrimetry When a Prcess Des Occur in the System ( bringing multiple ideas tgether ) H O(l) H O (g) ; H = 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

5 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 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 Zn(s) + HCl(aq) ZnCl (aq) + H (g) ; H =??? When 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 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

6 Answer Key, Prblem Set 8b Executin f Strategy: Hsys (4.18 J/g C x 50. g x (3.7 C.5 C)) = J (Nte: T = 1. C SF) ml Zn g Zn g/ml Zn J per ml f Zn reacted = ml Zn J ml Zn ,000 J -160 kj Wrk vs Heat, Sign Cnventins, and Relatin t Change in Internal Energy (E, q, w relatinships) 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 (+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 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 (-394) + (-34) -10 kj PS8b-6

7 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)) = 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 )) = kj/ml (d) CH3OH(l): ) NOTE: Physical states are critical in these prblems. The H f fr CH 3 OH(g) is greater than kj (it is 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)) = 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

8 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 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: 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

9 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)] (59.76) kj 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): 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 Hw much heat is required t warm 1.50 kg f sand frm 5.0 C t 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) 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 J 95 kj g x 1.50 kg x x ( ) J r C kg 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

10 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 = 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: 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)) J = -( ) J/g x mwater mwater J J/g 34.7 g = 35 g Meaning f a Thermchemical Equatin / Stichimetry with Energy What mass f natural gas (CH 4 ) must burn t emit 67 kj f heat? CH4(g) + O(g) CO(g) + HO(g) Hrxn = 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 = 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 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: (1.008) = 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)

11 Answer Key, Prblem Set 8b 1ml CH4 reacted g 67 kj prduced x x g CH kj prduced ml CH 4 Wrk vs Heat, Sign Cnventins, and Relatin t Change in Internal Energy (E, q, w relatinships) 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

12 Answer Key, Prblem Set 8b 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? 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 (-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) kj ( -½ x 155 kj). Prcess is exactly ne-half as much reactin as the reverse f the riginal 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 Cl CCl HCl (all gases) C(s) + H CH 4 ; H 1 = kj C(s) + Cl CCl 4 ; H + Cl HCl; H = kj H 3 = -9.3 kj Answer: Hverall 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 = kj ) Cl is in equatins () and (3), s SKIP IT! PS8b-1

13 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 = 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 kj (Yu shuld really check t make sure these add up t the target, but I will mit that here fr nw.) Hverall (-95.7) + (-184.6) 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) kj ( [(-85.8) + (-96.8)] [(+-0.6) + 3(0)] ) (d) 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) 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

14 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 g 17 kj prduced ml CO 950 g x 10 kj x x 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 ( J) J/g C 1.1 J/g C g C 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).) 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

15 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

Unit 14 Thermochemistry Notes

Unit 14 Thermochemistry Notes Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm

More information

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither

More information

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY CHEMICAL REACTIONS INVOLVE ENERGY The study energy and its transrmatins is knwn as thermdynamics. The discussin thermdynamics invlve the cncepts energy, wrk, and heat. Types Energy Ptential energy is stred

More information

Chapter 17 Free Energy and Thermodynamics

Chapter 17 Free Energy and Thermodynamics Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics

More information

Thermochemistry. Thermochemistry

Thermochemistry. Thermochemistry 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

More information

CHEM 103 Calorimetry and Hess s Law

CHEM 103 Calorimetry and Hess s Law CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the

More information

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany

More information

Thermodynamics Partial Outline of Topics

Thermodynamics Partial Outline of Topics Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)

More information

Thermodynamics and Equilibrium

Thermodynamics and Equilibrium Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,

More information

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C? NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic

More information

Lecture 4. The First Law of Thermodynamics

Lecture 4. The First Law of Thermodynamics Lecture 4. The First Law f Thermdynamics THERMODYNAMICS: Basic Cncepts Thermdynamics: (frm the Greek therme, meaning "heat" and, dynamis, meaning "pwer") is the study f energy cnversin between heat and

More information

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is

More information

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin

More information

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review) CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis

More information

Matter Content from State Frameworks and Other State Documents

Matter Content from State Frameworks and Other State Documents Atms and Mlecules Mlecules are made f smaller entities (atms) which are bnded tgether. Therefre mlecules are divisible. Miscnceptin: Element and atm are synnyms. Prper cnceptin: Elements are atms with

More information

Lecture 16 Thermodynamics II

Lecture 16 Thermodynamics II Lecture 16 Thermdynamics II Calrimetry Hess s Law Enthalpy r Frmatin Cpyright 2013, 2011, 2009, 2008 AP Chem Slutins. All rights reserved. Fur Methds fr Finding H 1) Calculate it using average bnd enthalpies

More information

Chemistry 114 First Hour Exam

Chemistry 114 First Hour Exam Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)

More information

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system

More information

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3

More information

CHE 105 EXAMINATION III November 11, 2010

CHE 105 EXAMINATION III November 11, 2010 CHE 105 EXAMINATION III Nvember 11, 2010 University f Kentucky Department f Chemistry READ THESE DIRECTIONS CAREFULLY BEFORE STARTING THE EXAMINATION! It is extremely imprtant that yu fill in the answer

More information

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change? Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S

More information

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature

More information

NUMBERS, MATHEMATICS AND EQUATIONS

NUMBERS, MATHEMATICS AND EQUATIONS AUSTRALIAN CURRICULUM PHYSICS GETTING STARTED WITH PHYSICS NUMBERS, MATHEMATICS AND EQUATIONS An integral part t the understanding f ur physical wrld is the use f mathematical mdels which can be used t

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk

More information

Spontaneous Processes, Entropy and the Second Law of Thermodynamics

Spontaneous Processes, Entropy and the Second Law of Thermodynamics Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer

More information

AP Chemistry Assessment 2

AP Chemistry Assessment 2 AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND

More information

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition)

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition) Name Chem 163 Sectin: Team Number: ALE 24. Vltaic Cells and Standard Cell Ptentials (Reference: 21.2 and 21.3 Silberberg 5 th editin) What des a vltmeter reading tell us? The Mdel: Standard Reductin and

More information

General Chemistry II, Unit I: Study Guide (part I)

General Chemistry II, Unit I: Study Guide (part I) 1 General Chemistry II, Unit I: Study Guide (part I) CDS Chapter 14: Physical Prperties f Gases Observatin 1: Pressure- Vlume Measurements n Gases The spring f air is measured as pressure, defined as the

More information

**DO NOT ONLY RELY ON THIS STUDY GUIDE!!!**

**DO NOT ONLY RELY ON THIS STUDY GUIDE!!!** Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) I. UV-Vis Absrbance Spectrscpy Lab Beer s law Relates cncentratin f a chemical species in a slutin and the absrbance f that

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra Chem 115 POGIL Wrksheet - Week 8 Thermchemistry (Cntinued), Electrmagnetic Radiatin, and Line Spectra Why? As we saw last week, enthalpy and internal energy are state functins, which means that the sum

More information

Chemistry 1A Fall 2000

Chemistry 1A Fall 2000 Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur

More information

Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs

More information

General Chemistry II, Unit II: Study Guide (part 1)

General Chemistry II, Unit II: Study Guide (part 1) General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reactin Equilibrium in the Gas Phase General Chemistry II Unit II Part 1 1 Intrductin Sme chemical reactins have a significant amunt

More information

Chemistry 20 Lesson 11 Electronegativity, Polarity and Shapes

Chemistry 20 Lesson 11 Electronegativity, Polarity and Shapes Chemistry 20 Lessn 11 Electrnegativity, Plarity and Shapes In ur previus wrk we learned why atms frm cvalent bnds and hw t draw the resulting rganizatin f atms. In this lessn we will learn (a) hw the cmbinatin

More information

Differentiation Applications 1: Related Rates

Differentiation Applications 1: Related Rates Differentiatin Applicatins 1: Related Rates 151 Differentiatin Applicatins 1: Related Rates Mdel 1: Sliding Ladder 10 ladder y 10 ladder 10 ladder A 10 ft ladder is leaning against a wall when the bttm

More information

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes. Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied

More information

Chapter 4 Thermodynamics and Equilibrium

Chapter 4 Thermodynamics and Equilibrium Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume

More information

GOAL... ability to predict

GOAL... ability to predict THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict

More information

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25 CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing

More information

2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS

2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS 2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS 6. An electrchemical cell is cnstructed with an pen switch, as shwn in the diagram abve. A strip f Sn and a strip f an unknwn metal, X, are used as electrdes.

More information

CHM 152 Practice Final

CHM 152 Practice Final CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the

More information

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics CHEM 116 Electrchemistry at Nn-Standard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam

More information

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures?

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures? Name: Perid: Unit 11 Slutins- Guided Ntes Mixtures: What is a mixture and give examples? What is a pure substance? What are allys? What is the difference between hetergeneus and hmgeneus mixtures? Slutins:

More information

Chemical Thermodynamics

Chemical Thermodynamics Chemical Thermdynamics Objectives 1. Be capable f stating the First, Secnd, and Third Laws f Thermdynamics and als be capable f applying them t slve prblems. 2. Understand what the parameter entrpy means.

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Introduction to Spacetime Geometry

Introduction to Spacetime Geometry Intrductin t Spacetime Gemetry Let s start with a review f a basic feature f Euclidean gemetry, the Pythagrean therem. In a twdimensinal crdinate system we can relate the length f a line segment t the

More information

CHEM 1001 Problem Set #3: Entropy and Free Energy

CHEM 1001 Problem Set #3: Entropy and Free Energy CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre.

More information

Lecture 12: Chemical reaction equilibria

Lecture 12: Chemical reaction equilibria 3.012 Fundamentals f Materials Science Fall 2005 Lecture 12: 10.19.05 Chemical reactin equilibria Tday: LAST TIME...2 EQUATING CHEMICAL POTENTIALS DURING REACTIONS...3 The extent f reactin...3 The simplest

More information

Solutions to the Extra Problems for Chapter 14

Solutions to the Extra Problems for Chapter 14 Slutins t the Extra Prblems r Chapter 1 1. The H -670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +

More information

188 CHAPTER 6 THERMOCHEMISTRY

188 CHAPTER 6 THERMOCHEMISTRY 188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101.

More information

Semester 2 AP Chemistry Unit 12

Semester 2 AP Chemistry Unit 12 Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved

More information

Chem 75 February 16, 2017 Exam 2 Solutions

Chem 75 February 16, 2017 Exam 2 Solutions 1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml

More information

Physics 262/266. George Mason University. Prof. Paul So

Physics 262/266. George Mason University. Prof. Paul So Physics 262/266 Gerge Masn University Prf. Paul S PHYS 262/266 Annuncements WELCOME TO A NEW SEMESTER! Curse Website - http://cmplex.gmu.edu/www-phys/phys262 - http://cmplex.gmu.edu/www-phys/phys266 Recitatins

More information

Unit 9: The Mole- Guided Notes What is a Mole?

Unit 9: The Mole- Guided Notes What is a Mole? Unit 9: The Mle- Guided Ntes What is a Mle? A mle is a name fr a specific f things Similar t a r a One mle is equal t 602 602,000,000,000,000,000,000,000 That s 602 with zers A mle is NOT an abbreviatin

More information

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s)

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s) Chapter 9 - Stichimetry Sectin 9.1 Intrductin t Stichimetry Types f Stichimetry Prblems Given is in mles and unknwn is in mles. Given is in mles and unknwn is in mass (grams). Given is in mass and unknwn

More information

Hess Law - Enthalpy of Formation of Solid NH 4 Cl

Hess Law - Enthalpy of Formation of Solid NH 4 Cl Hess Law - Enthalpy f Frmatin f Slid NH 4 l NAME: OURSE: PERIOD: Prelab 1. Write and balance net inic equatins fr Reactin 2 and Reactin 3. Reactin 2: Reactin 3: 2. Shw that the alebraic sum f the balanced

More information

A Chemical Reaction occurs when the of a substance changes.

A Chemical Reaction occurs when the of a substance changes. Perid: Unit 8 Chemical Reactin- Guided Ntes Chemical Reactins A Chemical Reactin ccurs when the f a substance changes. Chemical Reactin: ne r mre substances are changed int ne r mre new substances by the

More information

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 (Nte: questins 1 t 14 are meant t be dne WITHOUT calculatrs!) 1.Which f the fllwing is prbably true fr a slid slute with a highly endthermic heat

More information

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b . REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but

More information

This section is primarily focused on tools to aid us in finding roots/zeros/ -intercepts of polynomials. Essentially, our focus turns to solving.

This section is primarily focused on tools to aid us in finding roots/zeros/ -intercepts of polynomials. Essentially, our focus turns to solving. Sectin 3.2: Many f yu WILL need t watch the crrespnding vides fr this sectin n MyOpenMath! This sectin is primarily fcused n tls t aid us in finding rts/zers/ -intercepts f plynmials. Essentially, ur fcus

More information

CHM112 Lab Graphing with Excel Grading Rubric

CHM112 Lab Graphing with Excel Grading Rubric Name CHM112 Lab Graphing with Excel Grading Rubric Criteria Pints pssible Pints earned Graphs crrectly pltted and adhere t all guidelines (including descriptive title, prperly frmatted axes, trendline

More information

Name: Period: Date: BONDING NOTES HONORS CHEMISTRY

Name: Period: Date: BONDING NOTES HONORS CHEMISTRY Name: Perid: Date: BONDING NOTES HONORS CHEMISTRY Directins: This packet will serve as yur ntes fr this chapter. Fllw alng with the PwerPint presentatin and fill in the missing infrmatin. Imprtant terms

More information

Chem 111 Summer 2013 Key III Whelan

Chem 111 Summer 2013 Key III Whelan Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general

More information

/ / Chemistry. Chapter 1 Chemical Foundations

/ / Chemistry. Chapter 1 Chemical Foundations Name Chapter 1 Chemical Fundatins Advanced Chemistry / / Metric Cnversins All measurements in chemistry are made using the metric system. In using the metric system yu must be able t cnvert between ne

More information

Experiment #3. Graphing with Excel

Experiment #3. Graphing with Excel Experiment #3. Graphing with Excel Study the "Graphing with Excel" instructins that have been prvided. Additinal help with learning t use Excel can be fund n several web sites, including http://www.ncsu.edu/labwrite/res/gt/gt-

More information

Student Exploration: Cell Energy Cycle

Student Exploration: Cell Energy Cycle Name: Date: Student Explratin: Cell Energy Cycle Vcabulary: aerbic respiratin, anaerbic respiratin, ATP, cellular respiratin, chemical energy, chlrphyll, chlrplast, cytplasm, glucse, glyclysis, mitchndria,

More information

CHAPTER 13 Temperature and Kinetic Theory. Units

CHAPTER 13 Temperature and Kinetic Theory. Units CHAPTER 13 Temperature and Kinetic Thery Units Atmic Thery f Matter Temperature and Thermmeters Thermal Equilibrium and the Zerth Law f Thermdynamics Thermal Expansin Thermal Stress The Gas Laws and Abslute

More information

Physics 2010 Motion with Constant Acceleration Experiment 1

Physics 2010 Motion with Constant Acceleration Experiment 1 . Physics 00 Mtin with Cnstant Acceleratin Experiment In this lab, we will study the mtin f a glider as it accelerates dwnhill n a tilted air track. The glider is supprted ver the air track by a cushin

More information

Acids and Bases Lesson 3

Acids and Bases Lesson 3 Acids and Bases Lessn 3 The ph f a slutin is defined as the negative lgarithm, t the base ten, f the hydrnium in cncentratin. In a neutral slutin at 25 C, the hydrnium in and the hydrxide in cncentratins

More information

SPH3U1 Lesson 06 Kinematics

SPH3U1 Lesson 06 Kinematics PROJECTILE MOTION LEARNING GOALS Students will: Describe the mtin f an bject thrwn at arbitrary angles thrugh the air. Describe the hrizntal and vertical mtins f a prjectile. Slve prjectile mtin prblems.

More information

Pipetting 101 Developed by BSU CityLab

Pipetting 101 Developed by BSU CityLab Discver the Micrbes Within: The Wlbachia Prject Pipetting 101 Develped by BSU CityLab Clr Cmparisns Pipetting Exercise #1 STUDENT OBJECTIVES Students will be able t: Chse the crrect size micrpipette fr

More information

Making and Experimenting with Voltaic Cells. I. Basic Concepts and Definitions (some ideas discussed in class are omitted here)

Making and Experimenting with Voltaic Cells. I. Basic Concepts and Definitions (some ideas discussed in class are omitted here) Making xperimenting with Vltaic Cells I. Basic Cncepts Definitins (sme ideas discussed in class are mitted here) A. Directin f electrn flw psitiveness f electrdes. If ne electrde is mre psitive than anther,

More information

Physics 212. Lecture 12. Today's Concept: Magnetic Force on moving charges. Physics 212 Lecture 12, Slide 1

Physics 212. Lecture 12. Today's Concept: Magnetic Force on moving charges. Physics 212 Lecture 12, Slide 1 Physics 1 Lecture 1 Tday's Cncept: Magnetic Frce n mving charges F qv Physics 1 Lecture 1, Slide 1 Music Wh is the Artist? A) The Meters ) The Neville rthers C) Trmbne Shrty D) Michael Franti E) Radiatrs

More information

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry

More information

Medium Scale Integrated (MSI) devices [Sections 2.9 and 2.10]

Medium Scale Integrated (MSI) devices [Sections 2.9 and 2.10] EECS 270, Winter 2017, Lecture 3 Page 1 f 6 Medium Scale Integrated (MSI) devices [Sectins 2.9 and 2.10] As we ve seen, it s smetimes nt reasnable t d all the design wrk at the gate-level smetimes we just

More information

BASD HIGH SCHOOL FORMAL LAB REPORT

BASD HIGH SCHOOL FORMAL LAB REPORT BASD HIGH SCHOOL FORMAL LAB REPORT *WARNING: After an explanatin f what t include in each sectin, there is an example f hw the sectin might lk using a sample experiment Keep in mind, the sample lab used

More information

Fall 2013 Physics 172 Recitation 3 Momentum and Springs

Fall 2013 Physics 172 Recitation 3 Momentum and Springs Fall 03 Physics 7 Recitatin 3 Mmentum and Springs Purpse: The purpse f this recitatin is t give yu experience wrking with mmentum and the mmentum update frmula. Readings: Chapter.3-.5 Learning Objectives:.3.

More information

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command NUPOC SUDY GUIDE ANSWER KEY Navy Recruiting Cmmand CHEMISRY. ph represents the cncentratin f H ins in a slutin, [H ]. ph is a lg scale base and equal t lg[h ]. A ph f 7 is a neutral slutin. PH < 7 is acidic

More information

Name: Period: Date: BONDING NOTES ADVANCED CHEMISTRY

Name: Period: Date: BONDING NOTES ADVANCED CHEMISTRY Name: Perid: Date: BONDING NOTES ADVANCED CHEMISTRY Directins: This packet will serve as yur ntes fr this chapter. Fllw alng with the PwerPint presentatin and fill in the missing infrmatin. Imprtant terms

More information

5 th grade Common Core Standards

5 th grade Common Core Standards 5 th grade Cmmn Cre Standards In Grade 5, instructinal time shuld fcus n three critical areas: (1) develping fluency with additin and subtractin f fractins, and develping understanding f the multiplicatin

More information

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s .9 Kinetic Mlecular Thery Calculate the effective (rms) speeds f the He and Ne atms in the He-Ne gas laser tube at rm temperature (300 K). Slutin T find the rt mean square velcity (v rms ) f He atms at

More information

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013 CHEM-443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal

More information

37 Maxwell s Equations

37 Maxwell s Equations 37 Maxwell s quatins In this chapter, the plan is t summarize much f what we knw abut electricity and magnetism in a manner similar t the way in which James Clerk Maxwell summarized what was knwn abut

More information

We can see from the graph above that the intersection is, i.e., [ ).

We can see from the graph above that the intersection is, i.e., [ ). MTH 111 Cllege Algebra Lecture Ntes July 2, 2014 Functin Arithmetic: With nt t much difficulty, we ntice that inputs f functins are numbers, and utputs f functins are numbers. S whatever we can d with

More information

Work and Heat Definitions

Work and Heat Definitions Wrk and eat Deinitins FL- Surrundings: Everything utside system + q -q + System: he part S the rld e are bserving. Wrk, : transer energy as a result unbalanced rces - eat, q: transer energy resulting rm

More information

" 1 = # $H vap. Chapter 3 Problems

 1 = # $H vap. Chapter 3 Problems Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius

More information

Heat is energy and is measured in joules (J) or kilojoules (kj). The symbol for heat is H.

Heat is energy and is measured in joules (J) or kilojoules (kj). The symbol for heat is H. Causes f Change Calrimetry Hw Des Energy Affect Change? Heat vs. Temerature HEAT TEMPERATURE Definitin: Deends n: Examles: Heat is energy and is measured in jules (J) r kiljules (kj). The symbl fr heat

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

Compressibility Effects

Compressibility Effects Definitin f Cmpressibility All real substances are cmpressible t sme greater r lesser extent; that is, when yu squeeze r press n them, their density will change The amunt by which a substance can be cmpressed

More information

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium Lecture 17: 11.07.05 Free Energy f Multi-phase Slutins at Equilibrium Tday: LAST TIME...2 FREE ENERGY DIAGRAMS OF MULTI-PHASE SOLUTIONS 1...3 The cmmn tangent cnstructin and the lever rule...3 Practical

More information

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal Mass-Spring System A Hrizntal Mass-Spring System is where a mass is attached t a spring, riented hrizntally, and then placed

More information

Study Guide Physics Pre-Comp 2013

Study Guide Physics Pre-Comp 2013 I. Scientific Measurement Metric Units S.I. English Length Meter (m) Feet (ft.) Mass Kilgram (kg) Pund (lb.) Weight Newtn (N) Ounce (z.) r pund (lb.) Time Secnds (s) Secnds (s) Vlume Liter (L) Galln (gal)

More information

CHAPTER 6 THERMOCHEMISTRY. Questions

CHAPTER 6 THERMOCHEMISTRY. Questions CHAPTER 6 THERMOCHEMISTRY Questins 11. Path-dependent functins fr a trip frm Chicag t Denver are thse quantities that depend n the rute taken. One can fly directly frm Chicag t Denver, r ne culd fly frm

More information

More Tutorial at

More Tutorial at Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,

More information

Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol

Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol Recitatin 06 Mixture f Ideal Gases 1. Chapter 5: Exercise: 69 The partial pressure f CH 4 (g) is 0.175 atm and that f O 2 (g) is 0.250 atm in a mixture f the tw gases. a. What is the mle fractin f each

More information

Process Engineering Thermodynamics E (4 sp) Exam

Process Engineering Thermodynamics E (4 sp) Exam Prcess Engineering Thermdynamics 42434 E (4 sp) Exam 9-3-29 ll supprt material is allwed except fr telecmmunicatin devices. 4 questins give max. 3 pints = 7½ + 7½ + 7½ + 7½ pints Belw 6 questins are given,

More information

A Few Basic Facts About Isothermal Mass Transfer in a Binary Mixture

A Few Basic Facts About Isothermal Mass Transfer in a Binary Mixture Few asic Facts but Isthermal Mass Transfer in a inary Miture David Keffer Department f Chemical Engineering University f Tennessee first begun: pril 22, 2004 last updated: January 13, 2006 dkeffer@utk.edu

More information

lecture 5: Nucleophilic Substitution Reactions

lecture 5: Nucleophilic Substitution Reactions lecture 5: Nuclephilic Substitutin Reactins Substitutin unimlecular (SN1): substitutin nuclephilic, unimlecular. It is first rder. The rate is dependent upn ne mlecule, that is the substrate, t frm the

More information