188 CHAPTER 6 THERMOCHEMISTRY

Size: px
Start display at page:

Download "188 CHAPTER 6 THERMOCHEMISTRY"

Transcription

1 188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101. J L atm 101. J L atm = 1060 J = 1810 J ΔE = q + w = 107 J 1810 J = 770 J 5. w = PΔV;; we need the final vlume f the gas. Because T and n are cnstant, P 1 V 1 = P V. V V1 P L(15.0 atm) = 75.0 L P.00atm w = PΔV =.00 atm(75.0 L 10.0 L) = 10. L atm 6. w = 10. J = PΔV, 10 J = P(5 L 10. L), P = 14 atm 7. In this prblem, q = w = 950. J J 1 L atm = 9.8 L atm f wrk dne by the gases 101. J 101. J L atm 1kJ 1000J = 1. kj = wrk w = PΔV, 9.8 L atm = ΔE = q + w, 10.5 J = 5.5 J + w, w = J atm (V f L), V f = 11.0 L, V f = 11.0 L 1 L atm 101. J w = PΔV, 1.50 L atm = atm ΔV, ΔV =.06 L ΔV = V f V i,.06 L = 58.0 L V i, V i = 54.9 L = initial vlume 9. q = mlar heat capacity ml ΔT = w = PΔV = 1.00 atm (998 L 876 L) = 1 L atm ΔE = q + w = 0.9 kj + (1.4 kj) = 18.5 kj = 1.50 L atm 0.8 J 9.1 ml ( ) C = 0,900 J C ml = 0.9 kj 101. J L atm = 1,400 J = 1.4 kj

2 CHAPTER 6 THERMOCHEMISTRY H O(g) H O(l); ΔE = q + w; q = kj; w = PΔV Vlume f 1 ml H O(l) = ml H O(l) 18.0g ml 1cm = 18.1 cm = 18.1 ml 0.996g w = PΔV = 1.00 atm ( L 0.6 L) = 0.6 L atm ΔE = q + w = kj +.10 kj = 7.56 kj 101. J L atm = J =.10 kj Prperties f Enthalpy 41. This is an endthermic reactin, s heat must be absrbed in rder t cnvert reactants int prducts. The high-temperature envirnment f internal cmbustin engines prvides the heat. 4. One shuld try t cl the reactin mixture r prvide sme means f remving heat because the reactin is very exthermic (heat is released). The H SO 4 (aq) will get very ht and pssibly bil unless cling is prvided. 4. a. Heat is absrbed frm the water (it gets clder) as KBr disslves, s this is an endthermic prcess. b. Heat is released as CH 4 is burned, s this is an exthermic prcess. c. Heat is released t the water (it gets ht) as H SO 4 is added, s this is an exthermic prcess. d. Heat must be added (absrbed) t bil water, s this is an endthermic prcess. 44. a. The cmbustin f gasline releases heat, s this is an exthermic prcess. b. H O(g) H O(l); heat is released when water vapr cndenses, s this is an exthermic prcess. c. T cnvert a slid t a gas, heat must be absrbed, s this is an endthermic prcess. d. Heat must be added (absrbed) in rder t break a bnd, s this is an endthermic prcess Fe(s) + O (g) Fe O (s) ΔH = 165 kj; nte that 165 kj f heat is released when 4 ml Fe reacts with ml O t prduce ml Fe O. 165kJ a ml Fe = 1650 kj; 1650 kj f heat released 4 mlfe 165kJ b ml Fe O = 86 kj; 86 kj f heat released mlfe O

3 CHAPTER 6 THERMOCHEMISTRY H = E + PV; frm this equatin, H > E when V > 0, H < E when V < 0, and H = E when V = 0. Cncentrate n the mles f gaseus prducts versus the mles f gaseus reactants t predict V fr a reactin. a. There are mles f gaseus reactants cnverting t mles f gaseus prducts, s V = 0. Fr this reactin, H = E. b. There are 4 mles f gaseus reactants cnverting t mles f gaseus prducts, s V < 0 and H < E. c. There are 9 mles f gaseus reactants cnverting t 10 mles f gaseus prducts, s V > 0 and H > E. Calrimetry and Heat Capacity 51. Specific heat capacity is defined as the amunt f heat necessary t raise the temperature f ne gram f substance by ne degree Celsius. Therefre, H O(l) with the largest heat capacity value requires the largest amunt f heat fr this prcess. The amunt f heat fr H O(l) is: energy = s m ΔT = 5.0 g (7.0 C 15.0 C) =.0 10 J The largest temperature change when a certain amunt f energy is added t a certain mass f substance will ccur fr the substance with the smallest specific heat capacity. This is Hg(l), and the temperature change fr this prcess is: ΔT = energy s m 10.7 kj 0.14J 1000J kj 550. g = 140 C 5. a. s = specific heat capacity = 0.4J 0.4J since ΔT(K) = ΔT( C) K g Energy = s m ΔT = 0.4J g (98 K 7 K) = J b. Mlar heat capacity = 0.4J g Ag mlag 6J C ml c. 150 J = 0.4J m (15. C 1.0 C), m = = g Ag 5. s = specific heat capacity = q m T 1J = J/ C g 5.00g ( ) C Frm Table 6.1, the substance is slid aluminum.

4 19 CHAPTER 6 THERMOCHEMISTRY 54. s = 585J 15.6 g ( ) C = 0.19 J/ C g Mlar heat capacity = 0.19J 00.6 g mlhg 7.9 J C ml 55. Heat lss by ht water = heat gain by cler water The magnitudes f heat lss and heat gain are equal in calrimetry prblems. The nly difference is the sign (psitive r negative). T avid sign errrs, keep all quantities psitive and, if necessary, deduce the crrect signs at the end f the prblem. Water has a specific heat capacity = s = 4.18 J/ C g = 4.18 J/K g (ΔT in C = ΔT in K). Heat lss by ht water = s m ΔT = Heat gain by cler water = 09 J 15 J (0. K Tf ) = (Tf 80. K) K K 50.0 g (0. K T f ) K g 0.0 g (T f 80. K); heat lss = heat gain, s: K g T f = 15T f , 4T f = , T f = 11 K Nte that the final temperature is clser t the temperature f the mre massive ht water, which is as it shuld be. 56. Heat lss by ht water = heat gain by cld water; keeping all quantities psitive helps t avid sign errrs: m ht (55.0 C 7.0 C) = 90.0 g (7.0 C.0 C) m ht = 90.0g 15.0 C 18.0 C = 75.0 g ht water needed 57. Heat lss by Al + heat lss by Fe = heat gain by water; keeping all quantities psitive t avid sign errr: 0.89J 5.00 g Al (100.0 C T f ) J g Fe (100.0 T f ) = 97. g H O (T f.0 C) 4.5(100.0 T f ) + 4.5(100.0 T f ) = 407(T f.0), 450 (4.5)T f (4.5)T f = 407T f T f = 9850, T f =.7 C

5 CHAPTER 6 THERMOCHEMISTRY Heat released t water = 5.0 g H 10.J g H g methane 50.J = J g methane Heat gain by water = J = 50.0 g T T = 5.6 C, 5.6 C = T f 5.0 C, T f = 0. C 59. Heat gain by water = heat lss by metal = s m ΔT, where s = specific heat capacity. Heat gain = g (18. C 15.0 C) = 100 J A cmmn errr in calrimetry prblems is sign errrs. Keeping all quantities psitive helps t eliminate sign errrs. Heat lss = 100 J = s g (75.0 C 18. C), s = 100J g 56.7 = 0.5 J/ C g C 60. Heat gain by water = heat lss by Cu; keeping all quantities psitive helps t avid sign errrs: mass (4.9 C. C) = 0.0J 110. g Cu (8.4 C 4.9 C) 11(mass) = 100, mass = 10 g H O L ml/l = ml f bth AgNO and HCl are reacted. Thus ml f AgCl will be prduced because there is a 1 : 1 mle rati between reactants. Heat lst by chemicals = heat gained by slutin Heat gain = g (.40.60) C = 0 J Heat lss = 0 J; this is the heat evlved (exthermic reactin) when ml f AgCl is prduced. S q = 0 J and ΔH (heat per ml AgCl frmed) is negative with a value f: ΔH = 0J ml 1kJ 1000J = 66 kj/ml Nte: Sign errrs are cmmn with calrimetry prblems. Hwever, the crrect sign fr ΔH can be determined easily frm the ΔT data;; i.e., if ΔT f the slutin increases, then the reactin is exthermic because heat was released, and if ΔT f the slutin decreases, then the reactin is endthermic because the reactin absrbed heat frm the water. Fr calrimetry prblems, keep all quantities psitive until the end f the calculatin and then decide the sign fr ΔH. This will help eliminate sign errrs.

6 194 CHAPTER 6 THERMOCHEMISTRY 6. NaOH(aq) + HCl(aq) NaCl(aq) + H O(l) We have a stichimetric mixture. All f the NaOH and HCl will react L 1.0ml = 0.10 ml f HCl is neutralized by 0.10 ml NaOH. L Heat lst by chemicals = heat gained by slutin Vlume f slutin = = 00.0 ml Heat gain = 1.0 g 00.0 ml ml (1. 4.6)C = J = 5.6 kj Heat lss = 5.6 kj; this is the heat released by the neutralizatin f 0.10 ml HCl. Because the temperature increased, the sign fr ΔH must be negative, i.e., the reactin is exthermic. Fr calrimetry prblems, keep all quantities psitive until the end f the calculatin and then decide the sign fr ΔH. This will help eliminate sign errrs. ΔH = 5.6 kj 0.10ml = 56 kj/ml 6. Heat lst by slutin = heat gained by KBr; mass f slutin = 15 g g = 16 g Nte: Sign errrs are cmmn with calrimetry prblems. Hwever, the crrect sign fr ΔH can easily be btained frm the ΔT data. When wrking calrimetry prblems, keep all quantities psitive (ignre signs). When finished, deduce the crrect sign fr ΔH. Fr this prblem, T decreases as KBr disslves, s ΔH is psitive;; the disslutin f KBr is endthermic (absrbs heat). Heat lst by slutin = 16 g (4. C 1.1 C) = 1800 J = heat gained by KBr ΔH in units f J/g = 1800J 10.5 g KBr = 170 J/g ΔH in units f kj/ml = 170J g KBr g KBr mlkbr 1kJ 1000J = 0. kj/ml 64. NH 4 NO (s) NH 4 + (aq) + NO (aq) ΔH =?;; mass f slutin = 75.0 g g = 76.6 g Heat lst by slutin = heat gained as NH 4 NO disslves. T help eliminate sign errrs, we will keep all quantities psitive (q and ΔT) and then deduce the crrect sign fr ΔH at the end f the prblem. Here, because temperature decreases as NH 4 NO disslves, heat is absrbed as NH 4 NO disslves, s this is an endthermic prcess (ΔH is psitive). Heat lst by slutin = 76.6 g (5.00.4) C = 5 J = heat gained as NH 4 NO disslves

7 CHAPTER 6 THERMOCHEMISTRY 195 ΔH = 5J 1.60g NH NO g NH4NO 1kJ = 6.6 kj/ml NH 4 NO disslving ml NH NO 1000J Because ΔH is exthermic, the temperature f the slutin will increase as CaCl (s) disslves. Keeping all quantities psitive: L 1ml CaCl 81.5 kj heat lss as CaCl disslves = 11.0 g CaCl = 8.08 kj g CaCl ml CaCl heat gained by slutin = J = T f 5.0 C = mlHCl L ( ) g (T f 5.0 C) 10 = 14. C, T f = 14. C C = 9. C kJ heat released =.95 kj f heat released if HCl limiting mlhcl L 0.100mlBa(OH) L 118kJ heat released mlba(oh) =.54 kj heat released if Ba(OH) limiting Because the HCl reagent prduces the smaller amunt f heat released, HCl is limiting and.95 kj f heat are released by this reactin. Heat gained by slutin = J = ΔT = 1.76 C = T f T i = T f 5.0 C, T f = 6.8 C g ΔT 67. a. Heat gain by calrimeter = heat lss by CH 4 = 6.79 g CH 4 1mlCH 4 80kJ 16.04g ml = 40. kj 40. kj Heat capacity f calrimeter = = 1.5 kj/ C 10.8 C 1. 5 kj b. Heat lss by C H = heat gain by calrimeter = 16.9 C = 5 kj C A bmb calrimeter is at cnstant vlume, s the heat released/gained = q V = E: 5kJ ΔE cmb = 1.6 g C H 6.04g mlc H = kj/ml 68. First, we need t get the heat capacity f the calrimeter frm the cmbustin f benzic acid. Heat lst by cmbustin = heat gained by calrimeter. Heat lss = g 6.4kJ g = kj

8 198 CHAPTER 6 THERMOCHEMISTRY 76. P 4 O 10 P O ΔH = (967. kj) 10 PCl + 5 O 10 Cl PO ΔH = 10(85.7 kj) 6 PCl 5 6 PCl + 6 Cl ΔH = 6(84. kj) P Cl 4 PCl ΔH = 15.6 P 4 O 10 (s) + 6 PCl 5 (g) 10 Cl PO(g) ΔH = kj Standard Enthalpies f Frmatin 77. The change in enthalpy that accmpanies the frmatin f 1 mle f a cmpund frm its elements, with all substances in their standard states, is the standard enthalpy f frmatin fr a cmpund. The reactins that refer t H f are: Na(s) + 1/ Cl (g) NaCl(s);; H (g) + 1/ O (g) H O(l) 6 C(graphite, s) + 6 H (g) + O (g) C 6 H 1 O 6 (s) Pb(s) + S(rhmbic, s) + O (g) PbSO 4 (s) 78. a. Aluminum xide = Al O ; Al(s) + / O (g) Al O (s) b. C H 5 OH(l) + O (g) CO (g) + H O(l) c. NaOH(aq) + HCl(aq) H O(l) + NaCl(aq) d. C(graphite, s) + / H (g) + 1/ Cl (g) C H Cl(g) e. C 6 H 6 (l) + 15/ O (g) 6 CO (g) + H O(l) Nte: ΔH cmb values assume 1 mle f cmpund cmbusted. f. NH 4 Br(s) NH 4 + (aq) + Br (aq) 79. In general, n p ΔH f n, prducts r ΔH f, and all elements in their standard, reactants state have Δ = 0 by definitin. H f a. The balanced equatin is NH (g) + O (g) + CH 4 (g) HCN(g) + 6 H O(g). ( ml HCN ΔH f, HCN + 6 ml H O(g) H f, HO Δ ) ( ml NH Δ + ml CH 4 H ) H f, NH f, CH 4 [(15.1) + 6(4)] [(46) + (75)] = 940. kj

9 CHAPTER 6 THERMOCHEMISTRY 199 b. Ca (PO 4 ) (s) + H SO 4 (l) CaSO 4 (s) + H PO 4 (l) 14kJ 167kJ mlcaso 4 (s) mlhpo4(l) ml ml mlca 68 kj (6568 kj) = 65 kj c. NH (g) + HCl(g) NH 4 Cl(s) 416kJ (PO4) (s) mlhso ml kJ (l) ml (1 ml NH 4 Cl Δ ) (1 ml NH H f, NH4Cl ΔH f, + 1 ml HCl H f, HCl NH Δ ) 14kJ 1ml ml 46kJ 9kJ 1ml 1ml ml ml 14 kj + 18 kj = 176 kj 80. a. The balanced equatin is C H 5 OH(l) + O (g) CO (g) + H O(g). 9.5 kj ml ml 4kJ ml ml 78kJ 1ml ml 151 kj (78 kj) = 15 kj b. SiCl 4 (l) + H O(l) SiO (s) + 4 HCl(aq) Because HCl(aq) is H + (aq) + Cl (aq), Δ H f = = 167 kj/ml. 167kJ 4 ml ml 911kJ 1ml ml 687kJ 86kJ 1ml ml ml ml 1579 kj (159 kj) = 0. kj c. MgO(s) + H O(l) Mg(OH) (s) 95kJ 1 ml ml 60kJ 86kJ 1ml 1ml ml ml 95 kj (888 kj) = 7 kj 81. a. 4 NH (g) + 5 O (g) 4 NO(g) + 6 H O(g); n p ΔH f n, prducts r ΔH f, reactants 90. kj 4kJ 46kJ 4 ml 6 ml 4 ml = 908 kj ml ml ml

10 00 CHAPTER 6 THERMOCHEMISTRY NO(g) + O (g) NO (g) 4kJ 90. kj ml ml = 11 kj ml ml NO (g) + H O(l) HNO (aq) + NO(g) Nte: All 07kJ 90. kj ml 1ml ml ml ΔH f values are assumed ±1 kj. 4kJ 86kJ ml 1ml ml ml 140. kj b. 1 NH (g) + 15 O (g) 1 NO(g) + 18 H O(g) 1 NO(g) + 6 O (g) 1 NO (g) 1 NO (g) + 4 H O(l) 8 HNO (aq) + 4 NO(g) 4 H O(g) 4 H O(l) 1 NH (g) + 1 O (g) 8 HNO (aq) + 4 NO(g) + 14 H O(g) The verall reactin is exthermic because each step is exthermic. 416kJ 8. 4 Na(s) + O (g) Na O(s) ml ml = 8 kj Na(s) + H O(l) NaOH(aq) + H (g) 470. kj 86kJ ml ml = 68 kj ml ml Na(s) + CO (g) Na O(s) + CO(g) 416kJ kj 9.5 kj 1 ml 1ml 1ml = 1 kj ml ml ml In Reactins and, sdium metal reacts with the "extinguishing agent." Bth reactins are exthermic, and each reactin prduces a flammable gas, H and CO, respectively. 8. Al(s) + NH 4 ClO 4 (s) Al O (s) + AlCl (s) + NO(g) + 6 H O(g) 4kJ 90. kj 6 ml ml ml ml 704kJ 1676kJ 1ml 1ml ml ml 95kJ ml = 677 kj ml

11 64 CHAPTER 10 LIQUIDS AND SOLIDS a. The plateau at the lwest temperature signifies the melting/freezing f the substance. Hence the freezing pint is 0C. b. The higher temperature plateau signifies the biling/cndensatin f the substance. The temperature f this plateau is 10C. c. X(s) X(l) H = H fusin ; X(l) X(g) H = H vaprizatin The heat f fusin and the heat f vaprizatin terms refer t enthalpy changes fr the specific phase changes illustrated in the equatins abve. In a heating curve, energy is applied at a steady rate. S the lnger, higher temperature plateau has a larger enthalpy change assciated with it as cmpared t the shrter plateau. The higher temperature plateau ccurs when a liquid is cnverting t a gas, s the heat f vaprizatin is greater than the heat f fusin. This is always the case because significantly mre intermlecular frces are brken when a substance bils than when a substance melts. 97. a. Many mre intermlecular frces must be brken t cnvert a liquid t a gas as cmpared with cnverting a slid t a liquid. Because mre intermlecular frces must be brken, much mre energy is required t vaprize a liquid than is required t melt a slid. Therefre, H vap is much larger than H fus. b g Na c g Na 1mlNa.99g 1mlNa.99g.60kJ = 0.11 kj = 11 J t melt 1.00 g Na mlna 97.0 kj = 4. kj = 40 J t vaprize 1.00 g Na mlna d. This is the reverse prcess f that described in part c, s the energy change is the same quantity but ppsite in sign. Therefre, q = 40 J; i.e., 40 f heat will be released.

12 CHAPTER 10 LIQUIDS AND SOLIDS 65 1mlC6H 98. Melt: 8.5 g C 6 H g 1mlC6H Vaprize: 8.5 g C 6 H g 6 9.9kJ = 1.05 kj mlc H kj =.4 kj mlc H As is typical, the energy required t vaprize a certain quantity f substance is much larger than the energy required t melt the same quantity f substance. A lt mre intermlecular frces must be brken t vaprize a substance as cmpared t melting a substance. 99. T calculate q ttal, break up the heating prcess int five steps. H O(s, 0. C) H O(s, 0 C), ΔT = 0. C; let s ice = specific heat capacity f ice: 6 6 q 1 = s ice m ΔT =.0J g 0. C = J = 0. kj g C H O(s, 0 C) H O(l, 0 C), q = g H O 1ml 18.0g 6.0kJ = 167 kj ml H O(l, 0 C) H O(l, 100. C), q = 4. J g 100. C = J = 10 kj g C H O(l, 100. C) H O(g, 100. C), q 4 = g 1ml 18.0g 40.7 kj = 110 kj ml.0 J H O(g, 100. C) H O(g, 50. C), q 5 = g 150. C = J g C = 150 kj q ttal = q 1 + q + q + q 4 + q 5 = = 1680 kj 100. H O(g, 15 C) H O(g, 100. C), q 1 =.0 J/g C 75.0 g ( 5 C) = 800 J =.8 kj H O(g, 100. C) H O(l, 100. C), q = 75.0 g 1ml 40.7 kj = 169 kj 18.0g ml H O(l, 100. C) H O(l, 0 C), q = 4. J/g C 75.0 g (100. C) =,000 J = kj T cnvert H O(g) at 15 C t H O(l) at 0 C requires (.8 kj 169 kj kj =) 05 kj f heat remved. T cnvert frm H O(l) at 0 C t H O(s) at 0 C requires: q 4 = 75.0 g 1ml 6.0kJ = 5.1 kj 18.0g ml This amunt f energy puts us ver the 15 kj limit ( 05 kj 5.1 kj = 0. kj). Therefre, a mixture f H O(s) and H O(l) will be present at 0 C when 15 kj f heat is remved frm the gas sample.

13 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY kj AB B A B A 1 kj AB B A A B 0 kj AB A B A B E ttal = 0 kj kj 4 kj 1 kj 1 kj kj kj kj kj The mst likely ttal energy is kj. 1. a. H at 100 C and 0.5 atm; higher temperature and lwer pressure means greater vlume and hence larger psitinal prbability. b. N ; N at STP has the greater vlume because P is smaller and T is larger. c. H O(l) has a larger psitinal prbability than H O(s).. Of the three phases (slid, liquid, and gas), slids are mst rdered (have the smallest psitinal prbability) and gases are mst disrdered (have the largest psitinal prbability). Thus a, b, and f (melting, sublimatin, and biling) invlve an increase in the entrpy f the system since ging frm a slid t a liquid r frm a slid t a gas r frm a liquid t a gas increases disrder (increases psitinal prbability). Fr freezing (prcess c), a substance ges frm the mre disrdered liquid state t the mre rdered slid state; hence, entrpy decreases. Prcess d (mixing) invlves an increase in disrder (an increase in psitinal prbability), while separatin (phase e) increases rder (decreases psitinal prbability). S, f all the prcesses, a, b, d, and f result in an increase in the entrpy f the system.. a. Biling a liquid requires heat. Hence this is an endthermic prcess. All endthermic prcesses decrease the entrpy f the surrundings (ΔS surr is negative). b. This is an exthermic prcess. Heat is released when gas mlecules slw dwn enugh t frm the slid. In exthermic prcesses, the entrpy f the surrundings increases (ΔS surr is psitive). 4. a. ΔS surr = ΔH T ( 1kJ) = 7.45 kj/k = J/K 98K b. ΔS surr = ΔH 11kJ = 0.76 kj/k = 76 J/K T 98K 5. ΔG = ΔH TΔS;; when ΔG is negative, then the prcess will be spntaneus. a. ΔG = ΔH TΔS = 5 10 J (00. K)(5.0 J/K) = 4,000 J; nt spntaneus b. ΔG = 5,000 J (00. K)(100. J/K) = 5000 J; spntaneus c. Withut calculating ΔG, we knw this reactin will be spntaneus at all temperatures. ΔH is negative and ΔS is psitive (TΔS < 0). ΔG will always be less than zer with these sign cmbinatins fr ΔH and ΔS. d. ΔG = J (00. K)( 40. J/K) = 000 J; spntaneus

14 664 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY 6. ΔG = ΔH TΔS;; a prcess is spntaneus when ΔG < 0. Fr the fllwing, assume ΔH and ΔS are temperature-independent. a. When ΔH and ΔS are bth negative, ΔG will be negative belw a certain temperature where the favrable ΔH term dminates. When ΔG = 0, then ΔH = TΔS. Slving fr this temperature: T = ΔH ΔS 18,000J =.0 10 K 60. J / K At T <.0 10 K, this prcess will be spntaneus (ΔG < 0). b. When ΔH and ΔS are bth psitive, ΔG will be negative abve a certain temperature where the favrable ΔS term dminates. T = ΔH 18,000J =.0 10 K ΔS 60. J / K At T >.0 10 K, this prcess will be spntaneus (ΔG < 0). c. When ΔH is psitive and ΔS is negative, this prcess can never be spntaneus at any temperature because ΔG can never be negative. d. When ΔH is negative and ΔS is psitive, this prcess is spntaneus at all temperatures because ΔG will always be negative. 7. At the biling pint, ΔG = 0, s ΔH = TΔS. ΔS = ΔH T 7.5 kj/ ml = 8.9 (7 5) K 8. At the biling pint, ΔG = 0, s ΔH = TΔS. T = 10 kj/k ml = 89. J/K ml ΔH ΔS J/ml = 69.7 K 9.9J/K ml 9. a. NH (s) NH (l);; ΔG = ΔH TΔS = 5650 J/ml 00. K (8.9 J/K ml) ΔG = 5650 J/ml 5780 J/ml = 10 J/ml Yes, NH will melt because ΔG < 0 at this temperature. b. At the melting pint, ΔG = 0, s T = ΔH ΔS 5650 J/ml = 196 K. 8.9J/K ml 40. C H 5 OH(l) C H 5 OH(g); at the biling pint, G = 0 and S univ = 0. Fr the vaprizatin prcess, S is a psitive value, whereas H is a negative value. T calculate S sys, we will determine S surr frm H and the temperature; then S sys = S surr fr a system at equilibrium.

15 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY 665 S surr = H T J/ml = 110. J/K ml 51K S sys = S surr = (110.) = 110. J/K ml Chemical Reactins: Entrpy Changes and Free Energy 41. a. Decrease in psitinal prbability;; ΔS will be negative. There is nly ne way t achieve 1 gas mlecules all in ne bulb, but there are many mre ways t achieve the gas mlecules equally distributed in each flask. b. Decrease in psitinal prbability; ΔS is negative fr the liquid t slid phase change. c. Decrease in psitinal prbability; ΔS is negative because the mles f gas decreased when ging frm reactants t prducts ( mles 0 mles). Changes in the mles f gas present as reactants are cnverted t prducts dictates predicting psitinal prbability. The gaseus phase always has the larger psitinal prbability assciated with it. d. Increase in psitinal prbability;; ΔS is psitive fr the liquid t gas phase change. The gas phase always has the larger psitinal prbability. 4. a. Decrease in psitinal prbability (Δn < 0);; ΔS () b. Decrease in psitinal prbability (Δn < 0);; ΔS () c. Increase in psitinal prbability;; ΔS (+) d. Increase in psitinal prbability;; ΔS (+) 4. a. raphite (s); diamnd has a mre rdered structure (has a smaller psitinal prbability) than graphite. b. C H 5 OH(g); the gaseus state is mre disrdered (has a larger psitinal prbability) than the liquid state. c. CO (g); the gaseus state is mre disrdered (has a larger psitinal prbability) than the slid state. 44. a. He (10 K); S = 0 at 0 K b. N O; mre cmplicated mlecule, s has the larger psitinal prbability. c. NH (l); the liquid state is mre disrdered (has a larger psitinal prbability) than the slid state. 45. a. H S(g) + SO (g) S rhmbic (s) + H O(g); because there are mre mlecules f reactant gases than prduct mlecules f gas (Δn = < 0), ΔS will be negative.

16 670 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY b. Because ΔG is psitive, this reactin is nt spntaneus at standard cnditins and 98 K. c. ΔG = ΔH TΔS, ΔS = H G T 100. kj 5kJ = 0.16 kj/k 98K We need t slve fr the temperature when ΔG = 0: ΔG = 0 = ΔH TΔS, TΔS, T = H S 100.kJ 0.16kJ/K = 60 K This reactin will be spntaneus (ΔG < 0) at T > 60 K, where the favrable entrpy term will dminate. 6. a. ΔG = (70. kj) (50 kj) = 464 kj b. Because ΔG is psitive, this reactin is nt spntaneus at standard cnditins at 98 K. c. ΔG = ΔH TΔS, ΔG + TΔS = 464 kj + 98 K(0.179 kj/k) = 517 kj We need t slve fr the temperature when ΔG = 0: ΔG = 0 = ΔH TΔS, T = ΔH 517 kj = 890 K ΔS 0.179kJ/ K This reactin will be spntaneus at standard cnditins (ΔG < 0) when T > 890 K. Here the favrable entrpy term will dminate. 6. CH 4 (g) + CO (g) CH CO H(l) 484 [75 + (9.5)] = 16 kj;; ΔS = 160. ( ) = 40. J/K ΔG = ΔH TΔS = 16 kj (98 K)(0.40 kj/k) = 56 kj At standard cncentratins, this reactin is spntaneus nly at temperatures belw T = ΔH /ΔS = 67 K (where the favrable ΔH term will dminate, giving a negative ΔG value). This is nt practical. Substances will be in cndensed phases and rates will be very slw at this extremely lw temperature. CH OH(g) + CO(g) CH CO H(l) 484 [ (01)] = 17 kj;; ΔS = 160. ( ) = 78 J/K ΔG = 17 kj (98 K)(0.78 kj/k) = 90. kj This reactin als has a favrable enthalpy and an unfavrable entrpy term. But this reactin, at standard cncentratins, is spntaneus at temperatures belw T = ΔH /ΔS = 6 K (a much higher temperature than the first reactin). S the reactin f CH OH and CO will be preferred at standard cncentratins. It is spntaneus at high enugh temperatures that the rates f reactin shuld be reasnable.

17 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY C H 4 (g) + H O(g) CH CH OH(l) 78 (5 4) = 88 kj;; ΔS = 161 ( ) = 47 J/K When ΔG = 0, TΔS, s T = ΔH ΔS J = 60 K. 47J/K Since the signs f ΔH and ΔS are bth negative, this reactin at standard cncentratins will be spntaneus at temperatures belw 60 K (where the favrable ΔH term will dminate). C H 6 (g) + H O(g) CH CH OH(l) + H (g) 78 (84.7 4) = 49 kj;; ΔS = ( ) = 17 J/K Bth ΔH and ΔS have unfavrable signs. ΔG can never be negative when ΔH is psitive and ΔS is negative. S this reactin can never be spntaneus at standard cnditins. Thus the reactin C H 4 (g) + H O(g) C H 5 OH(l) wuld be preferred at standard cnditins. Free Energy: Pressure Dependence and Equilibrium P 65. ΔG = ΔG + RT ln Q;; fr this reactin: ΔG = ΔG + RT ln P NO NO P P ΔG = 1 ml(5 kj/ml) + 1 ml(0) [1 ml(87 kj/ml) + 1 ml(16 kj/ml)] = 198 kj O O ΔG = 198 kj J/K ml (98K) ln 1000J/kJ ( ( )( ) ( ) ) ΔG = 198 kj kj = 188 kj 66. ΔG = (0) + (9) [(4) + 1(00.)] = 90. kj ΔG = ΔG + RT ln P H S P H O P SO (8.145)(98) (0.00) = 90. kj + kj ln ( )(0.010) ΔG = 90. kj kj = 50. kj PN 67. ΔG = ΔG + RT ln Q = ΔG + RT ln P NO O4 ΔG = 1 ml(98 kj/ml) ml(5 kj/ml) = 6 kj a. These are standard cnditins, s ΔG = ΔG because Q = 1 and ln Q = 0. Because ΔG is negative, the frward reactin is spntaneus. The reactin shifts right t reach equilibrium.

18 67 CHAPTER 17 SPONTANEITY, ENTROPY, AND FREE ENERGY b. ΔG = 6 10 J J/K ml (98 K) ln ΔG = 6 10 J J = (0.1) Because ΔG = 0, this reactin is at equilibrium (n shift). c. ΔG = J J/K ml (98 K) ln (0.9) ΔG = 6 10 J J = J = 1 10 J Because ΔG is psitive, the reverse reactin is spntaneus, and the reactin shifts t the left t reach equilibrium. PNH 68. a. ΔG = ΔG + RT ln P P N H ; ΔG =, NH ΔG f = (17) = 4 kj (8.145J/K ml)(98k) (50.) ΔG = 4 kj + ln 1000J/kJ (00.)(00.) ΔG =4 kj kj = 67 kj (8.145J/K ml)(98k) (00.) b. ΔG = 4 kj + ln 1000J/kJ (00.)(600.) ΔG = 4 kj 4.4 kj = 68 kj 69. NO(g) + O (g) NO (g) + O (g); ΔG = p f, prducts Σn ΔG Σn ΔG r f, reactants ΔG = 1 ml(5 kj/ml) [1 ml(87 kj/ml) + 1 ml(16 kj/ml)] = 198 kj ΔG = RT ln K, K ΔG exp RT 5 ( J) exp 8.145J/K ml(98k) e = Nte: When determining expnents, we will rund ff after the calculatin is cmplete. This helps eliminate excessive rund ff errr. 70. ΔG = ml(9 kj/ml) [ ml(4 kj/ml) + 1 ml(00. kj/ml)] = 90. kj K ΔG exp RT 4 ( J) exp 8.145J/K ml(98k) e 6. = ΔG = ΔH TΔS ;; because there is a decrease in the number f mles f gaseus particles, ΔS is negative. Because ΔG is negative, ΔH must be negative. The reactin will be spntaneus at lw temperatures (the favrable ΔH term dminates at lw temperatures).

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

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

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

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

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

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)

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

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

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

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

SPONTANEITY, ENTROPY, AND FREE ENERGY

SPONTANEITY, ENTROPY, AND FREE ENERGY CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Questins. Living rganisms need an external surce f energy t carry ut these prcesses. Green plants use the energy frm sunlight t prduce glucse frm carbn dixide and

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Advanced Chemistry Practice Problems

Advanced Chemistry Practice Problems Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with

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

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

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

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 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions? 1 CHEM 1032 FALL 2017 Practice Exam 4 1. Which f the fllwing reactins is spntaneus under nrmal and standard cnditins? A. 2 NaCl(aq) 2 Na(s) + Cl2(g) B. CaBr2(aq) + 2 H2O(aq) Ca(OH)2(aq) + 2 HBr(aq) C.

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

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

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

BIT Chapters = =

BIT Chapters = = BIT Chapters 17-0 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride

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

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

Chem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points

Chem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points +2 pints Befre yu begin, make sure that yur exam has all 7 pages. There are 14 required prblems (7 pints each) and tw extra credit prblems (5 pints each). Stay fcused, stay calm. Wrk steadily thrugh yur

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

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

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

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

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

33. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process.

33. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic process. 31. This is an endothermic reaction so heat must be absorbed in order to convert reactants into products. The high temperature environment of internal combustion engines provides the heat. 33. a. Heat

More information

Thermochemistry Heats of Reaction

Thermochemistry Heats of Reaction hermchemistry Heats f Reactin aa + bb cc + dd hermchemical Semantics q V = Heat f Rxn at [V] = U = Energy (change) f Rxn q P = Heat f Rxn at [P] = H = Enthalpy (change) f Rxn Exthermic rxns: q < 0 Endthermic

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

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

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

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces

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

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 eep this site up and bring yu even mre cntent cnsider dnating via the lin n ur site. Still having truble understanding the material? Chec ut ur Tutring

More information

Thermodynamics 1/16/2013. Thermodynamics

Thermodynamics 1/16/2013. Thermodynamics Thermdynamics http://www.chem.purdue.edu/gchelp/hwtslveit/hwtslveit.html 1 Thermdynamics Study f energy changes and flw f energy Answers several fundamental questins: Is it pssible fr reactin t ccur? Will

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

" 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

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

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry Chapter 19 lectrchemistry Part I Dr. Al Saadi 1 lectrchemistry What is electrchemistry? It is a branch f chemistry that studies chemical reactins called redx reactins which invlve electrn transfer. 19.1

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

Tuesday, 5:10PM FORM A March 18,

Tuesday, 5:10PM FORM A March 18, Name Chemistry 153-080 (3150:153-080) EXAM II Multiple-Chice Prtin Instructins: Tuesday, 5:10PM FORM A March 18, 2003 120 1. Each student is respnsible fr fllwing instructins. Read this page carefully.

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

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O WYSE Academic Challenge Sectinal Chemistry Exam 2008 SOLUTION SET 1. Crrect answer: B. Use PV = nrt t get: PV = nrt 2. Crrect answer: A. (2.18 atm)(25.0 L) = n(0.08206 L atm/ml K)(23+273) n = 2.24 ml Assume

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

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

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

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

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia: University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,

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

Entropy and Gibbs energy

Entropy and Gibbs energy 14 Entrpy and Gibbs energy Answers t wrked examples WE 14.1 Predicting the sign f an entrpy change (n p. 658 in Chemistry 3 ) What will be the sign f the value f S fr: (a) crystallizatin f salt frm a slutin;

More information

Chapter 9 Chemical Reactions NOTES

Chapter 9 Chemical Reactions NOTES Chapter 9 Chemical Reactins NOTES Chemical Reactins Chemical reactin: Chemical change 4 Indicatrs f Chemical Change: (1) (2) (3) (4) Cnsist f reactants (starting materials) and prducts (substances frmed)

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

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

CHAPTER PRACTICE PROBLEMS CHEMISTRY

CHAPTER PRACTICE PROBLEMS CHEMISTRY Chemical Kinetics Name: Batch: Date: Rate f reactin. 4NH 3 (g) + 5O (g) à 4NO (g) + 6 H O (g) If the rate f frmatin f NO is 3.6 0 3 ml L s, calculate (i) the rate f disappearance f NH 3 (ii) rate f frmatin

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

Review Material for Exam #2

Review Material for Exam #2 Review Material fr Eam # 1. a. Calculate the mlarity f a slutin made with 184.6 mg sample f ptassium dichrmate disslved in enugh water t give 500.0 ml f slutin. 1 g 184.6 mg K Cr O 1 mle K Cr O 7 7 1000

More information

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS CHPTER 6 / HRVEY. CHEMICL B. THERMODYNMICS ND C. MNUPULTING CONSTNTS D. CONSTNTS FOR CHEMICL RECTIONS 1. Precipitatin Reactins 2. cid-base Reactins 3. Cmplexatin Reactins 4. Oxidatin-Reductin Reactins

More information

In the half reaction I 2 2 I the iodine is (a) reduced (b) oxidized (c) neither of the above

In the half reaction I 2 2 I the iodine is (a) reduced (b) oxidized (c) neither of the above 6.3-110 In the half reactin I 2 2 I the idine is (a) reduced (b) xidized (c) neither f the abve 6.3-120 Vitamin C is an "antixidant". This is because it (a) xidizes readily (b) is an xidizing agent (c)

More information

Chapter 3 Homework Solutions

Chapter 3 Homework Solutions Chapter Hmewrk Slutins. n = ml = 5 C = 98 K = 5 C = 98 K p = atm p = 5 atm. Cpm = (5/)R he entrpy changes fr the heating and cmpressin can e calculated separately and added. Heat at cnstant pressure S

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

Heat Effects of Chemical Reactions

Heat Effects of Chemical Reactions eat Effects f hemical Reactins Enthalpy change fr reactins invlving cmpunds Enthalpy f frmatin f a cmpund at standard cnditins is btained frm the literature as standard enthalpy f frmatin Δ (O (g = -9690

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

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate.

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate. CEM1405 2007-J-2 June 2007 In the spaces prvided, explain the meanings f the fllwing terms. Yu may use an equatin r diagram where apprpriate. 5 (a) hydrgen bnding An unusually strng diple-diple interactin

More information

A. Lattice Enthalpies Combining equations for the first ionization energy and first electron affinity:

A. Lattice Enthalpies Combining equations for the first ionization energy and first electron affinity: [15.1B Energy Cycles Lattice Enthalpy] pg. 1 f 5 CURRICULUM Representative equatins (eg M+(g) M+(aq)) can be used fr enthalpy/energy f hydratin, inizatin, atmizatin, electrn affinity, lattice, cvalent

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

Ch 10 Practice Problems

Ch 10 Practice Problems Ch 10 Practice Problems 1. Which of the following result(s) in an increase in the entropy of the system? I. (See diagram.) II. Br 2(g) Br 2(l) III. NaBr(s) Na + (aq) + Br (aq) IV. O 2(298 K) O 2(373 K)

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

KEY POINTS: NOTE: OCR A

KEY POINTS: NOTE: OCR A KEY: A.J.F.S DEFINITION: KEY POINTS: NOTE: OCR A Chemistry Mdule 3- Energy 2.3 (1) Enthalpy What is chemical energy? - Chemical energy is a special frm f ptential energy that lies within chemical bnds

More information

CHAPTER SIX THERMOCHEMISTRY. Questions

CHAPTER SIX THERMOCHEMISTRY. Questions CHAPTER SIX Questions 9. A coee-cup calorimeter is at constant (atmospheric) pressure. The heat released or gained at constant pressure is H. A bomb calorimeter is at constant volume. The heat released

More information

State of matter characteristics solid Retains shape and volume

State of matter characteristics solid Retains shape and volume **See attachment fr graphs States f matter The fundamental difference between states f matter is the distance between particles Gas Ttal disrder Much empty space Particles have cmpletely freedm f mtin

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

Chem 116 POGIL Worksheet - Week 4 Properties of Solutions

Chem 116 POGIL Worksheet - Week 4 Properties of Solutions Chem 116 POGIL Wrksheet - Week 4 Prperties f Slutins Key Questins 1. Identify the principal type f slute-slvent interactin that is respnsible fr frming the fllwing slutins: (a) KNO 3 in water; (b) Br 2

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

Chapter 8 Reduction and oxidation

Chapter 8 Reduction and oxidation Chapter 8 Reductin and xidatin Redx reactins and xidatin states Reductin ptentials and Gibbs energy Nernst equatin Disprprtinatin Ptential diagrams Frst-Ebswrth diagrams Ellingham diagrams Oxidatin refers

More information

Name:. Correct Questions = Wrong Questions =.. Unattempt Questions = Marks =

Name:. Correct Questions = Wrong Questions =.. Unattempt Questions = Marks = Name:. Crrect Questins = Wrng Questins =.. Unattempt Questins = Marks = 1. Which metal reacts mst vigrusly with water? (A) Al (B) Ca (C) Fe (D) K 2. Which are strng acids? I. HI II. HNO 3 III. H 2 SO 3

More information

CHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (40 pts.)

CHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (40 pts.) CHEMISTRY 202 Hour Exam II October 27, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 32 questions on 11 numbered pages. Check now to make sure you have a complete exam. You have two hours

More information

THE ANSWER KEY TO THIS EXAM WILL BE POSTED ON BULLETIN BOARD #4 IN THE HALLWAY EAST OF ROOM 1002 GILMAN AND ON THE CHEM 167 WEBSITE.

THE ANSWER KEY TO THIS EXAM WILL BE POSTED ON BULLETIN BOARD #4 IN THE HALLWAY EAST OF ROOM 1002 GILMAN AND ON THE CHEM 167 WEBSITE. PROF. JOHN VERKADE SPRING 2005 THIS EXAM CONSISTS OF 3 QUESTIONS ON 9 PAGES CHEM 67 HOUR EXAM II FEBRUARY 28, 2005 SEAT NO. NAME RECIT. INSTR. RECIT. SECT. GRADING PAGE Page 2 Page 3 Page 4 Page 5 Page

More information