Chapter 17 Free Energy and Thermodynamics

Size: px
Start display at page:

Download "Chapter 17 Free Energy and Thermodynamics"

Transcription

1 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

2 First Law f Thermdynamics yu can t win! First Law f Thermdynamics: Energy cannt be Created r Destryed the ttal energy f the universe cannt change thugh yu can transfer it frm ne place t anther E universe = 0 = E system + E surrundings Tr, Chemistry: A Mlecular Apprach 2

3 First Law f Thermdynamics Cnservatin f Energy Fr an exthermic reactin, lst heat frm the system ges int the surrundings tw ways energy lst frm a system, cnverted t heat, q used t d wrk, w Energy cnservatin requires that the energy change in the system equal the heat released + wrk dne E = q + w E = H + P V E is a state functin internal energy change independent f hw dne Tr, Chemistry: A Mlecular Apprach 3

4 Energy Tax yu can t break even! t recharge a battery with 100 kj f useful energy will require mre than 100 kj every energy transitin results in a lss f energy cnversin f energy t heat which is lst by heating up the surrundings Tr, Chemistry: A Mlecular Apprach 4

5 Heat Tax fewer steps generally results in a lwer ttal heat tax Tr, Chemistry: A Mlecular Apprach 5

6 Thermdynamics and Spntaneity thermdynamics predicts whether a prcess will prceed under the given cnditins spntaneus prcess nnspntaneus prcesses require energy input t g spntaneity is determined by cmparing the free energy f the system befre the reactin with the free energy f the system after reactin. if the system after reactin has less free energy than befre the reactin, the reactin is thermdynamically favrable. spntaneity fast r slw Tr, Chemistry: A Mlecular Apprach 6

7 Cmparing Ptential Energy The directin f spntaneity can be determined by cmparing the ptential energy f the system at the start and the end. Tr, Chemistry: A Mlecular Apprach 7

8 Reversibility f Prcess any spntaneus prcess is irreversible it will prceed in nly ne directin a reversible prcess will prceed back and frth between the tw end cnditins equilibrium results in n change in free energy if a prcess is spntaneus in ne directin, it must be nnspntaneus in the ppsite directin Tr, Chemistry: A Mlecular Apprach 8

9 Thermdynamics vs. Kinetics Tr, Chemistry: A Mlecular Apprach 9

10 Diamnd Graphite Graphite is mre stable than diamnd, s the cnversin f diamnd int graphite is spntaneus but dn t wrry, it s s slw that yur ring wn t turn int pencil lead in yur lifetime (r thrugh many f yur generatins). Tr, Chemistry: A Mlecular Apprach 10

11 Factrs Affecting Whether a Reactin Is Spntaneus The tw factrs that determine the thermdynamic favrability are the enthalpy and the entrpy. The enthalpy is a cmparisn f the bnd energy f the reactants t the prducts. bnd energy = amunt needed t break a bnd. H The entrpy factrs relates t the randmness/rderliness f a system S The enthalpy factr is generally mre imprtant than the entrpy factr Tr, Chemistry: A Mlecular Apprach 11

12 Enthalpy related t the internal energy H generally kj/ml strnger bnds = mre stable mlecules if prducts mre stable than reactants, energy released exthermic H = negative if reactants mre stable than prducts, energy absrbed endthermic H = psitive The enthalpy is favrable fr exthermic reactins and unfavrable fr endthermic reactins. Hess Law H rxn = Σ( H prd ) - Σ( H react ) Tr, Chemistry: A Mlecular Apprach 12

13 Substance H kj/ml Substance H kj/ml Al(s) 0 Al 2 O Br 2 (l) 0 Br 2 (g) C(diamnd) C(graphite) 0 CO(g) CO 2 (g) Ca(s) 0 CaO(s) Cu(s) 0 CuO(s) Fe(s) 0 Fe 2 O 3 (s) H 2 (g) 0 H 2 O 2 (l) H 2 O(g) H 2 O(l) HF(g) HCl(g) HBr(g) HI(g) I 2 (s) 0 I 2 (g) N 2 (g) 0 NH 3 (g) NO(g) NO 2 (g) Na(s) 0 O 2 (g) 0 S(s) 0 SO 2 (g)

14 Entrpy entrpy is a thermdynamic functin that increases as the number f energetically equivalent ways f arranging the cmpnents increases, S S generally J/ml S = k ln W k = Bltzmann Cnstant = 1.38 x J/K W is the number f energetically equivalent ways, unitless Randm systems require less energy than rdered systems Tr, Chemistry: A Mlecular Apprach 14

15 W Energetically Equivalent States fr the Expansin f a Gas Tr, Chemistry: A Mlecular Apprach 15

16 Macrstates Micrstates These micrstates all have the same macrstate S there are 6 different This particle macrstate can be achieved thrugh arrangements several different that arrangements f the particles result in the same macrstate Tr, Chemistry: A Mlecular Apprach 16

17 Macrstates and Prbability There is nly ne pssible arrangement that gives State A and ne that gives State C There are 6 pssible arrangements that give State B Therefre State B has higher entrpy than either State A r State B The macrstate with the highest entrpy als has the greatest dispersal f energy Tr, Chemistry: A Mlecular Apprach 17

18 Changes in Entrpy, S entrpy change is favrable when the result is a mre randm system. S is psitive Sme changes that increase the entrpy are: reactins whse prducts are in a mre disrdered state. (slid > liquid > gas) reactins which have larger numbers f prduct mlecules than reactant mlecules. increase in temperature slids dissciating int ins upn disslving Tr, Chemistry: A Mlecular Apprach 18

19 Increases in Entrpy Tr, Chemistry: A Mlecular Apprach 19

20 The 2 nd Law f Thermdynamics the ttal entrpy change f the universe must be psitive fr a prcess t be spntaneus fr reversible prcess S univ = 0, fr irreversible (spntaneus) prcess S univ > 0 S universe = S system + S surrundings if the entrpy f the system decreases, then the entrpy f the surrundings must increase by a larger amunt when S system is negative, S surrundings is psitive the increase in S surrundings ften cmes frm the heat released in an exthermic reactin Tr, Chemistry: A Mlecular Apprach 20

21 Entrpy Change in State Change when materials change state, the number f macrstates it can have changes as well fr entrpy: slid < liquid < gas because the degrees f freedm f mtin increases slid liquid gas Tr, Chemistry: A Mlecular Apprach 21

22 Entrpy Change and State Change Tr, Chemistry: A Mlecular Apprach 22

23 Heat Flw, Entrpy, and the 2 nd Law Heat must flw frm water t ice in rder fr the entrpy f the universe t increase Tr, Chemistry: A Mlecular Apprach 23

24 Temperature Dependence f S surrundings when a system prcess is exthermic, it adds heat t the surrundings, increasing the entrpy f the surrundings when a system prcess is endthermic, it takes heat frm the surrundings, decreasing the entrpy f the surrundings the amunt the entrpy f the surrundings changes depends n the temperature it is at riginally the higher the riginal temperature, the less effect additin r remval f heat has S surrundings H Tr, Chemistry: A Mlecular Apprach 24 = T system

25 Gibbs Free Energy, G maximum amunt f energy frm the system available t d wrk n the surrundings G = H T S G sys = H sys T S sys G sys = T S universe G reactin = Σ n G prd Σ n G react when G < 0, there is a decrease in free energy f the system that is released int the surrundings; therefre a prcess will be spntaneus when G is negative Tr, Chemistry: A Mlecular Apprach 25

26 Ex. 17.2a The reactin C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O (g) has H rxn = kj at 25 C. Calculate the entrpy change f the surrundings. Given: Find: Cncept Plan: Relatinships: H system = kj, T = 298 K S surrundings, J/K T, H S surr = H T sys S Slutin: S S surr surr = = H T 6.86 sys kj K ( 2044 kj) = 298 K 3 = J K Check: cmbustin is largely exthermic, s the entrpy f the surrunding shuld increase significantly

27 Free Energy Change and Spntaneity Tr, Chemistry: A Mlecular Apprach 27

28 Gibbs Free Energy, G prcess will be spntaneus when G is negative G will be negative when H is negative and S is psitive exthermic and mre randm H is negative and large and S is negative but small H is psitive but small and S is psitive and large r high temperature G will be psitive when H is + and S is never spntaneus at any temperature when G = 0 the reactin is at equilibrium Tr, Chemistry: A Mlecular Apprach 28

29 G, H, and S Tr, Chemistry: A Mlecular Apprach 29

30 Ex. 17.3a The reactin CCl 4(g) C (s, graphite) + 2 Cl 2(g) has H = kj and S = J/K at 25 C. Calculate G and determine if it is spntaneus. Given: H = kj, S = J/K, T = 298 K Find: G, kj Cncept Plan: T, H, S G Relatinships: G = H T S Slutin: Answer: G = H T S = ( 3 ) ( )( J J 298 K ) = J Since G is +, the reactin is nt spntaneus at this temperature. T make it spntaneus, we need t increase the temperature. K

31 Ex. 17.3a The reactin CCl 4(g) C (s, graphite) + 2 Cl 2(g) has H = kj and S = J/K. Calculate the minimum temperature it will be spntaneus. Given: H = kj, S = J/K, G < 0 Find: Τ, Κ Cncept Plan: G, H, S T Relatinships: Slutin: Answer: ( H T S) G = H T S G = < 0 ( ) ( J ) 3 ( )( J J T ) < 0 ( J K ) K 3 ( ) ( )( J J < T ) The temperature must be higher than 673K fr the reactin t be spntaneus K 673 K < T < T

32 The 3 rd Law f Thermdynamics Abslute Entrpy the abslute entrpy f a substance is the amunt f energy it has due t dispersin f energy thrugh its particles the 3 rd Law states that fr a perfect crystal at abslute zer, the abslute entrpy = 0 J/ml K therefre, every substance that is nt a perfect crystal at abslute zer has sme energy frm entrpy therefre, the abslute entrpy f substances is always + Tr, Chemistry: A Mlecular Apprach 32

33 Standard Entrpies S extensive entrpies fr 1 mle at 298 K fr a particular state, a particular alltrpe, particular mlecular cmplexity, a particular mlar mass, and a particular degree f disslutin Tr, Chemistry: A Mlecular Apprach 33

34 Substance S J/ml-K Substance S J/ml-K Al(s) 28.3 Al 2 O 3 (s) Br 2 (l) Br 2 (g) C(diamnd) 2.43 C(graphite) 5.69 CO(g) CO 2 (g) Ca(s) 41.4 CaO(s) Cu(s) CuO(s) Fe(s) Fe 2 O 3 (s) H 2 (g) H 2 O 2 (l) H 2 O(g) H 2 O(l) HF(g) HCl(g) HBr(g) HI(g) I 2 (s) I 2 (g) N 2 (g) NH 3 (g) NO(g) NO 2 (g) Na(s) O 2 (g) S(s) SO 2 (g) 248.5

35 Relative Standard Entrpies States the gas state has a larger entrpy than the liquid state at a particular temperature the liquid state has a larger entrpy than the slid state at a particular temperature Substance H 2 O (g) H 2 O (l) S, (J/ml K) Tr, Chemistry: A Mlecular Apprach 35

36 Relative Standard Entrpies Mlar Mass the larger the mlar mass, the larger the entrpy available energy states mre clsely spaced, allwing mre dispersal f energy thrugh the states Tr, Chemistry: A Mlecular Apprach 36

37 Relative Standard Entrpies the less cnstrained the structure f an alltrpe is, the larger its entrpy Alltrpes Tr, Chemistry: A Mlecular Apprach 37

38 Relative Standard Entrpies Mlecular Cmplexity larger, mre cmplex mlecules generally have larger entrpy Substance Ar (g) Mlar Mass S, (J/ml K) mre available energy states, allwing mre dispersal f energy thrugh the states NO (g) Tr, Chemistry: A Mlecular Apprach 38

39 Relative Standard Entrpies Disslutin disslved slids generally have larger entrpy distributing particles thrughut the mixture Substance KClO 3 (s) KClO 3 (aq) S, (J/ml K) Tr, Chemistry: A Mlecular Apprach 39

40 Ex Calculate S fr the reactin 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (l) Given: Find: Cncept Plan: Relatinships: Slutin: Check: standard entrpies frm Appendix IIB S, J/K S NH3, S O2, S NO, S H2O, S = = [4(S = ( ) ( S = Σn S ) p prducts Σn r S reactants ( ) ( Σ S ) prducts Σ S reactants NO( g) = [4(210.8 n p n r J K J K ) + 6(S H 2 ) + 6(188.8 O( g) J K )] [4(S Substance NH 3 (g) O 2 (g) NO(g) H 2 O(g) NH )] [4(192.8 S S is +, as yu wuld expect fr a reactin with mre gas prduct mlecules than reactant mlecules 3 ( g) J K ) + 5(S O 2 ) + 5(205.2 S, J/ml K ( g) J K )] )]

41 Calculating G at 25 C: G reactin = ΣnG f (prducts) - ΣnG f (reactants) at temperatures ther than 25 C: assuming the change in H reactin and S reactin is negligible G reactin = H reactin T S reactin Tr, Chemistry: A Mlecular Apprach 41

42 Substance G f kj/ml Substance G f kj/ml Al(s) 0 Al 2 O Br 2 (l) 0 Br 2 (g) C(diamnd) C(graphite) 0 CO(g) CO 2 (g) Ca(s) 0 CaO(s) Cu(s) 0 CuO(s) Fe(s) 0 Fe 2 O 3 (s) H 2 (g) 0 H 2 O 2 (l) H 2 O(g) H 2 O(l) HF(g) HCl(g) HBr(g) HI(g) I 2 (s) 0 I 2 (g) N 2 (g) 0 NH 3 (g) NO(g) NO 2 (g) Na(s) 0 O 2 (g) 0 S(s) 0 SO 2 (g)

43 Ex Calculate G at 25 C fr the reactin CH 4(g) + 8 O 2(g) CO 2(g) + 2 H 2 O (g) + 4 O 3(g) Given: Find: Cncept Plan: Relatinships: Slutin: G = = [( G = kj standard free energies f frmatin frm Appendix IIB G, kj G f f prd & react G = ( ) ( Σn G ) prducts Σn G reactants ( ) ( Σn G G ) p prducts Σn reactants f CO ) + = [( kj) + f 2( G f r p ) + ( G f f f r )] [( G Substance CH 4 (g) O 2 (g) CO 2 (g) H 2 O(g) O 3 (g) f G ) + 8( G 2 H2O O3 CH4 O2 2( kj) + ( kj)] [( 50.5 kj) + 8(0.0 kj)] f G f, kj/ml f )]

44 Ex The reactin SO 2(g) + ½ O 2(g) SO 3(g) has H = kj and S = J/K at 25 C. Calculate G at 125 C and determine if it is spntaneus. Given: H = kj, S = J/K, T = 398 K Find: G, kj Cncept Plan: T, H, S G Relatinships: G = H T S Slutin: Answer: G = = H ( ) ( )( ) 3 J J 398 K 94.0 = T S 3 J = 61.5 kj Since G is -, the reactin is spntaneus at this temperature, thugh less s than at 25 C K

45 G Relatinships if a reactin can be expressed as a series f reactins, the sum f the G values f the individual reactin is the G f the ttal reactin G is a state functin if a reactin is reversed, the sign f its G value reverses if the amunts f materials is multiplied by a factr, the value f the G is multiplied by the same factr the value f G f a reactin is extensive Tr, Chemistry: A Mlecular Apprach 45

46 Free Energy and Reversible Reactins the change in free energy is a theretical limit as t the amunt f wrk that can be dne if the reactin achieves its theretical limit, it is a reversible reactin Tr, Chemistry: A Mlecular Apprach 46

47 Real Reactins in a real reactin, sme f the free energy is lst as heat if nt mst therefre, real reactins are irreversible Tr, Chemistry: A Mlecular Apprach 47

48 G under Nnstandard Cnditins G = G nly when the reactants and prducts are in their standard states there nrmal state at that temperature partial pressure f gas = 1 atm cncentratin = 1 M under nnstandard cnditins, G = G + RTlnQ Q is the reactin qutient at equilibrium G = 0 G = RTlnK Tr, Chemistry: A Mlecular Apprach 48

49 Tr, Chemistry: A Mlecular Apprach 49

50 Example - G Calculate G at 427 C fr the reactin belw if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Q = P NH3 2 P N21 x P H2 3 (2.0 atm) 2 = (33.0 atm) 1 (99.0) 3 = 1.2 x 10-7 H = [ 2(-46.19)] - [0 +3( 0)] = kj = J S = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K G = J - (700 K)( J/K) G = J G = G + RTlnQ G = J + (8.314 J/K)(700 K)(ln 1.2 x 10-7 ) G = J = -46 kj 50

51 Tr, Chemistry: A Mlecular Apprach 51

52 Example - K Estimate the equilibrium cnstant and psitin f equilibrium fr the fllwing reactin at 427 C N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H = [ 2(-46.19)] - [0 +3( 0)] = kj = J S = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K G = J - (700 K)( J/K) G = J G = -RT lnk J = -(8.314 J/K)(700 K) lnk lnk = K = e = 3.45 x 10-4 since K is << 1, the psitin f equilibrium favrs reactants 52

53 Temperature Dependence f K fr an exthermic reactin, increasing the temperature decreases the value f the equilibrium cnstant fr an endthermic reactin, increasing the temperature increases the value f the equilibrium cnstant H ln K = R rxn 1 T + S R rxn Tr, Chemistry: A Mlecular Apprach 53

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 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

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

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

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

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

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

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

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

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

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

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

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

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. 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 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

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

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

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

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

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

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

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

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

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 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

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

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax.

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro First Law of Thermodynamics Chapter 17 Free Energy and Thermodynamics You can t win! First Law of Thermodynamics: Energy cannot be created or destroyed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

BIOLOGY 101. CHAPTER 8: An Introduction to Metabolism: Energy of Life

BIOLOGY 101. CHAPTER 8: An Introduction to Metabolism: Energy of Life BIOLOGY 101 CHAPTER 8: An Intrductin t Metablism: Energy f Life Energy f Life CONCEPTS: 8.1 An rganism's metablism transfrms matter and energy, subject t the laws f thermdynamics 8.2 The free-energy change

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

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 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

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

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 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

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

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

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

Electrochemistry. Reduction: the gaining of electrons. Reducing agent (reductant): species that donates electrons to reduce another reagent.

Electrochemistry. Reduction: the gaining of electrons. Reducing agent (reductant): species that donates electrons to reduce another reagent. Electrchemistry Review: Reductin: the gaining f electrns Oxidatin: the lss f electrns Reducing agent (reductant): species that dnates electrns t reduce anther reagent. Oxidizing agent (xidant): species

More information

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM 14 CHAPTER CHEMICAL EQUILIBRIUM 14.1 The Nature f Chemical Equilibrium 14. The Empirical Law f Mass Actin 14.3 Thermdynamic Descriptin f the Equilibrium State 14.4 The Law f Mass Actin fr Related and Simultaneus

More information

Chemistry 132 NT. Electrochemistry. Review

Chemistry 132 NT. Electrochemistry. Review Chemistry 132 NT If yu g flying back thrugh time, and yu see smebdy else flying frward int the future, it s prbably best t avid eye cntact. Jack Handey 1 Chem 132 NT Electrchemistry Mdule 3 Vltaic Cells

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

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 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

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

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

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

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

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 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

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

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

_J _J J J J J J J J _. 7 particles in the blue state; 3 particles in the red state: 720 configurations _J J J _J J J J J J J J _

_J _J J J J J J J J _. 7 particles in the blue state; 3 particles in the red state: 720 configurations _J J J _J J J J J J J J _ Dsrder and Suppse I have 10 partcles that can be n ne f tw states ether the blue state r the red state. Hw many dfferent ways can we arrange thse partcles amng the states? All partcles n the blue state:

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 17. Free Energy and Thermodynamics. Chapter 17 Lecture Lecture Presentation. Sherril Soman Grand Valley State University

Chapter 17. Free Energy and Thermodynamics. Chapter 17 Lecture Lecture Presentation. Sherril Soman Grand Valley State University Chapter 17 Lecture Lecture Presentation Chapter 17 Free Energy and Thermodynamics Sherril Soman Grand Valley State University First Law of Thermodynamics You can t win! The first law of thermodynamics

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

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

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

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

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

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

Lecture 13: Electrochemical Equilibria

Lecture 13: Electrochemical Equilibria 3.012 Fundamentals f Materials Science Fall 2005 Lecture 13: 10.21.05 Electrchemical Equilibria Tday: LAST TIME...2 An example calculatin...3 THE ELECTROCHEMICAL POTENTIAL...4 Electrstatic energy cntributins

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

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

Chem 116 POGIL Worksheet - Week 3 Intermolecular Forces, Liquids, Solids, and Solutions Chem 116 POGIL Wrksheet - Week 3 Intermlecular Frces, Liquids, Slids, and Slutins Why? Mst substances can exist in either gas, liquid, r slid phase under apprpriate cnditins f temperature and pressure.

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

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

3. Review on Energy Balances

3. Review on Energy Balances 3. Review n Energy Balances Objectives After cmpleting this chapter, students shuld be able t recall the law f cnservatin f energy recall hw t calculate specific enthalpy recall the meaning f heat f frmatin

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 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea. 1. The anti-cancer drug cis-platin is the complex: cis-[pt(nh ) (Cl) ]. In this complex, the

CHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea. 1. The anti-cancer drug cis-platin is the complex: cis-[pt(nh ) (Cl) ]. In this complex, the CHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea Sme useful cnstants: ln(10) = 2.303, R = 8.314 J/ml@K, F = 96,00 cul/ml, 2.303RT/F = 0.0916 V at 2EC. Assume a temperature f 2EC unless tld therwise.

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

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

ChE 471: LECTURE 4 Fall 2003

ChE 471: LECTURE 4 Fall 2003 ChE 47: LECTURE 4 Fall 003 IDEL RECTORS One f the key gals f chemical reactin engineering is t quantify the relatinship between prductin rate, reactr size, reactin kinetics and selected perating cnditins.

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

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

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

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