Chemistry 102 Spring 2017 Discussion #13, Chapter 17 Student name Key TA name Section. Things you should know when you leave Discussion today: ( G o f
|
|
- Dale Nicholson
- 6 years ago
- Views:
Transcription
1 Chemistry 0 Spring 07 Discussion #, Chapter 7 Student name Key TA name Section Things you should know when you leave Discussion today:. ΔSsys = ΔrS = Σ [ni (S )product Σ [nj (S )reactants. ΔGº = T ΔSnet = ΔH sys TΔS sys a. ΔG rn ( at 5 C only) = n P (G o f ) product b. ΔGº = - RT lnk =. RT log K ( ΔGº/RT.) c. K= 0 d. ΔG = ne F E = -q E Q. ΔG = RT ln( ) =. RT log K reactants ( K Q ) = ΔG + RT ln Q (ΔG -ΔGº)/(RT.) a. Q =0 b. ΔG = T ΔSnet = ΔHsys TΔSsys c. ΔG= ne FE = q E i. ΔG<0 process is Spontaneous ii. ΔG>0 process is Not Spontaneous iii. ΔG=0 process is at Equilibrium n R (G f o ). Circle the value of ΔG when both products and reactants are in their standard states. (Note: Q=) Circle the value of ΔG when Q = K. Circle the value of ΔG when Q = K. ΔG = ΔG = 0 ΔG = ΔG None of these ΔG = ΔG = 0 ΔG = ΔG None of these ΔG = ΔG = 0 ΔG = -RT lnk None of these Circle the value of ΔG when Q > K. When would ΔG = 0? K= ΔG = ΔG = 0 ΔG = -RT lnk None of these. a) If ΔG= 5 kj/mol and ΔGº= 4 kj/mol circle everything that must be true: Q= Q> Q< Q<K Q=K Q>K b) ( at home ) Assuming temperature of 98 K calculate the values of K and Q. (Useful Hint: at T=98K value of.rt =. 8.4(J.(mol K) 98K =5698J/mol 6kJ/mol) K=0 ( ΔGº/RT.) =0. Q= 0 (ΔG -ΔGº)/(RT.) =.5. a) If ΔG= -5kJ/mol and ΔGº= kj/mol circle everything that must be true: Q = Q > Q < Q < K Q = K Q > K
2 b) At home : Assuming temperature of 98 K, calculate the values of K and Q. K=0 ( ΔGº/RT.) =0. Q= 0 (ΔG -ΔGº)/(RT.) = The standard free energy change for the process A (aq) + B (aq) C (aq) is ΔG = +54 kj/mol. If the concentration of B is decreased from M to 0.50 M (but concentrations of A and C stay at standard states) circle the value of the free energy change, ΔG, relative to ΔG. ΔG < ΔG ΔG = ΔG ΔG > ΔG 5. Remember that ΔGº = - RT ln K = ΔHºsys TΔSºsys and so we can derive ln K= RT H sys + S sys R ln Keq Reaction Reaction 0 High T region /T 0 T Reaction Low T region /T T 0 /T Reaction 4 H sys S sys Using that lnk= + and that equations of the line is y = m + b answer the RT R following questions: H sys m= is the slope of the line RT If m>0 then ΔHºsys<0 and the process is Eothermic. If m<0 then ΔHºsys>0 and the process is Endothermic. S sys b= is the y-intercept of the line R If b>0 then ΔSºsys>0 and the reaction increases positional entropy. If b<0 then ΔSºsys<0 and the reaction decreases positional entropy
3 For Reactions,, and 4 in the plot above, answer the following questions. a. Is the reaction endothermic or eothermic and at what temperatures? a. Reaction : Eothermic. At all T b. Reaction : Eothermic. At all T c. Reaction : Endothermic At all T d. Reaction 4: Endothermic At all T b. If the reaction above are redo reactions, which ones will have positive standard voltages and at what temperatures? a. Reaction : Eº always positive b. Reaction : Eº only positive at low T c. Reaction : Eº only positive at High T d. Reaction 4: never positive c. Does the reaction increase or decrease entropy of the system and at what temperatures? a. Reaction : ΔSºsys>0 At all T b. Reaction : ΔSºsys<0 At all T c. Reaction : ΔSºsys>0 At all T d. Reaction 4: ΔSºsys<0 At all T d. For each reaction choose at what temperatures the ratio of product to reactants will be greater than or smaller than? Reaction : Products: High T Low T never always Reactants: High T Low T never always Reaction : Products: High T Low T never always Reactants: High T Low T never always Reaction : Products: High T Low T never always Reactants: High T Low T never always Reaction 4: Products: High T Low T never always Reactants: High T Low T never always e. It is a well-known fact that miing any group metal and water leads to an eplosion. The plot of which reaction corresponds to the reaction between Na and water: Na(s) + HO(l) Na + (aq) + OH - (aq) + H(g) Reaction Reaction Reaction Reaction 4 f. The plot of which reaction corresponds to freezing of water? Reaction Reaction Reaction Reaction 4
4 6. For the following combustion reaction calculate ΔG at 5 C. : ΔG f(kj/mol) (only at 5 C) O(g) 0 C6HO6 (s) CO (g) HO(l) -40. C6HO6 (s) + 6O (g) 6 CO (g) + 6 HO (l) ΔG = -940kJ/mol Without doing a calculation, at what temperatures will this process cease to be spontaneous? NEVER ΔHºsys<0 (eothermic) ΔSºsys>0 7. For autoionization of water at 0 K, Kw=0 - and at 00K, Kw=0-4. Calculate the value of ΔH and ΔS. Do your calculations to significant figure. ln K= ln K= H sys RT H sys RT + + S sys R S sys R H sys.(logk log K)= + RT J J S 57 = 60 molk molk H sys RT T T = H sys ( ) ; RT T H = -6997J=-60kJ 8. You have a buffer with a ph = 5 at two different temperatures: T = 00 K and T = 0 K HA(aq) + HO (l) A - (aq) + HO + (aq) You need to have the following ratios of weak acid and conjugate base to make this happen: at T [A - /[HA = / at T [A - /[HA = /4 To answer the following questions you need to draw the plot of lnk verses /T. (Hint:lnK=.0logK) Ka==[HO + [A - / [HA =0-5 / Ka==[HO + [A - / [HA =0-5 /4 log Ka= -5. at T=00K logka= -5.6 at T=0K 4
5 Now that you have logka, and.log Ka = lnka, you plot the data... /0 /00 T. *( -5.). *( -5.6) a. Will this reaction be eothermic, endothermic or neither. ΔH < 0 b. The system entropy change will be: Circle one: S < 0 9. Your body needs glutamine (an amino acid used to make proteins) for your survival. While glutamate is useless to you as-is, you need to make glutamine out of glutamate and ammonium ions through the following reaction: Glutamate + NH 4 + glutamine (ΔG rn=+4.kj/mol) What problem(s) do you see with this process? Not spontaneous at standard conditions( or reactants favorable) A wise man once said, Don t panic! You also make glutamine synthetase, an enzyme that helps this process along. If the process of : ATP ADP + PO 4 - (ΔG rn= -0kJ/mol): were somehow linked to the glutamine synthesis above, what would the new reaction be? (Don t worry about balancing.) What would the ΔG rn for this new reaction be? ΔG rn= ΔG+ ΔG = -5.8kJ/mol 0. Given the two reactions below, find the relationships between ΔGº and ΔGº. A(aq) + B(g) C(s) + D(aq) with Keq=K and ΔGºrn=ΔGº C(s) + D(aq) A(aq) + B(g) with Keq=K and ΔGºrn=ΔGº K= K ΔGº= - ΔGº. Given the two reactions below, find the relationships between ΔGº and ΔGº. A(aq) + B(g) C(s) + D(aq) with Keq=K and ΔGº=ΔGº A(aq) + B(g) C(s) + D(aq) with Keq=K and ΔGº=ΔGº K= K ΔGº= ΔGº 5
6 . For the following reaction at 5ºC using information below. H(g) + N(g) NH(g) ΔGº kj/mol P (atm) H(g) N(g) Calculate value of ΔG = - kj/mol NH(g) Calculate value of K=0 5. = 0 5 Calculate value of Q= 0 6 Calculate value of ΔG=4kJ/mol Based on your calculation did reaction pass an equilibrium point? yes Based on your calculation will the reverse reaction be spontaneous or not? Spontaneous. For the following reaction: Br(l) Br(g) For the reaction above at 00K, 5K, 40 K use the information given, to calculate ΔHsys, ΔSsys, ΔSsur, ΔSuniv, ΔG and the temperature at which process will cease to be spontaneous. ( Do your calculations to significant figures. Assume that ΔH rn and ΔS rn are independent of temperature.) ΔH rn= mol*δh f (Br(g))- mol*δh f (Br(l))=0.9kJ Endothermic ΔS rn = mol*s f (Br(g))- mol*s f (Br(l))= 9. J/ K; ΔSsys>0 ΔSnet(7K) = ΔSrn + ΔSsur(at 7K)= 9. J/K -. J/K = -9.9J/ K not spontaneous ΔSnet(98 K)= ΔSrn + ΔSsur(at 98 K)= 9. J/K -0.6 J/K=-0.4J/ K not spontaneous ΔSnet(4 K) = ΔSrn +ΔSsur(at 4 K)= 9. J/K J/K=. J/ K spontaneous ΔSsys>0 and ΔSsur<0 that means that process is spontaneous only when : ΔSsys > ΔSsur and it is possible only at T>Tequilibrium ΔH f (kj/mol) S (J/ Kmol) ΔG f(kj/mol) (only at 5 C) Br(g) Br(l) Teq = ΔHrn/ ΔSrn = 0900J/ (9. J/ K) = K Teq= K - 7=58 C Above 58 C process is spontaneous. ΔG rn(at any Temperature) = -T ΔSnet = ΔHsys - T ΔSsys ΔSnet >0 then ΔG rn< 0 process is spontaneous ΔSnet <0 then ΔG rn > 0 process is not spontaneous ΔSnet=0 then ΔG rn= 0 process is at equilibrium ΔG rn (at 5 C only) = mol*δg f (Br(g))- mol*δg f (Br(l))=.kJ not spontaneous ΔG rn(at any Temperature) = -T ΔSnet ΔG rn (7 K)= -7K* (-9.9J/ K )= 5.4k J not spontaneous ΔG rn (98 K)= -98 K*(-0.4J/ K)= 099. J=.kJ not spontaneous ΔG rn (4 K) = -4 K*. J/ K = -.kj spontaneous OR ΔG rn(at any Temperature) = ΔHsys - T ΔSsys ΔG rn (7 K)= 0.9*0 J -7 K *9. 9 J/ K =5.4k J not spontaneous ΔG rn (98 K)= 0.9*0 J - 98 K *9. 9 J/ K =.kj not spontaneous ΔG rn (4 K) =0.9*0 J - 4 K*9. 9 J/ K =-.kj spontaneouus 6
7 Useful information and preparation for chapter 8 not part of eam : [ A. Rate = ; units: M t s. Rate of chemical reactions depends on: a. Temperature b. Catalysts c. Initial concentrations d. Magnitude of the rate constant. Initial Reaction Rates: a. For a reaction: aa + bb cc + dd b. Initial Reaction Rate = c. Where [ A t [ A a t = [ B b t is an individual rate of loss of A, [ B b [ A ; [ C c [ A t a t t a t 4. Rate Law: a. Differential Rate Law: rate versus concentration aa + bb cc + dd = [ C c t = [ D d t rate Rate = k [A [B y Where k is a rate constant k= [A [B y Where is the order of the reaction with respect to species A, and y is the order of the reaction with respect to species B. b. Comparing initial rates: rate rate k[a y [B k[a y [B c. Units of the rate constant depend on over all reaction order. M ( y) M k = s where (+y) is the order of the overall reaction. y M M s 5. Integrated Rate Laws: see attached table for more information a. Zero-Order: [R t [R 0 kt b. First-Order: ln [R t kt [R 0 c. Second-Order: kt [R t [R 0 Questions on the material in chapter 8 not part of eam : 7
8 . A study of the reaction below gave the following result: Fill in the missing information in the table [NO [O NO(g)+O(g)NO(g) To find we need to compare the rate of eperiment to the rate of eperiment : NOTE: The rate constant does not change with eperimental conditions: it is constant. Concentration of NO changes but concentration of Ostays the same. [O=[ O What order is the reaction in respect to [NO = To find y we need to compare the rate of eperiment to the rate of eperiment (or and4): NOTE: The rate constant does not change with eperimental conditions: it is constant. Concentration of Ochanges but concentration of A stays the same. [NO=[ NO What order is the reaction with respect to [O y = Write the rate law: rate = k [NO [O - d [ NO dt [M/s E E E E Reaction rate Overall rate order = What are the units of this k? M ( y) M s y M M s = M s rate Calculate rate constant: k = [NO [ O =.5*0 M s What is the reaction rate when [NO = [O = 0.00M rate=k [NO [ O =.5*0 *(0.00M) *(0.00M) =.8*0 - M/s M s For each reactant, circle which plot correlating concentration with time will give you a straight line, and what the sign of the slope will be. [NO vs. time /[NO vs. time ln[no vs. time Slope: positive = k negative zero [O vs. time / [O vs. time ln[o vs. time Slope: positive negative = -k zero 8
9 . A study of the reaction below gave the following result: NO(g) + Cl (g) NOCl(g) What order is the reaction with respect to [NO To find we need to compare the rate of eperiment to the rate of eperiment : E NOTE: The rate constant does not change with eperimental conditions: it is constant. Concentration of NO changes but concentration of Cl stays the same. [Cl=[ Cl rate rate k[ A [ B k[ A [ B y y [ A A [. 4.8 log = log= log(/4) = = log log 4 = - = -0.5 What order is the reaction with respect to [Cl To find y we need to compare the rate of eperiment to the rate of eperiment : NOTE: The rate constant does not change with eperimental conditions: it is constant. Concentration of Clchanges but concentration of A stays the same. [NO=[ NO rate rate y k[ A [ B [ B 5.0 y k[ A [ B B ; = [.5 Fill in the table for reaction rate with respect to NO Write the rate law: rate = k [NO -0.5 [ Cl.6 Overall rate order: + y=.6+(-0.5)=. What are the units of this k? M -0. /s y y y [NO [Cl d[no dt ; log()=y log; y=.6 [M/s RXN rate E E Consider the chemical reaction: A + 4B C + D. The eperimentally determined rate law is Rate = k[a [B. When the initial concentrations are [Ai = 0.5 M and [Bi = 0.0 M, the rate of consumption of B is M/s. Give the value of the rate constant, k, including the units. Rate = d[b = k[a [B 4 dt 4 ( 80 5 M ) = k[50 M [0 M s k = 0 5 M/s 50 4 M = 80 M s 4. Consider the chemical reaction: A + 4B C + D. The eperimentally determined rate law is Rate = k [A [B. When the initial concentrations are [Ai = 0. M and [Bi = 0.0 M, the rate of consumption of B is M/s. Give the value of the rate constant, k, including the units. (Answer: 5 M - s - ) 9
10 5. Consider the reaction A + B D + C. Three eperiments were conducted and the initial reaction rate of each reaction was determined. For each of the reactants, find the appropriate order.(answer:,0.5) Initial conc. (mol/l) Initial Rate Ep [A [B (mol/l sec) # # # Consider the comproportionation reaction of bromate and bromide to make bromine gas. Four eperiments were conducted and the initial reaction rate was determined. H + (aq) + BrO (aq) + 0 Br (aq) 6 Br (g) + 6 HO (l) Initial Concentrations (mol/l) [BrO [Br [HO + Initial Reaction Rate (mol/l sec) Ep. # # # # a) Using the data, determine the order of the reaction with respect to HO +. rate4 rate k[ H O H O 4 [ k H O H O = [ [ 0.0 b) What is the initial rate of formation of bromine gas in eperiment #? Br (g) d [ Br rn rate = 6 dt d[ Br 6 Initial Reaction Rate =6 mol/l se dt In Peroration for net week: Half Life N N n 0 n ; tpass = n t½ ; t½ = k 7. In a first order chemical reaction, the concentration of a reactant is observed to fall from.000 M to 0.50 M between t = 6 s and t = 0 s. What is the concentration (in M) at t = s? One half-life M0.5M Two half life s 0.5M 0.50M time spend was 0s-6s=4s for half life s t½ = s at time s one more half-life must pass so the concentration is 0.5M 0
11 8. The first order reaction is 60% complete after 00 seconds. What is rate constant k and t½ for this reaction? 60% complete 40% remains N N n 0 n n 4 0 n n=. ; t½ = t/n= 76.9s; k = 0.69 t =.0 /s 9. Fluorine-8 has a half-life of 0 minutes. After 4 half-lives, how much time has elapsed? What percentage of your original F-8 remains? What percentage has decayed? (Answer: tpass=440minutes, remains=6.5%, decayed=9.75) t½ =0min n=4 t=440s N n N 0 4 =0.065 remains 0. You are trying to find the half-life of a new, meta-stable compound (X) created in your lab. You analyze the amount of X you have 0 minutes after initial production of X. You find you have 0% of the original amount. Approimately, what is t½? 6 m. <t½< 7.5 m. 7.5 m. <t½< 0 m. 0 m. <t½< 5 m. 5 m. <t½< 0 m.. For another compound (Y), 94% of an original 00.0 g has decayed in 60.0 minutes. Epress your answer to ONE SIGNIFICANT FIGURE. (a) How many grams of Y is remaining? (Answer: 6g) (b) How many half-lives have elapsed? (Answer: n=4 half-lives pass) (c) What is the half-life of this compound? (Answer: t½=5 min)
12 Reaction Order Zero order reaction = 0, y = 0, + y = 0 Differential Rate Law Conc. Dependence Rate = k[a 0 [B 0 Rate = k Rate is Constant Rate is independent of initial concentration. Units of the rate constant, k, depend on the reaction order. (M/s) = units of k k = (M/s) are the units of the rate In general the units of k are a good indicator of the reaction order. Half life, t/, Integrated Rate Law [Afinal = [Ainitial - ak t When: [Afinal = [ A initial [ A initial = [Ain- akt/ [ t/ = A initial ak [A [A0 Graph Concentration versus time [A0 is [A at t = 0 slope= -ak = -k t(s) First order reaction Eample: =, y = 0, + y = Rate = k[a [B 0 Rate = k[a Rate depends linearly on initial concentration (e. doubling [A doubles the rate) (M/s) = units of k * (M) k = (/s) ln [Afinal = ln [Ainitial - ak t [ A final ln = -ak t [ A initial [ A initial When: [Afinal = t/ = ln/ak =.log/ak (half life is independent of initial concentration) t/ is constant ln [A ln [A0 slope= -ak = -k t(s) Second order reaction Eample: =, y = 0, + y = Rate = k[a [B 0 Rate = k[a (M/s) = units of k * (M) k = M s /[Afinal = /[Ainitial + ak t [ A initial When : [Afinal = t/ = ak [ A initial /[A /[A0 slope= ak = k t(s) aa + bb cc [ A [ B [ C rate(m/s) = = = = k [A [B y a t b t c t and y are independent of stoichiometry and can only be determined from eperimental data
Chemistry 102 Spring 2017 Discussion #13, Chapter 17 Student name TA name Section. Things you should know when you leave Discussion today: ( G o f
Chemistry 0 Spring 07 Discussion #3, Chapter 7 Student name TA name Section Things you should know when you leave Discussion today:. ΔSsys = ΔrS = Σ [ni (S )]product - Σ [nj (S )]reactants. ΔGº = -T ΔSnet
More informationChemistry 102 Spring 2016 Discussion #12, Chapter 17 Student name TA name Section. Things you should know when you leave Discussion today: ( G o f
Chemistry 10 Spring 016 Discussion #1, Chapter 17 Student name TA name Section Things you should know when you leave Discussion today: 1. ΔS sys = Δ r S = Σ [n i (S )] product - Σ [n j (S )] reactants.
More informationLecture #14. Chapter 17 Free Energy and Equilibrium Constants
Lecture #14 Chapter 17 Free Energy and Equilibrium Constants Josiah W. Gibbs 1839-1903 Gibbs Free Energy (G) The maximum energy released by a system occurring at constant temperature and pressure that
More informationUnit 12. Thermochemistry
Unit 12 Thermochemistry A reaction is spontaneous if it will occur without a continuous input of energy However, it may require an initial input of energy to get it started (activation energy) For Thermochemistry
More informationENTROPY HEAT HEAT FLOW. Enthalpy 3/24/16. Chemical Thermodynamics. Thermodynamics vs. Kinetics
Chemical Thermodynamics The chemistry that deals with energy exchange, entropy, and the spontaneity of a chemical process. HEAT The energy that flows into or out of system because of a difference in temperature
More informationSecond Law of Thermodynamics
Second Law of Thermodynamics First Law: the total energy of the universe is a constant Second Law: The entropy of the universe increases in a spontaneous process, and remains unchanged in a process at
More informationCh 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 informationB 2 Fe(s) O 2(g) Fe 2 O 3 (s) H f = -824 kj mol 1 Iron reacts with oxygen to produce iron(iii) oxide as represented above. A 75.
1 2004 B 2 Fe(s) + 3 2 O 2(g) Fe 2 O 3 (s) H f = -824 kj mol 1 Iron reacts with oxygen to produce iron(iii) oxide as represented above. A 75.0 g sample of Fe(s) is mixed with 11.5 L of O 2 (g) at 2.66
More informationCH Practice Exam #2
CH1810 - Practice Exam #2 Part I - Multiple Choice - Choose the best answer and place the letter corresponding to the answer in the space provided. 1. Consider the following reaction and its equilibrium
More informationVanden Bout/LaBrake. Important Information. HW11 Due T DECEMBER 4 th 9AM. End of semester attitude survey closes next Monday
UNIT4DAY6-LaB Page 1 UNIT4DAY6-LaB Thursday, November 29, 2012 8:13 AM Vanden Bout/LaBrake CH301 The 2 nd Law of Thermodynamics GIBBS FREE ENERGY UNIT 4 Day 6 Important Information HW11 Due T DECEMBER
More informationChem 401 Unit 1 (Kinetics & Thermo) Review
KINETICS 1. For the equation 2 H 2(g) + O 2(g) 2 H 2 O (g) How is the rate of formation of H 2 O mathematically related to the rate of disappearance of O 2? 1 Δ [H2O] Δ[O 2] = 2 Δt Δt 2. Determine the
More informationCHE 107 Spring 2018 Exam 2
CHE 107 Spring 2018 Exam 2 Your Name: Your ID: Question #: 1 Which substance has the smallest standard molar entropy ( S)? A He(g) B H2O(g) C CH4(g) D F2(g) Question #: 2 Phosgene, a chemical weapon used
More informationChapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase
Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase changes Apply the second law of thermodynamics to chemical
More informationThermodynamic Fun. Quick Review System vs. Surroundings 6/17/2014. In thermochemistry, the universe is divided into two parts:
Thermodynamic Fun Quick Review System vs. Surroundings In thermochemistry, the universe is divided into two parts: The tem: The physical process or chemical reaction in which we are interested. We can
More informationName AP CHEM / / Collected AP Exam Essay Answers for Chapter 16
Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16 1980 - #7 (a) State the physical significance of entropy. Entropy (S) is a measure of randomness or disorder in a system. (b) From each of
More informationA proposed mechanism for the decomposition of hydrogen peroxide by iodide ion is: slow fast (D) H 2 O
Chemistry 112, Spring 2007 Prof. Metz Exam 2 Practice Use the following information to answer questions 1 through 3 A proposed mechanism for the decomposition of hydrogen peroxide by iodide ion is: H 2
More informationCHEM Dr. Babb s Sections Lecture Problem Sheets
CHEM 116 - Dr. Babb s Sections Lecture Problem Sheets Kinetics: Integrated Form of Rate Law 61. Give the integrated form of a zeroth order reaction. Define the half-life and find the halflife for a general
More informationWksht Rate Laws
Wksht 1.1 - Rate Laws Iowa State University Leader: Deborah Course: CHEM 178 Instructor: Bonaccorsi/Vela Date: 1/16/18 1. Describe the difference between average rate and instantaneous rate. Average rate-
More informationLecture 2. Review of Basic Concepts
Lecture 2 Review of Basic Concepts Thermochemistry Enthalpy H heat content H Changes with all physical and chemical changes H Standard enthalpy (25 C, 1 atm) (H=O for all elements in their standard forms
More informationChemistry 112, Spring 2007 Prof. Metz Exam 2 Solutions April 5, 2007 Each question is worth 5 points, unless otherwise indicated
Chemistry 11, Spring 007 Prof. Metz Exam Solutions April 5, 007 Each question is worth 5 points, unless otherwise indicated 1. A proposed mechanism for the reaction of NO with Br to give BrNO is NO + NO
More informationThermodynamics. Chem 36 Spring The study of energy changes which accompany physical and chemical processes
Thermodynamics Chem 36 Spring 2002 Thermodynamics The study of energy changes which accompany physical and chemical processes Why do we care? -will a reaction proceed spontaneously? -if so, to what extent?
More information1. Calculating arrangements due to distribution of molecules: Wpos(a,b,c,.) =
Chemistry 102 Spring 2018 Discussion #11, Chapter 17 Student name A name Section Review and hings you should know when you leave Discussion today: http://quantum.bu.edu/courses/ch102-spring-2016/notes/colligativeproperties-2014.pdf
More information7. a. A spontaneous process is one that occurs without any outside intervention.
CHAPTER SIXTEEN SPONTANEITY, ENTROPY, AND FREE ENERGY Questions 7. a. A spontaneous process is one that occurs without any outside intervention. b. Entropy is a measure of disorder or randomness. c. The
More informationThermodynamic Connection
Slide 1 Thermodynamic Connection Q and G and letters other than K Slide 2 Reaction Quotient What do you call an equilibrium constant when you aren t at equilibrium? A Reaction Quotient! (Q) Slide 3 Consider
More informationDisorder and Entropy. Disorder and Entropy
Disorder and Entropy Suppose I have 10 particles that can be in one of two states either the blue state or the red state. How many different ways can we arrange those particles among the states? All particles
More informationBCIT Fall Chem Exam #2
BCIT Fall 2017 Chem 3310 Exam #2 Name: Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially
More informationA is the frequency factor (related to the number of collisions)
Chemistry Week 10 Worksheet Notes Oregon State University Ea RT 1. Discuss k e k is the rate is the frequency factor (related to the number of collisions) Ea is the activation energy R is the gas constant
More informationSparks CH301 GIBBS FREE ENERGY. UNIT 4 Day 8
Sparks CH301 GIBBS FREE ENERGY UNIT 4 Day 8 What are we going to learn today? Quantify change in Gibbs Free Energy Predict Spontaneity at Specific Temperatures QUIZ: iclicker Questions S H2 = 131 J/K mol
More informationGibbs Free Energy Study Guide Name: Date: Period:
Gibbs Free Energy Study Guide Name: Date: Period: The basic goal of chemistry is to predict whether or not a reaction will occur when reactants are brought together. Ways to predict spontaneous reactions
More informationCalculating Rates of Substances. Rates of Substances. Ch. 12: Kinetics 12/14/2017. Creative Commons License
Ch. 2: Kinetics An agama lizard basks in the sun. As its body warms, the chemical reactions of its metabolism speed up. Chemistry: OpenStax Creative Commons License Images and tables in this file have
More information4. [7 points] Which of the following reagents would decrease the solubility of AgCl(s)? NaOH HCl NH 3 NaCN
1. [7 points] It takes 0.098 g of silver iodate, AgIO 3, to make 1.00-L of a saturated solution saturated at 25 C. What is the value of the solubility product, K sp? a. 3.5 10 4 b. 1.2 10 7 c. 9.8 10 2
More informationQuestions 1-3 relate to the following reaction: 1. The rate law for decomposition of N2O5(g) in the reaction above. B. is rate = k[n2o5] 2
Questions 1-3 relate to the following reaction: 2N2O5(g) 4NO2(g) + O2(g) 1. The rate law for decomposition of N2O5(g) in the reaction above A. is rate = k[n2o5] B. is rate = k[n2o5] 2 C. is rate = [NO2]
More information1) Define the following terms: a) catalyst; b) half-life; c) reaction intermediate
Problems - Chapter 19 (without solutions) 1) Define the following terms: a) catalyst; b) half-life; c) reaction intermediate 2) (19.10) Write the reaction rate expressions for the following reactions in
More informationFind 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. Still having trouble understanding the material? Check
More information5.111 Lecture Summary #30 Monday, November 24, 2014
5.111 Lecture Summary #30 Monday, November 24, 2014 Reading for Today: 14.114.5 in 5 th ed and 13.113.5 in 4 th ed. Reading for Lecture #31: 14.6, 17.7 in 5 th ed and 13.6, 17.7 in 4 th ed. Topic: Introduction
More informationEntropy, Free Energy, and Equilibrium
Entropy, Free Energy, and Equilibrium Chapter 17 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Spontaneous Physical and Chemical Processes A waterfall runs
More informationChemical Thermodynamics. Chapter 18
Chemical Thermodynamics Chapter 18 Thermodynamics Spontaneous Processes Entropy and Second Law of Thermodynamics Entropy Changes Gibbs Free Energy Free Energy and Temperature Free Energy and Equilibrium
More informationThermochemistry Lecture
Thermochemistry Lecture Jennifer Fang 1. Enthalpy 2. Entropy 3. Gibbs Free Energy 4. q 5. Hess Law 6. Laws of Thermodynamics ENTHALPY total energy in all its forms; made up of the kinetic energy of the
More informationChemistry 102 Spring 2018 Discussion #6, Chapter 13 & 14 Student name TA name Section
Chemistry 102 Spring 2018 Discussion #6, Chapter 13 & 14 Student name TA name Section 1. What is true if a forward reaction is spontaneous? (Chose all that apply) > 1 ; >1 = ; > 1 = 1 < 1
More informationChem 1B Dr. White 1 Chapter 17: Thermodynamics. Review From Chem 1A (Chapter 6, section 1) A. The First Law of Thermodynamics
Chem 1B Dr. White 1 Chapter 17: Thermodynamics Review From Chem 1A (Chapter 6, section 1) A. The First Law of Thermodynamics 17.1 Spontaneous Processes and Entropy A. Spontaneous Change Chem 1B Dr. White
More informationChem 1B, Test Review #2
1. The following kinetics data were obtained for the reaction: Expt.# 2NO(g) + Cl 2 (g) 2NOCl(g) [NO] 0 (mol/l) [Cl 2 ] 0 (mol/l) Initial Rate, (mol/l.s) 1 0.20 0.10 6.3 x 10 3 2 0.20 0.30 1.9 x 10 2 3
More informationChemistry 1A Fall Midterm Exam 3
Chemistry 1A Fall 2017 Name Student ID Midterm Exam 3 You will have 120 minutes to complete this exam. Please fill in the bubble that corresponds to the correct answer on the answer sheet. Only your answer
More informationChem 401 Unit 1 (Kinetics & Thermo) Review
KINETICS 1. For the equation 2 H 2(g) + O 2(g) 2 H 2 O (g) How is the rate of formation of H 2 O mathematically related to the rate of disappearance of O 2? 2. Determine the relative reaction rates of
More informationCHM 112 Chapter 16 Thermodynamics Study Guide
CHM 112 Chapter 16 Thermodynamics Study Guide Remember from Chapter 5: Thermodynamics deals with energy relationships in chemical reactions Know the definitions of system, surroundings, exothermic process,
More informationChapter 20: Thermodynamics
Chapter 20: Thermodynamics Thermodynamics is the study of energy (including heat) and chemical processes. First Law of Thermodynamics: Energy cannot be created nor destroyed. E universe = E system + E
More informationThe Laws of Thermodynamics
Entropy I. This, like enthalpy, Thus, II. A reaction is ( more on this later) if: (H, enthalpy) (S, entropy) III. IV. Why does entropy happen? Probability It s harder to keep things in order (look at my
More informationChapter 15 Chemical Equilibrium
Equilibrium To be in equilibrium is to be in a state of balance: Chapter 15 Chemical Equilibrium - Static Equilibrium (nothing happens; e.g. a tug of war). - Dynamic Equilibrium (lots of things happen,
More informationExam 2 Sections Covered: (the remaining Ch14 sections will be on Exam 3) Useful Information Provided on Exam 2:
Chem 101B Study Questions Name: Chapters 12,13,14 Review Tuesday 2/28/2017 Due on Exam Thursday 3/2/2017 (Exam 2 Date) This is a homework assignment. Please show your work for full credit. If you do work
More informationUnit #10. Chemical Kinetics
Unit #10 Chemical Kinetics Zumdahl Chapter 12 College Board Performance Objectives: Express the rate of a reaction in terms of changes in the concentration of a reactant or a product per time. Understand
More informationEnergy Ability to produce change or do work. First Law of Thermodynamics. Heat (q) Quantity of thermal energy
THERMOCHEMISTRY Thermodynamics Study of energy and its interconversions Energy is TRANSFORMED in a chemical reaction (POTENTIAL to KINETIC) HEAT (energy transfer) is also usually produced or absorbed -SYSTEM:
More information7/19/2011. Models of Solution. State of Equilibrium. State of Equilibrium Chemical Reaction
Models of Solution Chemistry- I State of Equilibrium A covered cup of coffee will not be colder than or warmer than the room temperature Heat is defined as a form of energy that flows from a high temperature
More informationChemistry Discussion #9, Chapter 16 Student name TA name Section. Things you should know when you leave Discussion today:
Chemistry 102 2016 Discussion #9, Chapter 16 Student name TA name Section Things you should know when you leave Discussion today: 1. Potential E cell at standard conditions.(all concentrations are at 1M)
More informationCh 17 Free Energy and Thermodynamics - Spontaneity of Reaction
Ch 17 Free Energy and Thermodynamics - Spontaneity of Reaction Modified by Dr. Cheng-Yu Lai spontaneous nonspontaneous Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous
More informationChem. 1B Final Practice Test 2 Solutions
First letter of last name Chem. 1B Final Practice Test 2 Solutions Name Print Neatly. You will lose 1 point if I cannot read your name or perm number. Student Number If you are sitting next to someone
More informationCHM 2046 Test #4 Review: Chapter 17 & Chapter 18
1. Which of the following is true concerning a nonspontaneous reaction? a. It s impossible for the reaction to occur b. The reaction occurs, but very slowly c. It can be made spontaneous by adding a catalyst
More informationCHM 152 Exam 4 Review Ch KEY
CHM 152 Exam 4 Review Ch. 18 19 KEY 1. Predict whether the entropy change will be positive or negative for the following: a. H 2 O (g) H 2 O (l) S - b. C 6 H 12 O 6 (s) 2C 2 H 5 OH(l) + 2CO 2 (g) S +_
More information4/19/2016. 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 informationThermochemistry. Chapter 6. Dec 19 8:52 AM. Thermochemistry. Energy: The capacity to do work or to produce heat
Chapter 6 Dec 19 8:52 AM Intro vocabulary Energy: The capacity to do work or to produce heat Potential Energy: Energy due to position or composition (distance and strength of bonds) Kinetic Energy: Energy
More informationCHEMISTRY - CLUTCH CH CHEMICAL THERMODYNAMICS.
!! www.clutchprep.com CONCEPT: THERMOCHEMICAL PROCESSES is the branch of physical science concerned with heat and its transformations to and from other forms of energy. In terms of a chemical reaction,
More information2/18/2013. Spontaneity, Entropy & Free Energy Chapter 16. The Dependence of Free Energy on Pressure Sample Exercises
Spontaneity, Entropy & Free Energy Chapter 16 16.7 The Dependence of Free Energy on Pressure Why is free energy dependent on pressure? Isn t H, enthalpy independent of pressure at constant pressure? No
More informationAP CHEMISTRY SCORING GUIDELINES
Mean 5.64 out of 9 pts AP CHEMISTRY Question 1 CO(g) + 1 2 O 2 (g) CO 2 (g) 1. The combustion of carbon monoxide is represented by the equation above. (a) Determine the value of the standard enthalpy change,
More informationconcentrations (molarity) rate constant, (k), depends on size, speed, kind of molecule, temperature, etc.
#80 Notes Ch. 12, 13, 16, 17 Rates, Equilibriums, Energies Ch. 12 I. Reaction Rates NO 2(g) + CO (g) NO (g) + CO 2(g) Rate is defined in terms of the rate of disappearance of one of the reactants, but
More informationb. 2.8 x 10-3 M min -1 d. 1.5 x 10-3 M min -1 Answer C
Chem 30 Name Exam 3 November 8, 207 00 Points Please follow the instructions for each section of the exam. Show your work on all mathematical problems. Provide answers with the correct units and significant
More informationEnergy Ability to produce change or do work. First Law of Thermodynamics. Heat (q) Quantity of thermal energy
THERMOCHEMISTRY Thermodynamics Study of energy and its interconversions Energy is TRANSFORMED in a chemical reaction (POTENTIAL to KINETIC) HEAT (energy transfer) is also usually produced or absorbed -SYSTEM:
More informationChem 116 POGIL Worksheet - Week 12 - Solutions Second & Third Laws of Thermodynamics Balancing Redox Equations
Chem 116 POGIL Worksheet - Week 12 - Solutions Second & Third Laws of Thermodynamics Balancing Redox Equations Key Questions 1. Does the entropy of the system increase or decrease for the following changes?
More informationCH Practice Exam #2
H1810 - Practice xam #2 Multiple hoice - hoose the best answer and place the letter corresponding to the answer in the space provided. 1. Give the direction of the reaction, if K
More informationSpontaneity, Entropy, and Free Energy
Spontaneity, Entropy, and Free Energy A ball rolls spontaneously down a hill but not up. Spontaneous Processes A reaction that will occur without outside intervention; product favored Most reactants are
More informationPractice Questions for Exam 2 CH 1020 Spring 2017
Practice Questions for Exam 2 CH 1020 Spring 2017 1. Pick all of the statements which are true about a reaction mechanism?. A rate law can be written from the molecularity of the slowest elementary step..
More informationChapter 16. Spontaneity, Entropy and Free energy
Chapter 16 Spontaneity, Entropy and Free energy Contents Spontaneous Process and Entropy Entropy and the second law of thermodynamics The effect of temperature on spontaneity Free energy Entropy changes
More information1.8. ΔG = ΔH - TΔS ΔG = ΔG + RT ln Q ΔG = - RT ln K eq. ΔX rxn = Σn ΔX prod - Σn ΔX react. ΔE = q + w ΔH = ΔE + P ΔV ΔH = q p = m Cs ΔT
ThermoDynamics Practice Exam Thermodynamics Name (last) (First) Read all questions before you start. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper
More informationThermodynamics: Free Energy and Entropy. Suggested Reading: Chapter 19
Thermodynamics: Free Energy and Entropy Suggested Reading: Chapter 19 System and Surroundings System: An object or collection of objects being studied. Surroundings: Everything outside of the system. the
More information1 of 8 Class notes lectures 6a, b, c
1 of 8 Class notes lectures 6a, b, c Last time: 1) entropy calculations 2) Gibb s Free energy Today: Field Trip READ CHAPT 15.11 before coming to field trip. Be prepared to ask questions and take notes.
More informationThermodynamics II. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Thermodynamics II Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves
More informationPhysical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points
Physical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points Name: KEY Gas constant: R = 8.314 J mol -1 K -1 = 0.008314 kj mol -1 K -1. Boltzmann constant k = 1.381 10-23 J/K = 0.6950 cm -1 /K h =
More information2/23/2018. Familiar Kinetics. ...and the not so familiar. Chemical kinetics is the study of how fast reactions take place.
CHEMICAL KINETICS & REACTION MECHANISMS Readings, Examples & Problems Petrucci, et al., th ed. Chapter 20 Petrucci, et al., 0 th ed. Chapter 4 Familiar Kinetics...and the not so familiar Reaction Rates
More informationChem 112, Fall 05 Exam 2a
Name: YOU MUST: Put your name and student ID on the bubble sheet correctly. Put all your answers on the bubble sheet. Please sign the statement on the last page of the exam. Please make sure your exam
More informationThe Equilibrium State. Chapter 13 - Chemical Equilibrium. The Equilibrium State. Equilibrium is Dynamic! 5/29/2012
Chapter 13 - Chemical Equilibrium The Equilibrium State Not all chemical reactions go to completion; instead they attain a state of equilibrium. When you hear equilibrium, what do you think of? Example:
More informationNCERT THERMODYNAMICS SOLUTION
NCERT THERMODYNAMICS SOLUTION 1. Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine
More informationChapter 11 Spontaneous Change and Equilibrium
Chapter 11 Spontaneous Change and Equilibrium 11-1 Enthalpy and Spontaneous Change 11-2 Entropy 11-3 Absolute Entropies and Chemical Reactions 11-4 The Second Law of Thermodynamics 11-5 The Gibbs Function
More informationCollision Theory. Unit 12: Chapter 18. Reaction Rates. Activation Energy. Reversible Reactions. Reversible Reactions. Reaction Rates and Equilibrium
Collision Theory For reactions to occur collisions between particles must have Unit 12: Chapter 18 Reaction Rates and Equilibrium the proper orientation enough kinetic energy See Both In Action 1 2 Activation
More informationOKANAGAN UNIVERSITY COLLEGE FINAL EXAMINATION CHEMISTRY 121
Name (Print) Surname Given Names Student Number Centre OKANAGAN UNIVERSITY COLLEGE FINAL EXAMINATION CHEMISTRY 2 Professor: Nigel Eggers, Renee Van Poppelen, Stephen McNeil April 5, 2004 Duration: 3 hours
More informationGeneral Chemistry revisited
General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that G = H TS where, eg, H = f H(C) + f H(D) - f H(A) - f H(B) G < 0 implied spontaneous to right G > 0 implied spontaneous to left In a very
More informationMultiple Choices: The red color is the correct answer
Multiple Choices: The red color is the correct answer 1. 2. 3. 4. 5. Which one of the following thermodynamic quantities is not a state function? (a) Gibbs free energy (b) enthalpy (c) entropy (d) internal
More informationEntropy and Free Energy
Page 1 Entropy and Free Energy How to predict if a reaction can occur at a reasonable rate? KINEICS Chapter 17 How to predict if a reaction can occur, given enough time? HERMODYNAMICS 1 Objectives Spontaneity
More informationSolutions to Thermodynamics Problems
Solutions to Thermodynamics Problems Chem03 Final Booklet Problem 1. Solution: Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05
More information1.8. ΔG = ΔH - TΔS ΔG = ΔG + RT ln Q ΔG = - RT ln K eq. ΔX rxn = Σn ΔX prod - Σn ΔX react. ΔE = q + w ΔH = ΔE + P ΔV ΔH = q p = m Cs ΔT
ThermoDynamics Practice Exam Thermodynamics Name (last) (First) Read all questions before you start. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper
More informationThermodynamics Review 2014 Worth 10% of Exam Score
Thermodynamics Review 2014 Worth 10% of Exam Score Name: Period: 1. Base your answer to the following question on the diagram shown below. 4. The heat of combustion for is kcal. What is the heat of formation
More informationChemistry 122 Wrap-Up Review Kundell
Chapter 11 Chemistry 122 Wrap-Up Review Kundell 1. The enthalpy (heat) of vaporization for ethanol (C 2 H 5 OH) is 43.3 kj/mol. How much heat, in kilojoules, is required to vaporize 115 g of ethanol at
More informationChemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry S H 2 = S H 2 R ln P H2 P NH
N (g) + 3 H (g) NH 3 (g) S N = S H = S NH 3 = S N R ln P N S H R ln P H S NH 3 R ln P NH3 ΔS rxn = (S Rln P NH 3 NH3 ) (S N Rln P N ) 3 (S H Rln P H ) ΔS rxn = S S NH 3 N 3S H + Rln P P 3 N H ΔS rxn =
More informationChapter 13: Chemical Equilibrium
Chapter 13: Chemical Equilibrium 13.1 The Equilibrium Condition Equilibrium: a state in which no observable changes occur H 2 O (l) H 2 O (g) Physical equilibrium: no chemical change. N 2(g) + 3H 2(g)
More informationEnd of Year Review ANSWERS 1. Example of an appropriate and complete solution H = 70.0 g 4.19 J/g C T = 29.8 C 22.4 C 7.4 C
End of Year Review ANSWERS 1. Example of an appropriate and complete solution H = mc T mol HCl m = 70.0 g c = 4.19 J/g C T = 9.8 C.4 C = 7.4 C mol HCl = 3.00 mol/ 0.000 = 0.0600 mol H = 70.0 g 4.19 J/g
More informationFINAL EXAM REVIEW KEY
FINAL EXAM REVIEW KEY Leaders: Deborah Course: CHEM 178 Instructor: Date: 1. What is the rate-determining step and why is it important? The slowest step of the elementary steps; looks the most like the
More informationChpt 19: Chemical. Thermodynamics. Thermodynamics
CEM 152 1 Reaction Spontaneity Can we learn anything about the probability of a reaction occurring based on reaction enthaplies? in general, a large, negative reaction enthalpy is indicative of a spontaneous
More informationUnit 5: Spontaneity of Reaction. You need to bring your textbooks everyday of this unit.
Unit 5: Spontaneity of Reaction You need to bring your textbooks everyday of this unit. THE LAWS OF THERMODYNAMICS 1 st Law of Thermodynamics Energy is conserved ΔE = q + w 2 nd Law of Thermodynamics A
More informationFormulate an operational definition of reaction rate. Include: examples of chemical reactions that occur at different rates.
Kinetics 1 UNIT 2: KINETICS OUTCOMES All important vocabulary is in Italics and bold. Formulate an operational definition of reaction rate. Include: examples of chemical reactions that occur at different
More informationChem. 1B Midterm 2 Version B Feb. 24, 2016
Chem. B Midterm 2 Version B Feb. 24, 206 First initial of last name Name: Print Neatly. You will lose point if I cannot read your name or perm number. Perm Number: All work must be shown on the exam for
More informationSaturday Study Session 1 3 rd Class Student Handout Thermochemistry
Saturday Study Session 1 3 rd Class Student Handout Thermochemistry Multiple Choice Identify the choice that best completes the statement or answers the question. 1. C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g)
More informationTHERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system
THERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system and its surroundings. a. System = That part of universe
More informationThermodynamics: Entropy, Free Energy, and Equilibrium
Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium spontaneous nonspontaneous In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using
More informationHomework 11 - Second Law & Free Energy
HW11 - Second Law & Free Energy Started: Nov 1 at 9:0am Quiz Instructions Homework 11 - Second Law & Free Energy Question 1 In order for an endothermic reaction to be spontaneous, endothermic reactions
More information