Chemical standard state: 1 M solutes, pure liquids, 1 atm gases Biochemical standard state: ph 7, all species in the ionic form found at ph 7

Similar documents
Chemical standard state: 1 M solutes, pure liquids, 1 atm gases Biochemical standard state: ph 7, all species in the ionic form found at ph 7

K a = [H + ][A ]/[HA] ph = log([h + ]) K b = [HA][HO ]/[A ]

K a = [H + ][A ]/[HA] ph = log([h + ]) K b = [HA][HO ]/[A ]

Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Final Exam (150 points total) December 19, 2016

VIEWING: Monday, May 19, 9:30-10:30 a.m., Chem

Chemistry 271, Section 21xx University of Maryland, College Park

+3 for pka = 14 pkb = 9.25 [+1 for the idea that pka and pkb are related.]

Chemistry 271, Section 22xx University of Maryland, College Park. Final Exam (200 points total) May 14, 2010

General Chemistry and Energetics Your Section #: Exam I (100 points total) October 6, 2010

R = J/mole K = cal/mole K = N A k B ln k = ( E a /RT) + ln A

General Chemistry and Energetics Your Section #: Exam I (100 points total) October 6, 2010

R = J/mole K = cal/mole K = N A k B K a K b = K w

Exam II (100 points total) November 5, 2012

S H/T 0 ph = log([h + ]) E = mc 2 S = klnw G = H T S ph = pk a + log([a ]/[HA]) K a = [H + ][A ]/[HA] G = RTlnK eq e iπ + 1 = 0

Biochemistry and Physiology ID #:

Disorder and Entropy. Disorder and Entropy

Your Section # or time:

Chem 6 sample exam 1 (100 points total)

Prof. Jason Kahn Your Signature: Exams written in pencil or erasable ink will not be re-graded under any circumstances.

Chem 1A, Fall 2015, Midterm Exam 3. Version A November 17, 2015 (Prof. Head-Gordon) 2. Student ID: TA:

Chem 112, Fall 05 (Garman/Weis) Final Exam A, 12/20/2005 (Print Clearly) +2 points

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

Chem GENERAL CHEMISTRY II MIDTERM EXAMINATION

Prof. Jason D. Kahn Your Signature: Exams written in pencil or erasable ink will not be re-graded under any circumstances.

b. Free energy changes provide a good indication of which reactions are favorable and fast, as well as those that are unfavorable and slow.

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

Chem1B General Chemistry II Exam 1 Summer Read all questions carefully make sure that you answer the question that is being asked.

Exam 3, Ch 7, 19, 14 November 9, Points

Chemistry 1A, Spring 2009 Midterm 3 April 13, 2009

Chemistry 12 Dr. Kline 28 September 2005 Name

Name. Chem 116 Sample Examination #2

Thermodynamics. Chem 36 Spring The study of energy changes which accompany physical and chemical processes

Chem Hughbanks Final Exam, May 11, 2011

Chemistry 1A, Spring 2008 Midterm Exam III, Version A April 14, 2008 (90 min, closed book)

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase

This test is closed note/book. One 8.5 x 11 handwritten crib sheet (one sided) is permitted.

Name (Print) Section # or TA. 1. You may use a crib sheet which you prepared in your own handwriting. This may be

CHEMISTRY 102 FALL 2009 EXAM 2 FORM B SECTION 501 DR. KEENEY-KENNICUTT PART 1

Questions Points Score Grader

Chemistry 1A, Spring 2007 Midterm Exam 3 April 9, 2007 (90 min, closed book)

Chemistry 2000 Lecture 9: Entropy and the second law of thermodynamics

KEY. Chemistry 1A, Fall2003 Midterm Exam III, Version A November 13, 2003 (90 min, closed book)

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

Chem GENERAL CHEMISTRY II

Chapter 19. Spontaneous processes. Spontaneous processes. Spontaneous processes

CHEMISTRY 107 Section 501 Final Exam Version A December 12, 2016 Dr. Larry Brown

Chem 6 Sample exam 1 (150 points total) NAME:

Chem Hughbanks Final Exam, May 11, 2011

1. (24 points) 5. (8 points) 6d. (13 points) 7. (6 points)

Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16

Equations: q trans = 2 mkt h 2. , Q = q N, Q = qn N! , < P > = kt P = , C v = < E > V 2. e 1 e h /kt vib = h k = h k, rot = h2.

Chem 112, Fall 05 Exam 2a

Le Châtelier's Principle. Chemical Equilibria & the Application of Le Châtelier s Principle to General Equilibria. Using Le Châtelier's Principle

Contents and Concepts

Chemistry 425 September 29, 2010 Exam 1 Solutions

c. K 2 CO 3 d. (NH 4 ) 2 SO 4 Answer c

aa + bb ---> cc + dd

1.8. 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

Contents and Concepts

Contents and Concepts

Thermodynamic Connection

Remember: You can use notes on both sides of one sheet of paper.

Exam 2, Ch 4-6 October 12, Points

1.8. 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

FRONT PAGE FORMULA SHEET - TEAR OFF

Chemistry 1A, Fall 2007 KEY Midterm Exam III November 13, 2007 (90 min, closed book)

Thermodynamic Properties. Standard Reduction Potentials

Chemical Thermodynamics

Chapter 19 Chemical Thermodynamics Entropy and free energy

Reaction Rate and Equilibrium Chapter 19 Assignment & Problem Set

Ch 18 Free Energy and Thermodynamics:

CHEMISTRY 202 Practice Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (40 pts.) 22 (20 pts.)

Chemistry 10401, Sections H*, Spring 2015 Prof. T. Lazaridis Final examination, May 18, Last Name: First Name:

Exam 1, Ch October 12, Points

Chemistry 102b First Exam. CIRCLE the section in which you are officially registered:

Chemistry 112, Fall 2006, Section 1 (Garman and Heuck) Final Exam A (100 points) 19 Dec 2006

General Chemistry Exam 3 Chem 211 Section 03, Spring 2017

Exam 4, Ch 14 and 15 December 7, Points

AP CHEMISTRY 2009 SCORING GUIDELINES

ANSWER KEY. Chemistry 25 (Spring term 2016) Midterm Examination

Placement Exam for Exemption from Chemistry 120 Fall 2006

Chapter 19 Chemical Thermodynamics Entropy and free energy

AP* Electrochemistry Free Response Questions page 1

Chapter Eighteen. Thermodynamics

Spontaneous Change: Entropy and Free Energy. 2 nd and 3 rd Laws of Thermodynamics

1. Entropy questions: PICK TWO (6 each)

MCB100A/Chem130 MidTerm Exam 2 April 4, 2013

Spontaneous Change: Entropy and Free Energy

Chemistry 112, Spring 2007 Prof. Metz Exam 2 Solutions April 5, 2007 Each question is worth 5 points, unless otherwise indicated

Chemistry 1A, Fall 2006 Final Exam, Version B Dec 12, 2006 (180 min, closed book)

1. (4) For this reaction: NO 2. + H 3 AsO 3 D H 2 AsO 3

Last 4 Digits of USC ID:

Dr. White Chem 1B Saddleback College 1. Experiment 15 Thermodynamics of the Solution Process

Chemistry 1A, Fall 2006 Final Exam, Version B Dec 12, 2006 (180 min, closed book)

CHEMISTRY 102 EXAM 2 SECTIONS /10/2004

Chapter 20: Thermodynamics

Name: Score: /100. Part I. Multiple choice. Write the letter of the correct answer for each problem. 3 points each

Chem. 1B Final Practice Test 2 Solutions

Test #3 Last Name First Name November 13, atm = 760 mm Hg

Chapter 19 Chemical Thermodynamics

Transcription:

Chemistry 271, Section 22xx Your Name: Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Exam II (100 points total) Your Section #: November 4, 2009 You have 53 minutes for this exam. Exams written in pencil or erasable ink will not be re-graded under any circumstances. Explanations should be concise and clear. I have given you more space than you should need. There is extra space on the last page if you need it. You will need a calculator for this exam. No other study aids or materials are permitted. Generous partial credit will be given, i.e., if you don t know, guess. Useful Equations: K a = [H + ][A ]/[HA] ph = log([h + ]) K b = [HA][HO ]/[A ] K w = [H + ][HO ] ph = pk a + log [A ]/[HA] G = nf E R = 0.08206 L atm/mole K 0 C = 273.15 K lnk eq = H /(RT) + S /R S q/t 0 ph = log([h + ]) PV = nrt S = klnw G = H T S R = 8.314 J/mole K F = 96500 C(oulomb)/mole 1 V = 1 J/C G = RTlnK eq W = N!/( n i!) n i /n 0 = exp[ (E i -E 0 )/kt] T M = H /[ S + R ln(c T /4)] 2.303RT/F = 0.0592 Volts at 25 C Chemical standard state: 1 M solutes, pure liquids, 1 atm gases E = E 2.303(RT/nF)log 10 Q Biochemical standard state: ph 7, all species in the ionic form found at ph 7 Honor Pledge: At the end of the examination time, please write out the following sentence and sign it, or talk to me about it: I pledge on my honor that I have not given or received any unauthorized assistance on this examination. (3 pts extra credit for filling out this page accurately and completely)

Chemistry 271, Sections 22xx Exam II, 11/4/09 2/7 1. (20 pts) Multiple choice: Circle the single best answer for each question (a; 4 pts) The Boltzmann distribution refers to the distribution of (a) microstates over configurations. (b) particles within boxes. (c) energy among particles. (d) entropy among reactants. (e) None of the above. (b; 4 pts) Free energy is useful because it (a) tells us the amount of heat released by a reaction. (b) is maximized when reactions come to equilibrium. (c) is constant during isothermal expansion of a gas. (d) reduces the fundamental spontaneity condition S universe > 0 to state functions solely of the system. (e) All of the above. (c; 4 pts) Entropy is given as a logarithm of the number of microstates because (a) it must be an increasing and extensive function of W. (b) the number of microstates cannot be calculated. (c) the Ka and Kb equilibria add to give the water self-dissociation equilibrium. (d) Boltzmann chose the functional form arbitrarily. (e) None of the above. (d; 4 pts) Reduced carbon (like humans) and molecular oxygen can coexist because (a) our skin protects our contents. (b) the kinetics of human oxidation are slow at room temperature. (c) redox reactions require electrodes. (d) we breathe out oxygen. (e) (b) and (d). (e; 4 pts) The expansion of an ideal gas into a vacuum occurs with (a) no work being done. (b) no change in internal energy. (c) no heat exchange with the surroundings. (d) a positive entropy change. (e) All of the above.

Chemistry 271, Sections 22xx Exam II, 11/4/09 3/7 2. (40 pts) Redox chemistry Ammonium, NH + 4 (aq), can be converted to nitrate, NO 3 (aq), by molecular oxygen, O 2 (g), in acidic solution. (a; 3 pts) What are the oxidation numbers of nitrogen in ammonium, nitrogen gas, and nitrate? (b; 10 pts) The balanced reduction half-reaction for the conversion above is good old O 2 + 4 H + + 4 e > 2 H 2 O. What is the balanced oxidation half-reaction? What is the balanced overall reaction? (Hint: n = 8). (c; 5 pts) The E red values for the reduction and oxidation half-reactions are +1.23 V and +0.88 V respectively. Calculate E and G = 270 (units here:-> ). (d; 3 pts) The above is all at the chemical standard state. In general, what does this mean in terms of concentrations and partial pressures of reactants and products?

Chemistry 271, Sections 22xx Exam II, 11/4/09 4/7 (e; 12 pts) Calculate E at ph 7, po 2 = 0.21 atm, 1 M each [NH 4 + ] and [NO 3 ], and 25 C. Use the fact that E = 0 at equilibrium or the G from (c) above to calculate the ratio of [NH 4 + ]/[NO 3 ] at equilibrium at ph 7, po 2 = 0.21 atm, 25 C. There is no need to worry about K p vs. K c. (f; 3 pts) What industrial process is used to make ammonia? (g; 4 pts) Thinking about the oxidation states of N, and recognizing that NO 3 is nearly as good an oxidant as O 2, why do you think NH 4 NO 3 is extremely dangerous? Extra credit: name a city that had a very bad experience with it.

Chemistry 271, Sections 22xx Exam II, 11/4/09 5/7 3. (40 pts) Thermodynamics, DNA, and van t Hoff. (a; 4 pts) What does S = q/t mean in terms of the effect of adding energy to hot and cold systems? (b; 8 pts) Sketch a van t Hoff plot for an exothermic ordering reaction like DNA hybridization.

Chemistry 271, Sections 22xx Exam II, 11/4/09 6/7 (c; 6 pts) Rationalize the signs of H and S for the hybridization process. (d; 6 pts) In terms of LeChatelier s principle, why does the reaction become less favorable as temperature increases? In terms of the statistical nature of the universe, why does the reaction become less favorable as temperature increases? (e; 8 pts) Given H = 20 kj/mol and S = 53 J/mol K for a hybridization reaction, calculate G and K eq at 20 C and 60 C.

Chemistry 271, Sections 22xx Exam II, 11/4/09 7/7 (f; 8 pts) K eq = 2α/(1-α) 2 C t, where α is the fraction of double-stranded DNA and C t is the total strand concentration. Given that α = 0.5 at the melting temperature T m, use the two main expressions for G to show that T m = H /[ S + R ln (C t /4)]. Page Score 1 /3 2 /20 3 /21 4 /19 5 /12 6 /20 7 /8 Total