reduction kj/mol
|
|
- Myrtle Blair
- 6 years ago
- Views:
Transcription
1 1. Glucose is oxidized to water and CO 2 as a result of glycolysis and the TCA cycle. The net heat of reaction for the oxidation is kj/mol. a) How much energy is required to produce glucose from water and CO 2? Is this value thermodynamically favorable? b) Predict qualitatively, whether the oxidation or the reduction is spontaneous. Explain your answer. c) Plants efficiently fix CO 2 to form glucose. In this biosynthetic pathway, if photons of 700nm wavelength are absorbed by the chlorophyll photosystem synthetic complex, how many photons are required to fix one mole of CO 2? d) Given your answer to c and the constraints of the second law of thermodynamics comment on the number of photons actually used in CO 2 fixation. Explain your answer. a) The first thing is to write write and balance reaction: C 6 H 12 O 6 + 6H 2 O à 6CO H 2 O oxidation kj/mol Reverse reaction 6CO H 2 O à C 6 H 12 O 6 + 6H 2 O reduction kj/mol Formation of glucose is simply the reverse reaction so the energy is equivalent in magnitude but opposite in sign. The formation of glucose is not thermodynamically favorable b) The heat of reaction for the oxidation is favorable (-2870 kj/mol ). Since 7 moles are being converted to 12 moles and the products are 12 moles of gas entropy should be favorable in the direction of oxidation. Therefore the oxidation reaction should be spontaneous. c) E=hv = hc/λ convert 700nm wavelength to 700E-9m E= 6.626E-34 m 2 kg/s * 2.997E8m/s *1/700E-9 E=2.837E-19 J for a single photon 2870 kj/mol = 2870E3 J for one mole of glucose 2870E3 J/ 2.837E-19J = E25 photons for a mole of glucose a single mole of glucose corresponds to 6 moles of CO E25 /6 = 1.686E24 ( a bit more than a mole of photons for a mole of gas) d) Since the second law of thermodynamics prohibits perfect energy transfer much of the light energy would be lost (not used to fix CO 2 ). Therefore it would take more than the calculated E24 photons to fix a mole.
2 2. Giant sequoias, an indigenous species to California, live up to 3500 years and can reach heights in excess of 300 m. a) Calculate the work that must be done in transporting a single water molecule to the top of a 300 m sequoia. b) Show (explain) how your answer involves the first law of thermodynamics (use equations and words). a) W=mgh mass of one water molecule 18.0g/mol x 1mol NA b) first law U=q+w constant as work is done to the surroundings heat must be absorbed into the system a) w=mgh mass single water molecule 18g/mol * 1mol/6.023E23 = 2.99E-23 g = 2.99E-26 kg w = 2.99E-26 kg * 300 m * 9.8 m/s 2 = E-23 kgm 2 /s 2 = E-23 J actually = E-23 J as the work is done against gravity b)the conservation of energy indicates that energy is conserved forms can change but total energy is invariant. In this case the work involved in moving the molecule up the tree is balanced by the heat gained by the system. ΔU = q + w since ΔU is 0 (the water molecule does not change internal energy) 0 = q + w and -w = q
3 3. The entropy change for an nucleic acid was investigated upon thermal unfolding under constant pressure (d refers to denaturation). The following values were measured: ΔH 0 d = 9.2 kcal/mol, T m = 72 0 C, ΔC p = 1200 cal/mol K. a) Calculate the enthalpy of denaturation at T m b) Calculate ΔS d at T m - Is the unfolding at T m entropically favorable? Rationalize/explain c) Calculate ΔS d at 15 0 C- Is the unfolding entropically favorable? Rationalize/explain a) constant pressure q=h therefore ddh = dcp dt H (72) H (25) = 1200 cal/mol K * (72-25)K Solve for H (72) H (72) = 9.2kcal/mol cal/mol K * (72-25)K = kcal/mol b) all at 72C S = H/T S (72) = 65780cal / ( ) K = 177 cal/k this number is >0 indicating increase in entropy and therefore entropically favorable c) DS (15) = ds (72) + dcpln (288K/335) = 177 cal/k cal/mol * ln(288/345) = cal/molk this number is <0 indicating a decrease in entropy and therefore not entropically favorable. Since entropy is highly dependent upon temperature at low temperature the decrease in entropy is significant moving from a favorable process at higher temperatures to an unfavorable process when cold.
4 4. The following reaction is catalyzed by the enzyme creatine kinase: Creatine phosphate + ADP ----> Creatine + ATP Under the listed conditions (1 atm, 37 o C, ph 7.0, pmg 3.0 and an ionic strength of 0.25 M) with the concentrations of all reactants at 1mM and products equal to 10 mm: ΔG 335 for this reaction is kj/mole. ΔG = kj/mol a) What is the equilibrium constant for this reaction at 335K? Is this reaction favorable? Explain b) What is the standard free energy change, ΔG 0, for this reaction at 335 K? Is this reaction favorable under standard conditions? Explain c) What is the standard enthalpy change, ΔH 0, for this reaction? d) What is the standard entropy change, ΔS 0, for this reaction? Is this change favorable? Explain e) Given your answers is this reaction driven by entropy or enthalpy or both? Explain your rationale a) Keq= [creatine][atp]/[creatine-p][adp] 10mm(10mm)/1mM(1mM) = 100 (can keep in mm since units cancel) yup more products than reactants by a long shot so the Keq>>1 b) ΔG = ΔG 0 + RTlnKeq J/mol = ΔG J/molK * 335K *ln J/mol = ΔG J/mol ΔG 0 = -26,126 J/mol = kj/mol yes very favorable at 335 K (more so than at 298 K) c) Have G 0 values at more than one temperature and want the enthalpy - use Gibbs Helmholtz G2/T2 - G1/T1 = ΔH (1/T2 1/T1) /335 - (-12600/298) = ΔH (1/335 1/298) J/molK J/molK = ΔH 1/K J/mol = ΔΗ ΔH= 9633 J/mol or 9.63 kj/mol d) ΔS ΔG=ΔH-TΔS J/mol = 9633 J/mol K* ΔS ΔS = J/molK very favorable the contribution of entropy out weighs the contribution of enthalpy in this reaction.
5 5. A diatomic ideal gas (0.5 moles) is expanded from a volume of 1.5 L to 3 L. a) Under a constant external pressure of 1 atm, at a temperature of 25 0 C, calculate the work. Is the work done on the system or by the system? Explain practically and mathematically. b) If the expansion in part a) is carried out incrementally allowing the system to come to equilibrium at each step rather than in a single step expansion, calculate the work. Again is work done on the system or by the system? c) Compare the work in part a to part b. Explain any difference or if no difference explain why. d) If the expansion in part b is allowed to occur with a change in temperature (final temperature 40 0 C) under a pressure of 1 atm, calculate the work, heat, internal energy change and enthalpy change for the process. e) Can the values for ΔE and ΔH calculate in part c be applied to the process in part b? Why or why not? Explain. a) w = -P(V 2 -V 1 ) this is a single step transformation under a constant external pressure w= 1 atm(3-1.5 )L = -1.5 Latm *101.3 J/Latm = J The negative sign indicates work done by the system on the surroundings b) w= nrt ln V 2 /V 1 this is a reversible path maximum work = 0.5mol*8.314J/molK (298K) ln (3/1.5) = J The negative sign indicates work done by the system on the surrounding c) part b reversible path maximum work entire area under the curve value is larger sign agrees d) ΔU =q p pdv ideal gas 0.5 mol diatomic C v = df*nr/2 and datiomic df = 5 C v = 5*0.5mol*8.314J/molK/2 = J/K Cp = C v +nr = 10.39J/K +0.5mol (8.314J/molK) = J/K ΔU = C v (ΔT) = J/K (40-25)K = J in 0.5 mol or J/mol ΔH = C p (ΔT) = J/K (40-25)K =218 J in 0.5 mol or 436 J/mol Δq = ΔH = 436 J/mol (constant pressure) or 218 J ΔU = q + w = J/mol = 436 J/mol + w w = J/mol (expansion!!!) e) yes internal energy and enthalpy are state functions path independent cannot be applied for q and w and as you see in parts a and b w is different than in d
6 6. Given the standard enthalpy of combustions of: methane gas ( kj/mol), hydrogen gas ( kj/mol), C graphite ( kj/mol), the bond dissociation enthalpy of hydrogen gas (435.8 kj/mol), and the latent heat of sublimation of graphite (718.4 kj/mol) evaluate the bond enthalpy of the C-H bond in methane. Clearly explain using a brief diagram (description) of the path chosen. Hint: CH4 (g) à C(g) + 4H(g) would represent bond formation Combustion of methane all kj/mol CH4 + 2O2 -à CO2 + 2H2O H2 + O2à 2 H2O C (s)+ O2 à CO H2à2H C(s) à C(g) What is C-H? energy Want C(g) + 4 H(g) à CH4 (g) Path devised CH4 + 2O2 -à CO2 + 2H2O CO2 à C (s)+ O (reversed) C(s) à C(g) H2O à 2H2 + O (reversed) 2H2à4H x Net Reaction CH4 à C + 4H all gas This describes making methane from 1 C and 4 H Total for 4 bonds = 344 kj/mol for one C-H bond
a) Write the reaction that occurs (pay attention to and label ends correctly) 5 AGCTG CAGCT > 5 AGCTG 3 3 TCGAC 5
Chem 315 Applications Practice Problem Set 1.As you learned in Chem 315, DNA higher order structure formation is a two step process. The first step, nucleation, is entropically the least favorable. The
More informationChem 350 Problem Set 1 Thermodynamics Key
Chem 350 Problem Set 1 Thermodynamics 2018 - Key 1. A man at sea level with a 1.0 L resting lung capacity breathes deeply inhaling one mole of gas increasing his lung capacity to 1.6 L. Temperature remains
More informationEnergy is the capacity to do work
1 of 10 After completing this chapter, you should, at a minimum, be able to do the following. This information can be found in my lecture notes for this and other chapters and also in your text. Correctly
More informationClass XI Chapter 6 Thermodynamics Chemistry
Class XI Chapter 6 Chemistry Question 6.1: 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
More information10 NEET 31 Years 11. The enthalpy of fusion of water is kcal/mol. The molar entropy change for the melting of ice at
6 Thermodynamics. A gas is allowed to expand in a well insulated container against a constant external pressure of.5 atm from an initial volume of.50 L to a final volume of 4.50 L. The change in internal
More informationGibbs Free Energy. Evaluating spontaneity
Gibbs Free Energy Evaluating spontaneity Predicting Spontaneity An increase in entropy; Changing from a more structured to less structured physical state: Solid to liquid Liquid to gas Increase 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 informationThermochemistry Chapter 8
Thermochemistry Chapter 8 Thermochemistry First law of thermochemistry: Internal energy of an isolated system is constant; energy cannot be created or destroyed; however, energy can be converted to different
More informationThe energy of oxidation of 11 g glucose = kj = kg/m 2 s 2
Chem 350 thermo problems Key 1. How many meters of stairway could a 70kg man climb if all the energy available in metabolizing an 11 g spoonful of sugar to carbon dioxide and water could be converted to
More informationThermodynamics. For the process to occur under adiabatic conditions, the correct condition is: (iii) q = 0. (iv) = 0
Thermodynamics 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 pressure volume
More informationLecture 3: Thermodynamics
3 LAWS OF THERMODYNAMICS Lecture 3: Thermodynamics Matter and energy are conserved Margaret A. Daugherty Fall 2004 Entropy always increases Absolute zero is unattainable System and Surroundings 1st Law
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 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 informationThe Second Law of Thermodynamics (Chapter 4)
The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made
More informationEntropy. Spontaneity. Entropy. Entropy mol of N 2 at 1 atm or 1 mol of N 2 at atm. process a process that occurs without intervention
Entropy Spontaneity process a process that occurs without intervention can be fast or slow Entropy (s) the measure of molecular randomness or disorder Think of entropy as the amount of chaos Entropy Predict
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 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 informationChapter 7 Chemical Reactions: Energy, Rates, and Equilibrium
Chapter 7 Chemical Reactions: Energy, Rates, and Equilibrium Introduction This chapter considers three factors: a) Thermodynamics (Energies of Reactions) a reaction will occur b) Kinetics (Rates of Reactions)
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 informationChemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry
Recall the equation. w = -PΔV = -(1.20 atm)(1.02 L)( = -1.24 10 2 J -101 J 1 L atm Where did the conversion factor come from? Compare two versions of the gas constant and calculate. 8.3145 J/mol K 0.082057
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 informationFor more info visit
Basic Terminology: Terms System Open System Closed System Isolated system Surroundings Boundary State variables State Functions Intensive properties Extensive properties Process Isothermal process Isobaric
More informationSecond law of thermodynamics
Second law of thermodynamics It is known from everyday life that nature does the most probable thing when nothing prevents that For example it rains at cool weather because the liquid phase has less energy
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 informationBasics of Thermodynamics: Easy learning by Dr. Anjana Sen
Basics of Thermodynamics: Easy learning by Dr. Anjana Sen Part 1: Theory and concept Part 2: Definitions and equations Part 3: Laws of Thermodynamics Part 1: theory and concept Thermodynamics means conversion
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 informationThermodynamics. Thermodynamics of Chemical Reactions. Enthalpy change
Thermodynamics 1 st law (Cons of Energy) Deals with changes in energy Energy in chemical systems Total energy of an isolated system is constant Total energy = Potential energy + kinetic energy E p mgh
More informationHomework Problem Set 6 Solutions
Chemistry 360 Dr. Jean M. Standard Homework Problem Set 6 Solutions 1. Determine the amount of pressure-volume work performed by 50.0 g of liquid water freezing to ice at 0 C and 1 atm pressure. The density
More informationChapter 8 Thermochemistry: Chemical Energy
Chapter 8 Thermochemistry: Chemical Energy 國防醫學院生化學科王明芳老師 2011-11-8 & 2011-11-15 Chapter 8/1 Energy and Its Conservation Conservation of Energy Law: Energy cannot be created or destroyed; it can only be
More informationHomework Problem Set 8 Solutions
Chemistry 360 Dr. Jean M. Standard Homework roblem Set 8 Solutions. Starting from G = H S, derive the fundamental equation for G. o begin, we take the differential of G, dg = dh d( S) = dh ds Sd. Next,
More informationChemistry 1A, Spring 2007 Midterm Exam 3 April 9, 2007 (90 min, closed book)
Chemistry 1A, Spring 2007 Midterm Exam 3 April 9, 2007 (90 min, closed book) Name: KEY SID: TA Name: 1.) Write your name on every page of this exam. 2.) This exam has 34 multiple choice questions. Fill
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 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 informationCHAPTER THERMODYNAMICS
54 CHAPTER THERMODYNAMICS 1. If ΔH is the change in enthalpy and ΔE the change in internal energy accompanying a gaseous reaction, then ΔHis always greater than ΔE ΔH< ΔE only if the number of moles of
More informationMME 2010 METALLURGICAL THERMODYNAMICS II. Fundamentals of Thermodynamics for Systems of Constant Composition
MME 2010 METALLURGICAL THERMODYNAMICS II Fundamentals of Thermodynamics for Systems of Constant Composition Thermodynamics addresses two types of problems: 1- Computation of energy difference between two
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 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 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 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 informationChemical Thermodynamics
Page III-16-1 / Chapter Sixteen Lecture Notes Chemical Thermodynamics Thermodynamics and Kinetics Chapter 16 Chemistry 223 Professor Michael Russell How to predict if a reaction can occur, given enough
More informationPrinciples of Bioenergetics. Lehninger 3 rd ed. Chapter 14
1 Principles of Bioenergetics Lehninger 3 rd ed. Chapter 14 2 Metabolism A highly coordinated cellular activity aimed at achieving the following goals: Obtain chemical energy. Convert nutrient molecules
More informationLecture 2: Biological Thermodynamics [PDF] Key Concepts
Lecture 2: Biological Thermodynamics [PDF] Reading: Berg, Tymoczko & Stryer: pp. 11-14; pp. 208-210 problems in textbook: chapter 1, pp. 23-24, #4; and thermodynamics practice problems [PDF] Updated on:
More information1 A reaction that is spontaneous.
Slide 1 / 55 1 A reaction that is spontaneous. A B C D E is very rapid will proceed without outside intervention is also spontaneous in the reverse direction has an equilibrium position that lies far to
More informationASSIGNMENT SHEET #11 APQ ANSWERS
ASSIGNMENT SHEET #11 APQ ANSWERS #1 a. The unit for q must be an energy unit, typically Joules or calories. The unit for mass is the gram. The unit for specific heat is J per gram-degree or calorie per
More informationChapter Eighteen. Thermodynamics
Chapter Eighteen Thermodynamics 1 Thermodynamics Study of energy changes during observed processes Purpose: To predict spontaneity of a process Spontaneity: Will process go without assistance? Depends
More informationExam 2 Solutions. for a gas obeying the equation of state. Z = PV m RT = 1 + BP + CP 2,
Chemistry 360 Dr. Jean M. Standard Fall 016 Name KEY 1.) (14 points) Determine # H & % ( $ ' Exam Solutions for a gas obeying the equation of state Z = V m R = 1 + B + C, where B and C are constants. Since
More informationBCIT Fall Chem Exam #2
BCI Fall 2016 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 informationPhysical Biochemistry. Kwan Hee Lee, Ph.D. Handong Global University
Physical Biochemistry Kwan Hee Lee, Ph.D. Handong Global University Week 3 CHAPTER 2 The Second Law: Entropy of the Universe increases What is entropy Definition: measure of disorder The greater the disorder,
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 informationS and G Entropy and Gibbs Free Energy
Week 3 problem solving + equa'ons, + applica'ons S and G Entropy and Gibbs Free Energy Classical defini,on of ΔS = q/t Problem: A calorimeter measured heat of 120J absorbed by a drug specimen while T increased
More informationUNIVERSITY OF TORONTO FACULTY OF APPLIED SCIENCE AND ENGINEERING TERM TEST 2 17 MARCH First Year APS 104S
UNIERSIY OF ORONO Please mark X to indicate your tutorial section. Failure to do so will result in a deduction of 3 marks. U 0 U 0 FACULY OF APPLIED SCIENCE AND ENGINEERING ERM ES 7 MARCH 05 U 03 U 04
More informationChapter 19 Chemical Thermodynamics
Chapter 19 Chemical Thermodynamics Spontaneous Processes Entropy and the Second Law of Thermodynamics The Molecular Interpretation of Entropy Entropy Changes in Chemical Reactions Gibbs Free Energy Free
More informationUNIT 9 IB MATERIAL KINETICS & THERMODYNAMICS
UNIT 9 IB MATERIAL KINETICS & THERMODYNAMICS Name: ESSENTIALS: Know, Understand, and Be Able To State that combustion and neutralization are exothermic processes. Calculate the heat energy change when
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 information= 16! = 16! W A = 3 = 3 N = = W B 3!3!10! = ΔS = nrln V. = ln ( 3 ) V 1 = 27.4 J.
Answer key: Q1A Both configurations are equally likely because the particles are non-interacting (i.e., the energy does not favor one configuration over another). For A M = 16 N = 6 W A = 16! 0.9 101 =
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 informationChapter 10 Lecture Notes: Thermodynamics
Chapter 10 Lecture Notes: Thermodynamics During this unit of study, we will cover three main areas. A lot of this information is NOT included in your text book, which is a shame. Therefore, the notes you
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 information3/30/2017. Section 17.1 Spontaneous Processes and Entropy Thermodynamics vs. Kinetics. Chapter 17. Spontaneity, Entropy, and Free Energy
Chapter 17 Spontaneity, Entropy, and Thermodynamics vs. Kinetics Domain of Kinetics Rate of a reaction depends on the pathway from reactants to products. Thermodynamics tells us whether a reaction is spontaneous
More informationCHM 111 Dr. Kevin Moore
CHM 111 Dr. Kevin Moore Kinetic Energy Energy of motion E k 1 2 mv 2 Potential Energy Energy of position (stored) Law of Conservation of Energy Energy cannot be created or destroyed; it can only be converted
More informationThe Nature of Energy. Chapter Six: Kinetic vs. Potential Energy. Energy and Work. Temperature vs. Heat
The Nature of Energy Chapter Six: THERMOCHEMISTRY Thermodynamics is the study of energy and its transformations. Thermochemistry is the study of the relationship between chemical reactions and energy 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 informationEnergy and Cells. Appendix 1. The two primary energy transformations in plants are photosynthesis and respiration.
Energy and Cells Appendix 1 Energy transformations play a key role in all physical and chemical processes that occur in plants. Energy by itself is insufficient to drive plant growth and development. Enzymes
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (40 pts.)
CHEMISTRY 202 Hour Exam II October 27, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 32 questions on 11 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationModule 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics
Module 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics Objectives In this Lecture you will learn the following The need for studying thermodynamics to understand chemical and biological processes.
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 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 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 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 informationLecture 7 Enthalpy. NC State University
Chemistry 431 Lecture 7 Enthalpy NC State University Motivation The enthalpy change ΔH is the change in energy at constant pressure. When a change takes place in a system that is open to the atmosphere,
More informationChapter 17: Energy and Kinetics
Pages 510-547 S K K Chapter 17: Energy and Kinetics Thermochemistry: Causes of change in systems Kinetics: Rate of reaction progress (speed) Heat, Energy, and Temperature changes S J J Heat vs Temperature
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 informationThermodynamics: Study of heat and its relationship with other forms of energy
Unit 6 The 6 th planet in our solar system is Saturn Ch. 5: Thermodynamics: Study of heat and its relationship with other forms of energy Two types of energy: Kinetic: movement, active energy Potential:
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 informationI PUC CHEMISTRY CHAPTER - 06 Thermodynamics
I PUC CHEMISTRY CHAPTER - 06 Thermodynamics One mark questions 1. Define System. 2. Define surroundings. 3. What is an open system? Give one example. 4. What is closed system? Give one example. 5. What
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 information1. Use the Data for RNAse to estimate:
Chem 78 - - Spr 1 03/14/01 Assignment 4 - Answers Thermodynamic Analysis of RNAseA Denaturation by UV- Vis Difference Absorption Spectroscopy (and Differential Scanning Calorimetry). The accompanying excel
More informationChapter 19 Chemical Thermodynamics Entropy and free energy
Chapter 19 Chemical Thermodynamics Entropy and free energy Learning goals and key skills: Explain and apply the terms spontaneous process, reversible process, irreversible process, and isothermal process.
More informationTopic 05 Energetics : Heat Change. IB Chemistry T05D01
Topic 05 Energetics 5.1-5.2: Heat Change IB Chemistry T05D01 5.1 Exothermic and endothermic reactions - 1 hour 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change
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 informationPractice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set.
Practice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set. The symbols used here are as discussed in the class. Use scratch paper as needed. Do not give more than one answer for any question.
More informationChapter 5 Thermochemistry
Chapter 5 Thermochemistry Learning Outcomes: Interconvert energy units Distinguish between the system and the surroundings in thermodynamics Calculate internal energy from heat and work and state sign
More informationCHEM Thermodynamics. Entropy, S
hermodynamics Change in Change in Entropy, S Entropy, S Entropy is the measure of dispersal. he natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal
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 informationChemistry and the material world Unit 4, Lecture 4 Matthias Lein
Chemistry and the material world 123.102 Unit 4, Lecture 4 Matthias Lein Gibbs ree energy Gibbs ree energy to predict the direction o a chemical process. Exergonic and endergonic reactions. Temperature
More informationAP* Thermodynamics Free Response Questions page 1. Essay Questions
AP* Thermodynamics Free Response Questions page 1 Essay Questions 1991 The reaction represented above is a reversible reaction. BCl 3 (g) + NH 3 (g) Cl 3 BNH 3 (s) (a) Predict the sign of the entropy change,
More informationS = k log W CHEM Thermodynamics. Change in Entropy, S. Entropy, S. Entropy, S S = S 2 -S 1. Entropy is the measure of dispersal.
, S is the measure of dispersal. The natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal of matter: Thermodynamics We analyze the constraints on
More informationChapter 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 informationChem 1B Objective 12: Predict whether a reaction occurs using thermodynamics.
Chem 1B Objective 12: Predict whether a reaction occurs using thermodynamics. Key Ideas: Does a reaction occur? Some reactions occur spontaneously; others do not. G is the criterion used to predict whether
More informationAdditional Calculations: 10. How many joules are required to change the temperature of 80.0 g of water from 23.3 C to 38.8 C?
Additional Calculations: 10. How many joules are required to change the temperature of 80.0 g of water from 23.3 C to 38.8 C? q = m C T 80 g (4.18 J/gC)(38.8-23.3C) = 5183 J 11. A piece of metal weighing
More informationThermochemistry is the study of the relationships between chemical reactions and energy changes involving heat.
CHEM134- F18 Dr. Al- Qaisi Chapter 06: Thermodynamics Thermochemistry is the study of the relationships between chemical reactions and energy changes involving heat. Energy is anything that has the capacity
More informationMCAT General Chemistry Discrete Question Set 19: Thermochemistry & Thermodynamics
MCAT General Chemistry Discrete Question Set 19: Thermochemistry & Thermodynamics Question No. 1 of 10 1: A metal with a high heat capacity is put on a hot plate. What will happen? Question #01 A. The
More informationTHERMODYNAMICS. Topic: 5 Gibbs free energy, concept, applications to spontaneous and non-spontaneous processes VERY SHORT ANSWER QUESTIONS
THERMODYNAMICS Topic: 5 Gibbs free energy, concept, applications to spontaneous and non-spontaneous processes 1. What is Gibbs energy? VERY SHORT ANSWER QUESTIONS Gibbs energy (G): The amount of energy
More informationThermochemistry: the study of energy (in the from of heat) changes that accompany physical & chemical changes
Thermochemistry Thermochemistry: the study of energy (in the from of heat) changes that accompany physical & chemical changes heat flows from high to low (hot cool) endothermic reactions: absorb energy
More informationChapter 5. Thermochemistry
Chapter 5 Thermochemistry Dr. A. Al-Saadi 1 Preview Introduction to thermochemistry: Potential energy and kinetic energy. Chemical energy. Internal energy, work and heat. Exothermic vs. endothermic reactions.
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 informationTHERMODYNAMICS. Dr. Sapna Gupta
THERMODYNAMICS Dr. Sapna Gupta FIRST LAW OF THERMODYNAMICS Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes. First Law of Thermodynamics Energy cannot
More informationEquations: 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.
Constants: R = 8.314 J mol -1 K -1 = 0.08206 L atm mol -1 K -1 k B = 0.697 cm -1 /K = 1.38 x 10-23 J/K 1 a.m.u. = 1.672 x 10-27 kg 1 atm = 1.0133 x 10 5 Nm -2 = 760 Torr h = 6.626 x 10-34 Js For H 2 O
More information