Chapter 1 Thermodynamics and Equilibrium - Copy

October 27, 2017 | Author: JakeNK94 | Category: Gibbs Free Energy, Chemical Equilibrium, Entropy, Enthalpy, Chemical Reactions
Share Embed Donate

Short Description

Thermodynamics Chemistry Introduction...



 

Thermodynamics is an extensive and far-reaching scientific discipline that deals with the interconversion of heat and other forms of energy. Thermodynamics enables us to use information gained from experiments on a system to draw conclusions about other aspects of the same system without further experimentation. It is possible to calculate the enthalpy of reaction from the standard enthalpies of formation of the reactant and product molecules. This chapter introduces the second law of thermodynamics and the Gibbs freeenergy function. It also discusses the relationship between Gibbs free energy and chemical equilibrium.

GENERAL CHEMISTRY CHAPTER 1: THERMODYNAMIC AND EQUILIBRIUM 1.1 First Law of Thermodynamics; Enthalpy 1. The internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system. 2. Internal energy is a state function. That is, it is a property of a system that depends only on its present state, which is completely determined by variables such as temperature and pressure. 3. When a system changes from one state to another, its internal energy changes from one definite value to another.

4. Work, on the other hand, is the energy exchange that results when a force F moves an object through a distance d; work (w) equals F × d. 5. The system in Figure 18.2 gains internal energy from the heat absorbed and loses internal energy via the work done. 6. In general, the net change of internal energy equals heat plus work. 7. The first law of thermodynamics states that the change in internal energy of a system, U, equals q+ w. 8. Heat Of Reaction And Internal Energy: o You write qp = -152.4 kJ, where the subscript p indicates that the process occurs at constant pressure (the pressure of the atmosphere).


o o

The negative sign is given because w is work done by the system and represents energy lost by it. Note that F/A is the pressure, P, which equals that of the atmosphere. When the system expands, V is positive, so W is negative. The system does work on the surroundings. When the system contracts, V is negative, so W is positive. The surroundings do work on the system.


If you apply the first law to this chemical system, you can relate the change in internal energy of the system to the heat of reaction.

9. Enthalpy and Enthalpy Change: o The heat content of a chemical system is called the enthalpy (H). o The enthalpy change (H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.


Calculate enthalpy change for the following:

o o

Ho=[(-319.2-285.8)-[(-2)(45.9)-393.5)]kJ=-119.7kJ Negative sign means reaction is exothermic

1.2 Entropy and the Second Law of Thermodynamics 1. Entropy, S, is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy. 2. Suppose you place a hot cup of coffee on the table. 3. Heat energy from hot coffee flows slowly to the table and to air surrounding the cup. 4. In this process, energy spreads out or disperses. 5. The entropy of the system (coffee cup) and its surroundings (table and surrounding air) increases in this spontaneous process. 6. The SI unit of entropy is joules per kelvin (J/K). Entropy, like enthalpy, is a state function. 7. Why is the entropy of liquid water higher than that of solid water (ice)? 8. In an ice crystal, water molecules occupy regular fixed positions in the crystal lattice. 9. In the liquid state, water molecules can rotate as well as vibrate internally and can move around somewhat (though not as freely as molecules can move in the gaseous state).

10. The entropy increases, as you expect, because the energy of water becomes more dispersed when it melts.


11. Second Law of Thermodynamics: o Second law of thermodynamics, which states that the total entropy of a system and its surroundings always increases for a spontaneous process. o Energy can be neither created nor destroyed during a spontaneous, or natural, process—its total amount remains fixed. o But energy is dispersed in a spontaneous process, which means that entropy is produced (or created) during such a process.


If we delete “entropy created” from the right side of the equation for S, we know that the left side is then greater than the right side.

Second Law of Thermodynamics: For a spontaneous process at a given temperature T, the change in entropy of the system is greater than the heat divided by the absolute temperature, q/T. 12. Entropy and Molecular Disorder: o Entropy is related to energy dispersal. o The energy of a molecular system may at first be concentrated in a few energy states and then later dispersed among many more energy states. o In such a case, the entropy of the system increases. o The molecular system has increased in disorder in the sense. o Entropy is a measure of disorder. o Example: When ice melts, water molecules leave on ordered crystal and enter a more disordered liquid state. o The value of S increases in the direction that disorder increases. o


Entropy (J/mol.K)

H2O (l)


H2O (g)


13. Entropy Change for a Phase Transition

o o o

Consider the melting of ice. The heat absorbed is the heat of fusion, Hfus, which is known from experiment to be 6.0 kJ for 1 mol of ice. You get the entropy change for melting by dividing Hfus by the absolute temperature of the phase transition, 273 K.


14. Criterion for Spontaneous Reactions

o o o

If H - TS is negative for the reaction, you would predict that it is spontaneous. However, if H - TS is positive, the reaction is nonspontaneous in the direction written but spontaneous in the opposite direction. If H - TS is zero, the reaction is at equilibrium.

1.3 Standard Entropies and the Third Law of Thermodynamics 1. The third law of thermodynamics states that a substance that is perfectly crystalline at 0 K has an entropy of zero. 2. The standard entropy of a substance or ion, also called its absolute entropy, S°, is the entropy value for the standard state of the species (indicated by the superscript degree sign). 3. For a substance, the standard state is the pure substance at 1atm pressure. 4. Entropy Change for a Reaction: o S > 0 (positive), it is likely that  [1] A molecule is broken into two or more smaller molecules – creates more disorder and can move independently  The dinitrogen tetroxide dissociates into two nitrogen dioxide molecules.

GENERAL CHEMISTRY  [2] Increase in moles of gas  An increase in the number of moles. If the product side of the equation has more moles than the reactant side More particles moving about is a more random state than fewer particles moving about.  [3] Phase transitions like  solid to liquid  liquid to gas  solid to gas.  Liquids are more disordered than solids. WHY? - Solids have a more regular structure than liquids.  Gases are more disordered than their respective liquids. WHY? - Gases particles are in a state of constant random motion.  Generally speaking gases have higher entropy than liquids which have higher entropy than solids. Example 2a:  Predicting the sign of the entropy change of a reaction  The fermentation of glucose to ethanol  




o o

 Molecule (glucose) breaks into smaller molecular (carbon dioxide and ethanol)  Moreover, this results in a gas being released  So is positive and the entropy increases Example 2b:  Preparation of urea from ammonia and carbon dioxide  The moles of gas decrease (by 3 mol)  Which would decrease the entropy  So is negative Example 2c:  What is the sign for the following reaction?  Because there is no change in the number of moles of gas, you cannot predict the sign of So from the rules given It is useful to be able to predict the sign of S (qualitative work) For quantitative work, you need to find out the value of S

1.4 Free Energy and Spontaneity 1. The American physicist J. Willard Gibbs (1839–1903) introduced the concept of free energy, G, which is a thermodynamic quantity. 2. This quantity gives a direct criterion for spontaneity of reaction.

GENERAL CHEMISTRY 3. As a reaction proceeds at given temperature and pressure, reactants form products , these changes in H an S, result in a change in free energy. 4. At standard state, the standard free energy change is 5. Standard Free Energies of Formation o The standard free energy of formation, G°f, of a substance is defined similarly to the standard enthalpy of formation. o That is, G°f is the free-energy change that occurs when 1 mol of substance is formed from its elements in their stablest states at 1 atm and at a specified temperature (usually 25°C). 6. G° as a Criterion for Spontaneity o The standard free-energy change G° is still a useful guide to the spontaneity of reaction in these cases. o If the reactants are at standard conditions and give products at standard conditions, the free-energy change you need to look at is G°. o The following rules are useful in judging the spontaneity of a reaction:

1.5 Interpretation of Free Energy 1. In principle, if a reaction is carried out to obtain the maximum useful work, no entropy is produced. 2. It can be shown that the maximum useful work, wmax, for a spontaneous reaction is G.

3. The term free energy comes from this result. The free-energy change is the maximum energy available, or free, to do useful work. 4. Free-Energy Change During Reaction


1.6 Relating G to the Equilibrium Constant 1. The thermodynamic equilibrium constant (k) is the equilibrium constant in which the concentrations of gases are expressed in partial pressures in atmospheres whereas the concentrations of solutes in solutions are expressed in molarities. 2. For a reaction involving only solutes in liquid solution, K is identical to Kc . 3. For a reaction involving only gas, K is equal to Kp. 4. If you want the free-energy change when reactants under nonstandard conditions are changed to products under nonstandard conditions (G), you can obtain it from the standard free-energy change G using the following equation 5. At equilibrium, G=0, Q=K

1.7 Change of Free Energy with Temperature 1. How do you find G° or K at another temperature? 2. In this method, you assume that H° and S° are essentially constant with respect to temperature. (This is only approximately true.)

3. Remember that the superscript degree sign (°) refers to substances in standard states, that is, at 1 atm and at the specified temperature. 4. Although until now this was 25°C (298 K), we now consider other temperatures. 5. Note that in general, G° depends strongly on temperature. 6. Spontaneity and Temperature Change:


Questions and Answers FIRST LAW OF THERMODYNAMICS a) 18.31 A gas is cooled and loses 82 J of heat. The gas contracts as it cools, and work done on the system equal to 29 J is exchanged with the surroundings. What are q, w, and U?

b) 18.33 What is U when 1.00 mol of liquid water vaporizes at 100° C? The heat of vaporization, H° vap, of water at 100° C is 40.66 kJ/mol.

c) 18.34 What is U for the following reaction at 25_C?


ENTROPY CHANGES a) 18.35 Chloroform, CHCl3, is a solvent and has been used as an anesthetic. The heat of vaporization of chloroform at its boiling point (61.2°C) is 29.6 kJ/mol. What is the entropy change when 1.20 mol CHCl3 vaporizes at its boiling point?

b) 18.37 The enthalpy change when liquid methanol, CH3OH, vaporizes at 25°C is 38.0 kJ/mol. What is the entropy change when mol of vapor in equilibrium with liquid condenses to liquid at 25°C? The entropy of this vapor at 25°C is 255 J/(mol.K). What is the entropy of the liquid at this temperature?

c) 18.39 Predict the sign of S°, if possible, for each of the following reactions. If you cannot predict the sign for any reaction, state why.


d) 18.41 Calculate S° for the following reactions, using standard entropy values.

e) 18.44 What is the change in entropy, S° for the reaction. See Table 18.1 for values of standard entropies. Does the entropy of the chemical system increase or decrease as you expect? Explain.


FREE-ENERGY CHANGE AND SPONTANEITY a) 18.45 Using enthalpies of formation (Appendix C), calculate H° for the following reaction at 25°C. Also calculate S° for this reaction from standard entropies at 25_C. Use these values to calculate G° for the reaction at this temperature.

b) 18.47 The free energy of formation of one mole of compound refers to a particular chemical equation. For each of the following, write that equation.

c) 18.49 Calculate the standard free energy of the following reactions at 25°C, using standard free energies of formation.


d) 18.51 On the basis of G° for each of the following reactions, decide whether the reaction is spontaneous or nonspontaneous as written. Or, if you expect an equilibrium mixture with significant amounts of both reactants and products, say so.

e) 18.53 Calculate H° and G° for the following reactions at 25°C, using thermodynamic data from Appendix C; interpret the signs of H° and G°.


MAXIMUM WORK a) 18.55 Consider the reaction of 2 mol H2(g) at 25°C and 1 atm with 1 mol O2(g) at the same temperature and pressure to produce liquid water at these conditions. If this reaction is run in a controlled way to generate work, what is the maximum useful work that can be obtained? How much entropy is produced in this case?

b) 18.57 What is the maximum work that could be obtained from 4.85 g of zinc metal in the following reaction at 25°C?

CALCULATION ON EQUILIBRIUM CONSTANTS a) 18.59 Give the expression for the thermodynamic equilibrium constant for each of the following reactions.


b) 18.60 Give the expression for the thermodynamic equilibrium constant for each of the following reactions.

c) 18.61 What is the standard free-energy change G° at 25°C for the following reaction? Obtain necessary information from Appendix C. What is the value of the thermodynamic equilibrium constant K?

d) 18.63 Calculate the standard free-energy change and the equilibrium constant Kp for the following reaction at 25°C. See Table 18.2 for data.


e) 18.65 Obtain the equilibrium constant Kc at 25°C from the free energy change for the reaction. See Appendix C for data.

FREE ENERGY AND TEMPERATURE CHANGE a) 18.67 Use data given in Tables 6.2 and 18.1 to obtain the value of Kp at 1000°C for the reaction. Carbon monoxide is known to form during combustion of carbon at high temperatures. Do the data agree with this? Explain.

GENERAL CHEMISTRY b) 18.68 Use data given in Tables 6.2 and 18.1 to obtain the value of Kp at 2100°C for the reaction. Nitric oxide is known to form in hot flames in air, which is a mixture of N2 and O2. It is present in auto exhaust from this reaction. Are the data in agreement with this result? Explain.

c) 18.69 Sodium carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3. Estimate the temperature at which NaHCO3 decomposes to products at 1 atm. See Appendix C for data.

GENERAL CHEMISTRY GENERAL PROBLEMS a) 18.71 Find the sign of S° for the reaction. The reaction is endothermic and spontaneous at 25°C. Explain the spontaneity of the reaction in terms of enthalpy and entropy changes.

b) 18.73 Estimate the value of H° for the following reaction from bond energies (Table 9.5). Is the reaction exothermic or endothermic? Note that the reaction involves the breaking of symmetrical molecules (H2 and Cl2) and the formation of a less symmetrical product (HCl). From this, would you expect S° to be positive or negative? Comment on the spontaneity of the reaction in terms of the changes in enthalpy and entropy.

c) 18.75 Acetic acid, CH3COOH, freezes at 16.6°C. The heat of fusion, Hfus, is 69.0 J/g. What is the change of entropy, S, when 1 mol of liquid acetic acid freezes to the solid?

d) 18.77 Without doing any calculations, decide what the sign of S° will be for each of the following reactions.


e) 18.79 The following equation shows how nitrogen dioxide reacts with water to produce nitric acid: Predict the sign of S° for this reaction.

f) 18.81 Acetic acid in vinegar results from the bacterial oxidation of ethanol. What is S° for this reaction? Use standard entropy values. (See Appendix C for data.)

g) 18.83 Is the following reaction spontaneous as written? Explain. Do whatever calculation is needed to answer the question.

h) 18.85 The reaction is nonspontaneous at room temperature but becomes spontaneous at a much higher temperature. What can you conclude from this about the signs of H° and S°, assuming that the enthalpy and entropy changes are not greatly affected by the temperature change? Explain your reasoning.

i) 18.87 Calculate G° at 25°C for the reaction. See Appendix C for values of G°f. What is the value of the solubility product constant, Ksp, for this reaction at 25°C?


j) 18.89 Consider the decomposition of phosgene, COCl2. Calculate H°, and S° at 25°C for this reaction. See Appendix C for data. What is G° at 25°C? Assume that H° and S° are constant with respect to a change of temperature. Now calculate G° at 800°C. Compare the two values of G°. Briefly discuss the spontaneity of the reaction at 25°C and at 800°C.

k) 18.93 For the reaction the value of G° is -702.2 kJ at 25°C. Other data are as follows:

Calculate the absolute entropy, S°, per mole of O2(g).


l) 18.95 Tin(IV) oxide can be reacted with either hydrogen or carbon to form tin and water vapor or carbon dioxide, respectively. a. See Appendix C for data. Calculate H° and S° at 25°C for the reaction of SnO2 with H2 and for the reaction of SnO2 with C. b. At what temperature will each of these processes become spontaneous? c. Industrially, which process is preferred? Why?

m) 18.97 For the decomposition of formic acid,

GENERAL CHEMISTRY H° = +29 kJ/mol at 25°C. a. Does the tendency of this reaction to proceed to a state of minimum energy favor the formation of water and carbon monoxide or formic acid? Explain. b. Does the tendency of this reaction to proceed to a state of maximum entropy favor the formation of products or reactants? Explain.

n) 18.99 For the reaction H° and G° are negative and S° is positive. a. At equilibrium, will reactants or products predominate? Why? b. Why must the reaction system be heated in order to produce copper(I) sulfide?

o) 18.101 When 1.000 g of gaseous butane, C4H10, is burned at 25°C and 1.00 atm pressure, H2O(l) and CO2(g) are formed with the evolution of 49.50 kJ of heat. Calculate the molar enthalpy of formation of butane. (Use enthalpy of formation data for H2O and CO2.) b. G°f of butane is -17.2 kJ/mol. What is G° for the combustion of 1 mol butane? c. From a and b, calculate S° for the combustion of 1 mol butane.

p) 18.103 a. Calculate K1 at 25°C for phosphoric acid:

b. Which thermodynamic factor is the most significant in accounting for the fact that phosphoric acid is a weak acid? Why?


View more...


Copyright ©2017 KUPDF Inc.