Chapter 18: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

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1 Chapter 8: hermodynamics: Spontaneous and Nonspontaneous eactions and Processes Problems: , , , Why do some reactions occur but others don t? We can answer these questions using thermodynamics, the study of energy transformations. st Law of hermodynamics: Energy is neither created nor destroyed. Essentially the law of conservation of energy While energy is converted from one form to another, the energy of the universe is constant. system: that part of the universe being studied surroundings: the rest of the universe outside the system In an exothermic reaction like methane burning, CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) the bonds formed in the products are stronger than the bonds broken in the reactants. he difference in energy between the bonds broken and the bonds formed is released to the surroundings. Heat flows from the system to the surroundings. In an endothermic reaction like the following, N 2 (g) + O 2 (g) 2 NO(g) more energy is required to break the stronger bonds in the reactants compared to the bonds formed in the reactants. he difference in energy must be absorbed from the surroundings for the reaction to occur. Heat flows from the surroundings to the system when the reaction occurs. CHEM 62: Gilbert Chapter 4 page

2 8. SPONANEOUS POCESSES AND ENOPY spontaneous process: takes place naturally, without continuous outside intervention e.g., a drop of food coloring will spread in a glass of water or once ignited H 2 gas burns with O 2 present in air. nonspontaneous process: only takes place as a result of continuous intervention e.g., electricity is required to convert water to H 2 gas and O 2 gas Keep in mind that spontaneous processes may be fast or slow. a. An explosion is fast and spontaneous. b. he process of rusting is slow and spontaneous. Consider the following spontaneous processes at room temperature: A drop of food coloring will spread in a glass of water. Methane (CH 4 ) burns in O 2 gas. Ice melts in your hand. Ammonium chloride dissolves in a test tube with water, making the test tube colder. Example: What do these processes or reactions have in common? Are they all exothermic? Are they all endothermic? CHEM 62: Gilbert Chapter 4 page 2

3 he result of all of these processes/reactions is an increase in entropy. Entropy (S): a measure of the molecular randomness or disorder of a system A thermodynamic function that describes the number of arrangements (positions and/or energy levels) available to a system at a given instant. he more ways a particular state can be achieved, the more likely or the greater the probability of finding the system in that state. For example, imagine a brand new deck of cards with the jokers removed. If the deck is divided and shuffled, the deck goes from being ordered to disordered; the more it s shuffled, the more disordered it becomes. But why? In 877 Ludwig Boltzmann, an Austrian physicist, introduced the concept of entropy to explain why cards become more disordered the more they re shuffled. He determined there were ways a deck of cards can be organized compared to only one way for it to be perfectly ordered like a new pack of cards. here are more states for a disordered deck of cards. When shuffled a deck of cards is more likely to be disordered. Nature proceeds towards the states that have the highest probabilities of existing. hus, the universe tends towards the most probable disordered states. 8.0 MICOSAES Ex. Consider the ideal gas in the following image. Explain what will happen when the valve between the bulbs is opened. How many arrangements resulting in about the same number of atoms in each of the two bulbs? How many arrangements have almost all of the atoms in one bulb and one a few or no atoms in the other? If we consider a particular arrangement of particles a microstate, then many more microstates exist where particles are evenly distributed in both bulbs and very few microstates exist where the atoms are all in one bulb or mostly in one bulb. hus, if a gas is placed in one bulb of an empty container, and the connecting valve is opened, the gas will spontaneously expand to fill the entire vessel. However, the opposite process (the gaseous particles filling a vessel all moving to one bulb leaving the other bulb completely or mostly empty), while not completely impossible, is highly improbable. CHEM 62: Gilbert Chapter 4 page 3

4 he expansion of a gas also demonstrates the idea of positional probability, a type of probability that depends on the number of arrangements in space that yield a particular state. he greater the space or volume available to particles, the higher the positional entropy. For example, positional entropy increases from solid to liquid and from liquid to gas. Solids have the smallest volume and relatively few positions available for particles. Gases have much greater volume with many positions available for gas particles. Positional Entropy and Solutions he formation of a solution when a solid dissolves (e.g., NaCl dissolving in water) or two liquids mix increases the positional entropy of a system. he overall volume available to each component increases. hus, the number of positions available to the particles in the system increases. CHEM 62: Gilbert Chapter 4 page 4

5 Entropy and Energy States Positional entropy results from the different energy levels available to a system. o understand entropy we must recognize how atoms and molecules behave at the molecular level. For example, the kinetic energy of a molecule can be in any of the following forms: translational: motion through space rotational: motion about its center of mass vibrational: stretching, compression, bending and twisting of chemical bonds he more complex the molecule, the more energy levels available to it. he higher its entropy. Compare the limited vibrational energy available for the NO molecule compared to the various forms of vibrational energy available for the NO 2 molecule below. Consider an O 2 molecule and the different kinds of motion available to it: CHEM 62: Gilbert Chapter 4 page 5

6 Quantum mechanics teaches that energy at the atomic level is not continuous but discrete. Only certain energy levels are available for electrons within an atom. he distribution of electrons in those energy levels is given by an electron configuration. In the same way, the motion of atoms and molecules or their energy is also quantized. ranslational energy levels are very close together. he quantized difference between energy levels is so small that for temperatures above absolute zero, especially for room temperature and above, the energy levels for translational energy form a continuum. Very little energy is required for a gaseous atom or molecule to move through space (i.e., experience translational motion). Consider where vibrational and rotational energy levels fall on the Electromagnetic Spectrum: he rotational and vibrational energy levels for atoms and molecules are also quantized. An O 2 molecule absorbing energy in the microwave region rotational energy An O 2 molecule absorbing more energy in the infrared region vibrational energy he difference between rotational and vibrational energy levels is also small but greater than the difference between different translational energy levels. At low temperatures, O 2 molecules may not have enough energy to rotate and vibrate. At higher temperature like room temperature O 2 molecules are rotating, vibrating, and moving in space. Note that at a given instant, an O 2 molecule occupies a specific energy state or energy level. At lower temperatures, only certain energy states are available or accessible. As temperature increases, more energy states become accessible. he O 2 molecules moves in space and also rotates and vibrates. It can even gain enough energy for its electrons to be excited to higher energy levels. he molecule reaches its first electronic excited state. At very high temperatures, the molecule vibrates enough to dissociate into two O atoms. his is the dissociation energy for the O 2 molecule. CHEM 62: Gilbert Chapter 4 page 6

7 We can represent these different energy states or levels in the following way: Note that rotational energy levels (orange) are much closer together than the vibrational energy levels (green). Each energy level corresponds to an O 2 molecule at a given temperature and state.. An O 2 molecule may be in the electronic ground state (all electrons in the lowest energy levels) with only certain vibrational modes available (v ). 2. An O 2 molecule may be in an electronic excited state (with one or more electrons in a higher energy level even if the lowest levels are not full) with more vibrational modes available (v 2 ). Example: Consider the differences between an O 2 molecule and an O 3 molecule. a. Explain the difference between v 3 for the Electronic ground state O 2 molecule and v 3 for the Electronic excited state O 2 molecule. b. Would the O 3 molecule have more or fewer vibrational energy levels? Explain why. CHEM 62: Gilbert Chapter 4 page 7

8 Now imagine a system consisting of a mole of O 2 molecules at the same temperature. Not only do the molecules have these available energy levels, but now they re also colliding with one another, so all of these energy levels are superimposed on one another. More energy levels are available for all of the O 2 molecules in the system. More microstates are available for the system. While the average speed for a molecule at a given temperature is constant, many gas molecules can move at a range of speeds characterized by Maxwell-Boltzmann distribution curves. Consider H 2 at two different temperatures. Consider the distribution of molecular speeds for equal numbers of H 2 gas molecules at (a) 300K and (b) 373K at the right. Instead of using bars, a curve shows the distribution of speeds. hese curves are generally called Maxwell-Boltzmann distribution curves, as Maxwell was the first to deduce the pattern theoretically and Ludwig Boltzmann was the first to effectively substantiate them. Next, consider O 2 at three different temperatures. Consider the Maxwell-Boltzmann distribution for O 2 at 300K, 000K, and 2000K. Again these curves actually correspond to the distribution of speeds that would be given by the sets of bars indicating the number of molecules with each speed at each temperature. Consider many O 2 molecules moving and colliding at a given temperature. hey experience translational motion while also rotating and vibrating. Because kinetic energy is KE= ½ mv 2, a particle s velocity (speed and directionality) is directly related to its kinetic energy. he different energy states (the combination of the translational, rotational, and vibrational energies) available to an O 2 molecule also have a Maxwell-Boltzmann distribution. he higher the system s temperature, the more energy states available to the system. More energy states available more microstates available to a system. he higher the entropy of the system CHEM 62: Gilbert Chapter 4 page 8

9 Ex. : Explain which one of the following has the greater entropy: mole of H 2 O at room temperature or mole of O 2 gas at room temperature. 8.2 HEMODYNAMIC ENOPY Isothermal and Nonisothermial Processes Isothermal processes occur at a constant temperature (e.g. melting and boiling). Entropy is a state function (or state property), so it depends only on the present state. he change in state functions depends only on the initial state and the final state, not the pathway from the initial to the final state. hus, unlike reaction rates that depend on the reaction pathway (i.e., whether or not a catalyst is used), state functions like entropy are the same regardless of pathway. he entropy change (ΔS) can be determined using only initial and final states: ΔS = S final S initial Example: Check all of the following for which the entropy change (ΔS) is positive: o a. Dry ice, CO 2 (s), undergoes sublimation at room temperature. o b. Sugar dissolves in water. o c. Water freezes at 0 C. o d. A supersaturated solution of sodium acetate recrystallizes. Entropy Changes in Chemical eactions with Gases he change in positional entropy is dominated by the relative number of gaseous molecules in a chemical reaction. If a reaction increases the number of gaseous molecules, ΔS is positive. If a reaction decreases the number of gaseous molecules, ΔS is negative. Example: Check all of the following for which the entropy change (ΔS) is positive: o a. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) o b. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) o c. (NH 4 ) 2 CO 3 (s) 2 NH 3 (g) + CO 2 (g) + H 2 O(g) o d. N 2 O 4 (g) 2 NO 2 (g) CHEM 62: Gilbert Chapter 4 page 9

10 Spontaneous Combustion: Early studies in thermodynamics led scientists to propose that a process would be spontaneous if it were exothermic. A system tends to undergo changes that lower its potential energy (i.e., make it more stable). Exothermicity is a driving force for spontaneous reactions and processes. Example: Does ice melt at room temperature? Is this process endothermic or exothermic? hus, some endothermic processes also occur spontaneously. ENOPY AND HE SECOND LAW OF HEMODYNAMICS 2 nd Law of hermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe. Essentially, the entropy of the universe is increasing. Increasing the entropy of the universe (S univ ) is the driving force for a spontaneous process. ΔS univ = ΔS sys + ΔS surr where ΔS sys and ΔS surr represent the changes in the entropy of the system and surroundings, respectively. hus, if ΔS universe is positive for a process or reaction, it is spontaneous. 4.3 ABSOLUE ENOPY, HE 3 rd LAW OF HEMODYNAMICS, AND SUCUE In Chapter 5 on hermochemistry, because enthalpy (H) is also a state function, the enthalpy change (ΔH) was determined using the initial and final states for a reaction. he enthalpy change (ΔH) for a reaction was calculated using the enthalpy change (ΔH) for the formation of reactants and products from their elements in their standard states. However, since enthalpy is heat flow, there is no inherent heat flow of a substance, only the heat flow associated with forming the substance from its component elements. he absolute enthalpy of a substance cannot be determined. CHEM 62: Gilbert Chapter 4 page 0

11 he 3 rd Law of hermodynamics: At absolute zero (= 0 K), the entropy is zero (S=0) for a perfect crystal (in which all particles are perfectly aligned, with no gaps, in a crystal structure). Consider the images below showing a perfect crystal with S=0 versus imperfect crystals with higher entropy (S>0). Example: Explain why a crystal would become more disordered as temperature increases. By setting a value of zero for entropy at absolute zero to be the reference point, scientists can estimate the entropy value for pure substances at 25 C. he absolute entropy (S ) of a compound or element can be determined. If we consider that entropy is a thermodynamic function that describes the number of microstates (positions and energy levels) available to a system in a given state. he more microstates available to an atom or molecule, the higher its entropy. he more microstates available to a system with those atoms or molecules he higher the entropy of the system Consider again that the kinetic energy of a molecule can be in any of the following forms: translational (motion through space), rotational (motion about its center of mass), and vibrational (stretching, compression, bending and twisting of chemical bonds) CHEM 62: Gilbert Chapter 4 page

12 he more complex the molecule, the more energy levels available to it. he higher its entropy. he additional vibrational energy levels in NO 2 molecules increase its entropy. S =2 J/mol K for NO and S =240 J/mol K for NO 2 Ex. he absolute entropy would be higher for mole of nitrogen or mole of neon at 25 C? Explain. Ex. 2 Use the element symbols and chemical formulas to rank mole of each of the following substances in terms of increasing entropy at 25 C: C diamond, water, ice, oxygen gas, ammonia, mercury, and helium. < < < < < < lowest entropy highest entropy Standard Entropy Values (S ) are given for substances at 25 C and atm. superscript denotes at P= atm and =25 C Just like the calculation of ΔH from standard enthalpies of formation, the standard entropy change (ΔS ) can be calculated for a general reaction, a A + b B c C + d D where a,b,c,d = stoichiometric coefficients ΔS = Σ n S (products) Σ m S (reactants) = [c S (C) + d S (D)] [a S (A) + b S (B)] CHEM 62: Gilbert Chapter 4 page 2

13 C 2 H 5 OH(l) O 2 (g) H 2 (g) CO 2 (g) H 2 O(g) H 2 O(l) N 2 (g) NO(g) 6 J/mol K 205 J/mol K 3 J/mol K 24 J/mol K 89 J/mol K 70 J/mol K 92 J/mol K 2 J/mol K Ex. : For each of the following, calculate the standard entropy change (ΔS ) and the standard free energy change (ΔG ), then indicate if the reaction is spontaneous at 25 C. If the reaction is not spontaneous at 25 C, calculate at what temperatures the reaction would be spontaneous. (Assume the given standard entropy values, S, are valid over the required temperature ranges for these reactions.) a. N 2 (g) + O 2 (g) 2 NO(g) ΔH = kj b. 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ΔH = 572 kj CHEM 62: Gilbert Chapter 4 page 3

14 c. C 2 H 5 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) ΔH = 79 kj 8.6 EMPEAUE AND SPONANEIY Consider the vaporization of mole of water to steam, H 2 O(l) H 2 O(g) a. If mole of water is the system, is ΔS sys for the vaporization of water positive or negative? Explain why. Consider that the entropy changes for the surroundings (ΔS surr ) are determined primarily by heat flow. We can determine the heat flow and hence, the ΔS surr, if we know whether the reaction is endothermic or exothermic. b. Is the process of water vaporizing endothermic or exothermic? Explain how heat flows to/from the system and the surroundings. c. Does the heat flow described in part b increase or decrease the entropy of the particles in the surroundings? CHEM 62: Gilbert Chapter 4 page 4

15 d. Given the ΔS sys from part a and the ΔS surr from part b, is the vaporization of water spontaneous or not i.e., is ΔS univ positive or negative? From practical experience, does water always vaporize? From practical experience, we know water only vaporizes spontaneously for temperatures above water s boiling point (00 C) at normal atmospheric pressure (about atm). emperature is another key factor in determining the significance of ΔS surr and the overall sign for ΔS univ i.e., whether or not a process or reaction is spontaneous. 8.5 FEE ENEGY (G) hus, two driving forces causing chemical reactions to occur are:. he formation of more stable (low-energy) products from less stable (high-energy) reactants in exothermic reactions (ΔH<0). 2. he formation of products that have more entropy than the reactants (ΔS>0). Determining the Entropy Change in the Universe he American mathematical physicist J. Willard Gibbs ( ) is credited with establishing the mathematical foundation of modern thermodynamics. He developed a way to use ΔH sys and ΔS sys to predict whether or not a reaction would be spontaneous at constant temperature and pressure. Gibbs defined a new state function, free energy (G), which is the energy available to do useful work i.e., that portion of energy that can be dispersed into the universe. We can express ΔS surr in terms of the enthalpy change for the system, ΔH, as follows: ΔS surr = ΔH with ΔH in joules (J), and absolute in Kelvins Note, because ΔH is the enthalpy change for the system, the sign convention is critical. In exothermic reactions, the system releases heat to the surroundings. Because ΔH is negative ΔS surr is positive. In endothermic reactions, the system absorbs heat from the surroundings. Because ΔH is positive ΔS surr is negative. CHEM 62: Gilbert Chapter 4 page 5

16 Ex. a. Calculate ΔS surr and ΔS univ for the vaporization of.00 mole of water at 25 C. (ΔH=40.7 kj/mol, and ΔS sys =9 J/K for mole of water vaporizing.) b. Calculate ΔS surr and ΔS univ for the vaporization of.00 mole of water at 25 C. hus, endothermic reactions will only be spontaneous at high temperatures. Exothermicity as a driving force is only relevant at low temperatures. hus, if ΔS surr = ΔH, then ΔS univ is ΔS univ = ΔS sys + ΔS surr = ΔS sys ΔH sys hen multiplying both sides by yields: ΔS univ = ΔS sys ΔH sys = (ΔH sys ΔS sys ) hen multiplying both sides by - yields: his leads to Gibb s free energy (G), and Gibb s free energy change (ΔG): ΔS univ = ΔH sys ΔS sys G = H S ΔG = ΔS univ Gibb s equation: ΔG = ΔH ΔS where the sys subscript can be dropped since each term is based on the system. Because is always positive and the sign of ΔS univ indicates if a process/reaction is spontaneous, for any process/reaction occurring at a constant temperature, the sign of ΔG indicates if the process/reaction is spontaneous. If ΔS univ >0, the process is spontaneous ΔG = (+) = ve (a negative value). If ΔS univ <0, the process is nonspontaneous ΔG = ( ) = +ve (a positive value). If ΔG<0 (negative), the reaction/process is spontaneous. If ΔG>0 (positive), the reaction/process is nonspontaneous (or the reverse reaction/process is spontaneous). If ΔG=0, the reaction/process is at equilibrium (neither the forward nor the reverse process is favored). CHEM 62: Gilbert Chapter 4 page 6

17 Combining this with what we determined previously yields the following temperaturedependence for the Gibb s equation and spontaneous reactions: ΔH < 0 ΔS > 0 Spontaneous at all temperatures ΔH < 0 ΔS < 0 Spontaneous at low temperatures ΔS ΔH > 0 ΔS > 0 Spontaneous at high temperatures ΔH > 0 ΔS < 0 Nonspontaneous ΔH 8.5 FEE ENEGY AND FEE ENEGY CHANGE he Meaning of Free Energy Let s consider that the energy released by a spontaneous chemical reaction at constant temperature and pressure is the maximum amount of energy available to do useful work. i.e., this is the free energy from the chemical reaction. earranging ΔG = ΔH ΔS and solving for ΔH yields: ΔH = ΔG + ΔS emember that ΔH results from bonds being broken and made in chemical reactions. he ΔG part can be used for motion, light, heat, etc. ie., to do useful work. Why ΔG is called free energy and is considered the energy to do useful work. However, keep in mind that the work obtained from ΔG is never 00% efficient. Next, the ΔS part is the temperature-dependent change in entropy, that part of the energy change that simply increases the entropy of the universe It is not usable Calculating Free Energy Changes Unlike the standard enthalpy change (ΔH ), which is heat flow accompanying a reaction and can be determined experimentally using a calorimeter, the standard free energy change (ΔG ) cannot be measured directly. However, ΔG can be determined from other measurements and values. his is important because knowing ΔG for different conditions will allow us to compare the relative tendency for reactions to occur. CHEM 62: Gilbert Chapter 4 page 7

18 Note: ΔG f =0 for elements in their standard states (their naturally occurring form at 25 C) just like H =0 for elements in their standard states. Δ f Consider again the following definitions for standard states: Definitions of standard state (or reference form). A gasesous substance with P= atm 2. An aqueous solution with a concentration of M at a pressure of atm 3. Pure liquids and solids 4. he most stable form of elements at atm and 25 C standard free energy of formation ( ΔG f ) of a substance is the free-energy change for the formation of mole of the substance from its elements in their standard states, hus, the ΔG f of a substance is a measure of its thermodynamic stability with respect to its constituent elements. Substances with negative values for ΔG f are stable and do not decompose into their constituent elements. e.g. water s ΔG f = kj/mol while CO 2 s ΔG f = kj/mol Consider that the standard free energy change (ΔG ) is the free-energy change for a reaction when the reactants in their standard states are converted to the products in their standard states. hus, we can use a table of standard free energy of formation values ( ΔG f ) for reactants and products to calculate standard free energy changes (ΔG ) for various chemical reactions, just like we did for ΔH and ΔS. hus, the standard free energy change (ΔG ) can be calculated as ΔG = (sum of the ΔG f for all products) (sum of the ΔG f for all reactants) so for a general reaction, a A + b B c C + d D where a,b,c,d = stoichiometric coefficients ΔG = Σ n ΔG f (products) Σ m ΔG (reactants) = [c ΔG f (C) + d ΔG f (D)] [a ΔG f (A) + b ΔG f (B)] CHEM 62: Gilbert Chapter 4 page 8

19 Ex. : CaCl 2 dissolves in water as follows: CaCl 2 (s) Ca 2+ (aq) + 2 Cl (aq) a. Calculate ΔG for this process at 25 C using the following: CaCl 2 (s) s ΔG = 748. kj/mol; Ca +2 (aq) s ΔG = kj/mol; Cl (aq) s ΔG = 3.2 kj/mol. f f f b. Does this occur spontaneously? Explain. Ex. 2: Calculate the free energy change (ΔG ) for the following reactions using the standard free energy of formation values given below, then indicate if the reaction is spontaneous. ΔG f (NO)=+90.3 kj/mol and ΔG f (NO 2 )=+33.2 kj/mol a. N 2 (g) + O 2 (g) 2 NO(g) b. 2 NO(g) + O 2 (g) 2 NO 2 (g) CHEM 62: Gilbert Chapter 4 page 9

20 8.7 FEE ENEGY AND CHEMICAL EQUILIBIUM he standard free energy change (ΔG ) only applies to standard state conditions, where temperature is 25 C and all gas pressures are atm. hus, ΔG only allows us to predict if a reaction is spontaneous at these conditions. However, many reactions do not occur at these conditions A more general free energy change at nonstandard conditions (ΔG) allows us to predict if a reaction/process is spontaneous for all other conditions: ΔG = ΔG + ln Q where =8.345 J/mol K, =absolute temperature in Kelvins, and Q=reaction quotient indicating the state of a system at a given instant. Ex. 3 Consider again the vaporization of mole of water: H 2 O(l) H 2 O(g) ΔH =+40.7 kj a. Calculate the standard free energy (ΔG ) for the process given ΔS = 8.9 J/mol K, then explain whether or not the vaporization of water spontaneous under standard state conditions? hen how do we explain the presence of water vapor in the atmosphere at 25 C? b. Write the expression for the reaction quotient (Q) for this process then calculate the free energy for the process when the partial pressure of water vapor in the air is atm at 25 C. CHEM 62: Gilbert Chapter 4 page 20

21 c. Calculate the free energy for the process when the partial pressure of water vapor in the air is atm at 95 C. d. Compare the free energy values for the vaporization of water at 25 C and 95 C and explain the difference. What does the difference indicate about the spontaneity of the process as temperature increases? Is this expected? Ex. 4: a. Use the standard free energy change (ΔG ) you calculated on Ex. 2 part b on p. 9 to calculate ΔG for the reaction below at 95 C when the partial pressure of each reactant is 5.00 atm and the partial pressure of the NO 2 is 0.50 atm. 2 NO(g) + O 2 (g) 2 NO 2 (g) b. Is the reaction more or less spontaneous under these conditions? Explain if this is what you would expect based on Le Châtelier s Principle and the change from standard state conditions. CHEM 62: Gilbert Chapter 4 page 2

22 he Meaning of ΔG for a Chemical eaction Do the results from Ex. 4 on the previous page indicate that given partial pressures of 5.00 atm for NO and O 2 and a partial pressure of 0.50 atm for NO 2 at 95 C will insure the reaction will go to completion? Not necessarily. Although the reaction is more spontaneous at these conditions, indicating reactants are more likely to be converted to products, the lowest free energy for the system may not be one where only products are present. he system achieves the lowest possible free energy by going to equilibrium, not completion. he plot above shows ΔG for the following reaction: N 2 O 4 (g) 2 NO 2 (g) he upper plot shows what ΔG for a system consisting of: Point () only reactants, N 2 O 4,Point (2) only products, NO 2, and Point (3) the equilibrium mixture of N 2 O 4 and NO 2. hus, a reaction is spontaneous until the system achieves equilibrium. FEE ENEGY AND EQUILIBIUM Previously we defined equilibrium to be when the forward and reverse reactions are occurring at equal rates. We can now consider equilibrium from a thermodynamic point of view. he equilibrium point occurs at the lowest value of free energy available to a reaction system. CHEM 62: Gilbert Chapter 4 page 22

23 Consider the following hypothetical reaction: A(g) B(g) he figure below shows the stages when.00 mole of A is placed in a vessel with P=2.00 atm. (a) he initial free energy of A (G A ) is much higher than the initial free energy of B (G B ). (b) G A decreases and G B increases as reactant A is converted to product B. (c) Given enough time, the system reaches equilibrium where G A =G B. he plot of G versus Fraction of A reacted on the left shows G for the system changing as it approaches equilibrium. he second set shows the stages when.00 mole of B is placed in a vessel with P=2.00 atm. (d) he initial free energy of A (G A ) is much lower than the initial free energy of B (G B ). (e) G A increases and G B decreases as reactant B is converted to product A. (f) Given enough time, the system reaches equilibrium where G A =G B. Notice that in both examples where total number of moles in the container is.00 and the total pressure is 2.00 atm, and they both reach the same equilibrium point regardless of the initial amounts consisting of only reactants, only products, or a mixture of reactants and products. CHEM 62: Gilbert Chapter 4 page 23

24 Starting with ΔG = ΔG + ln Q we can solve for ΔG at equilibrium. (Hint: Consider the value for ΔG at equilibrium and Q=K at equilibrium.) hus, at equilibrium: ΔG = Ex. Consider again the vaporization of mole of water : H 2 O(l) H 2 O(g) ΔH =+40.7 kj a. Solve for the equilibrium constant for this process at 25 C using the ΔG you calculated on Ex. 3a on p. 20. b. Note that the vapor pressure for water is about 0.03 atm at 25 C. What is the meaning of vapor pressure of water? Explain how it is related to the K calculated in part a. c. Given the formula for K, explain how the equilibrium constant changes as temperature increases for this process? Is this expected? d. Explain the relationship between the standard free energy change and the equilibrium constant K for an endothermic process like the vaporization of water. CHEM 62: Gilbert Chapter 4 page 24

25 Ex. 2 Consider the following reaction: CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) a. Calculate ΔG for the reaction at 25 C using the standard values below: Substance CO(g) CO 2 (g) H 2 (g) H 2 O(g) ΔH f (kj/mol) ΔS (J/mol K) b. Calculate the equilibrium constant for this reaction. c. Calculate the free energy change for the reaction at 345 C where the partial pressure of the each reactant is 2.50 atm and the partial pressure of each product is.50 atm. Is the reaction spontaneous at these conditions? CHEM 62: Gilbert Chapter 4 page 25

26 8.8 INFLUENCE OF EMPEAUE ON EQUILIBIUM CONSANS Combining the following equations: ΔG = ln K and ΔG = ΔH ΔS yields ln K = ΔH ΔS and dividing both sides by yields which can be written in the form of a line equation: H ΔS ln K = -Δ + - ΔH ln K = ΔS + y = m x + b - ΔH his plot of ln K versus is a van t Hoff plot, where slope= and y-intercept= ΔS. Example: Consider the van t plot for the reaction, CO(g) + H 2 O(g) CO 2 (g) + H 2 (g), for temperatures ranging from 478K to 643K. Calculate ΔH and ΔS for the reaction using the data above. Compare these values of ΔH and ΔS with those calculated based on ΔH f and S values on the previous page. ln K van't Hoff plot for CO reacting with Steam y = 480.x E-03.60E-03.70E-03.80E-03.90E E E E-03 / (/K) CHEM 62: Gilbert Chapter 4 page 26

27 CHEM 62: Gilbert Chapter 4 page 27 Next, consider equilibrium constants K and K 2 at temperatures and 2, respectively: ΔS ΔH - ln K + = and ΔS ΔH - ln K = Now, solve for ln K 2 ln K = + ΔS ΔH ΔS ΔH - = ΔH - 2 ΔH - = - ΔH - 2 van t Hoff Equation: = ΔH - K K ln Example: Again consider the following reaction, CO(g) + H 2 O(g) CO 2 (g) + H 2 (g), a. Calculate the equilibrium constant for the reaction at 725K using the equilibrium constant for the reaction at 25 C you calculated using ΔH f values on page 25. b. How does the equilibrium constant at 725K compare to the equilibrium constant at 25 C? Is this expected given the nature of the reaction? Explain.

H 2 (g) + ½ O 2 (g) H 2 O(l) H o f [NO(g)] = 90.2 kj/mol; H o f [H 2 O(g)] = kj/mol H o f [NH 3 (g)] = kj/mol; H o f [O 2 (g)] =?

H 2 (g) + ½ O 2 (g) H 2 O(l) H o f [NO(g)] = 90.2 kj/mol; H o f [H 2 O(g)] = kj/mol H o f [NH 3 (g)] = kj/mol; H o f [O 2 (g)] =? Chapter 16 Thermodynamics GCC CHM152 Thermodynamics You are responsible for Thermo concepts from CHM 151. You may want to review Chapter 8, specifically sections 2, 5, 6, 7, 9, and 10 (except work ). Thermodynamics:

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