Part 1: (18 points) Define or explain three out of the 6 terms or phrases, below. Limit definitions to 200 words or less.

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Frank Wilkerson
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1 Chemistry 452/ July 24 Midterm Examination Key rofessor G. Drobny Boltzmann s constant=k B =1.38x1-23 J/K=R/N A, where N A is Avagadro s number and R is the Universal Gas Constant. Universal gas constant=r=8.31j/mole-k=.821l-atm/mole-k 1 Joule=1J=1 Nt-m=1kg-m 2 /s 2 11J=1 L-atm. art 1: (18 points) Define or explain three out of the 6 terms or phrases, below. Limit definitions to 2 words or less. a) Carnot cycle. Define and give an important result of this cycle. Answer: A hypothetical heat engine, consisting of a four step thermodynamic cycle. The cycle is composed of a reversible isothermal expansion, followed by a reversible adiabatic expansion, then a reversible isothermal contraction and a reversible adiabatic contraction back to the initial state. An important result is the limiting TH TL efficiency for which heat engines can convert heat to work ε = where T H is TH the temperature of the high T reservoir and T L is the temperature of the low T reservoir. (77 words) b) Maxwell Relations. Give mathematical expressions and explain how these equations are used in thermodynamics. T = from de = TdS dv V S S V T V = from dh = TdS + Vd S S S = from da = SdT dv V T T V S V = from dg = SdT + Vd T T These equations are used to convert all thermodynamic derivatives into functions of, T, V, α, κ, or c. E αt Example: = T = V T T V κ c) Define state function as the term is used in the field of thermodynamics. Give two examples. Answer: A property of a system at equilibrium that is dependent on the equilibrium values of the state variables i.e.,v,t,n. When a state function changes as a result of a change in,v, or T, the change is dependent only on the initial state of the system and the final state of the system, and not on the path followed between the intial and

2 final states. The internal energy E, enthalpy H, entropy S, and free energies A and G are state functions. d) Exact and inexact differentials. Define the terms as they are used in mathematics. Explain how exact and inexact differentials are used in thermodynamics. How are their properties physically relevant? Answer: A differential expresses infinitesimal change in one or more dimensions. For M xydx, + N xydy,. If the two dimensions a differential has the form ( ) ( ) Z Z differential is exact it can be expressed as dz = dx + dy x y and so an exact y x M N differential fulfills Euler s criterion i.e. y =. The line integrals of exact x x y differentials are independent of path, and this property is important for state functions, which can be represented as exact differentials. Therefore E, H, S, G, and A are exact differentials. Inexact differentials are path dependent when integrated and in thermodynamics include work w and heat q. e) Explain the difference between the Helmholtz and Gibbs free energies. In particular relate these two quantities to reversible work. Answer: The definitions of the Helmholtz free energy A and Gibbs free energy G are A=E-TS and G=H-TS. At equilibrium in a closed system doing only -V work, at constant V and T the Helmholtz free energy is minimum, whereas at constant and T and Gibbs free energy is minimum. da = d ( E TS) ) SdT + dw where the equality holds for reversible processes and the inequality holds for irreversibly processes. So at constant T and for reversible work, da=dw. Similarly dg = d( H TS) = d( E TS + V) = da+ d( V) SdT + dw + dv + Vd If work is the sum of reversible V work and other types of reversible work, dg = SdT + dw + dv + Vd = SdT dv + dwother + dv + Vd = SdT + Vd + dwother At constant and T and for reversible processes the change in G yields the non-v work. f) Define the term reversible as it is used in the field of thermodynamics. Solution: A process whereby a system is continually and infinitesimally displaced from equilibrium, then allowed to re-establish equilibrium, before being displaced again. The process can be changed, i.e. reversed, by an infinitesimal change in the applied force (36 words). art 2: (2 points) Answer two out of the four questions below. Answers should be no longer than 2 words.

3 a) From the point of view of the First Law of Thermodynamics, will the temperature of water that falls over a waterfall increase, decrease, or stay the same? Explain. Solution: From water falling E = mg h h<. To conserve energy potential energy is converted to heat. The temperature increases. b) Is a man who transforms into work through muscular effort the energy of the food he eats and the air he breaths a more or less efficient machine than a reversible steam engine, whose boiler temperature is equal to the man s normal body temperature, and whose condenser temperature is the temperature of the surrounding air? Explain your answer. Solution: Human is more efficient. Suppose T H =31K. T L =3K Such a reversible steam engine would have an efficiency of only about 3% because TH TL 31 3 ε = = =.32 TH 31 But this efficiency equation only governs the work production of heat engines. A human is not generating work from thermal gradients and the efficiency of the human is not governed by the same constraints. Note from rofessor: work production in humans, as measured by the efficiency of production of AT from glycolysis, Krebs cycle, and terminal electron transfer, etc. is about 45%. Work is produced by exploiting a chemical potential gradient in living organisms, not a thermal gradient. c).g. Tait was a contemporary of Lord Kelvin. In a lecture in 1876 Tait asked Why is it that if I have a quantity of work or potential energy I can convert the whole of it, if I please, into heat; but when I have got it converted into heat, I cannot convert the heat back again, except in part, into work or potential energy? Answer Tait s question. Solution: With heating comes increased thermal motion, the random nature of which cannot be reversed completely to produce work. To do so would require a large negative entropy change in the universe forbidden by the second law. There is no prohibition to going in the opposite direction, i.e. completely converting work to heat. Such a process increases entropy. d) Early in the 19 th century scientists and engineers made great efforts to quantify the functioning of heat engines. These efforts eventually produced the field of thermodynamics. An early theory proposed that the amount of heat, also called caloric was fixed in the universe, and that if the amount of caloric in a body increased or decreased, it must be as a result of an exchange of caloric with the surroundings. Comment on the validity of this caloric theory. In your answer, make specific reference to the relevant modern forms of the laws of thermodynamics.

4 Solution: The problem with the caloric idea is that heat is not conserved. Energy is conserved and energy is the sum of heat and work. Heat and work are different forms of energy which are exchanged between the system and the surroundings in order to achieve energy conservation. rofessor s Note: people became suspicious of the caloric theory in the early 19 th century after observing the behavior of artillery after loaded versus unloaded discharge. A loaded cannon heats up much less than an unloaded cannon. In the loaded cannon the energy released by the explosion of gun powder partly propels the cannon ball (work) and the rest of the energy is discharged as heat. In the unloaded cannon almost all of the energy is discharged as heat. You can damage a black powder weapon by repeatedly firing it unloaded. art 3: (3 points) erform two out of the four calculations given below. a) Calculate the work done when one mole of an ideal monatomic gas at an initial temperature of T=298K and an initial pressure of 1 atm expand isothermally and reversibly until the final pressure is 1 atm. Solution: At constant temperature V=constant. Therefore if the ratio of the final to the initial is 1/1, the ratio of the volumes is 1/1. Then V final w= nrtln = ( 1mole)( 8.31 J / mole K)( 298K) ln ( 1) = 572J Vinitial b) Suppose five moles of an ideal monatomic gas at an initial volume of 12.2L and temperature of 298K expand adiabatically and reversibly until the volume has reached 48.8L. Calculate the temperature change T, E, and the work done. C Solution: For a reversible adiabatic change V γ 5R = cons tan t., where γ = = 2 =. C R V 3 1 Because the gas is ideal (V=nRT) it follows TV γ = cons tan t. So γ 1 2/3 γ 1 γ 1 V L TV 1 1 = TV 2 2 T2 = T1 = ( 298K) = 119.4K V2 48.8L T = = 178.6K E = w = nc 3 V T = ( 5moles)( R 2 )( 178.6K ) = 11,131J c) Oxygen reacts with solid glycylglycine C4H8N 2O3 to form urea CH4N 2 O, carbon dioxide and water: 3O2( g) + C4H8N2O3( s) CH4N2O( s) + 3CO2( g) + 2H2O( ). At T=298K and 1 atm the standard enthalpy change is H = kj / mole. The heat capacities for the reactants and products are:

5 c bgh c 4 2bgh c 2b gh c 2b gh c 2 b g C C H N O s = J / mole K; C CH N O s = J / mole K; C O g = J / mole K; C CO g = J / mole K; C H O = 753. J / mole K Calculate the standard enthalpy change H at T=33K. b g b g c b gh c b gh c b gh b g b g b g b g hj mole K gj mole K J mole K Solution: H = H + C T T T2 T1 2 1 = C C urea C CO g C H O C O g C glycine c b = / = / = / H = H + C T T T2 T1 ( 2 1) ( )( ) = kj / mole J / mole K 33K 298K = kj / mole kj / mole = kj / mole h d) Calculate the entropy change when 5 g each of oxygen O 2, nitrogen N 2, and hydrogen H 2 are mixed at 1 atm pressure and 273K. Solution: 5gm 5gm moleso2 = = 1.56 moles; moles N2 = = 1.79moles 32 gm / mole 28 gm / mole 5gm moles H2 = = 25moles 2 gm / mole total = = 28.35moles χo = =.55; χ.63;.88 2 N = = χ 2 H = = S = R χiln χi = 8.31 J / moles K.55 ln ln ln.88 = 3.71 J / K i ( ) ( ) ( ) ( ) ( ) ( ) ( ) art 4 (32 points): erform one of the two multi-step calculations. a) Supercooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically unstable and tends to freeze into ice irreversibly. Freezing rain is the conversion of supercooled water to ice. Suppose we have two moles of supercooled water turning into ice at T=263K and at a constant pressure of 1 atm. Calculate the enthalpy change H and entropy change S of the water, the surroundings and the universe. The following data are useful: For water freezing at T=273K, H=-61 J/mole The heat capacity for liquid water is C =75.3 J/K-mole The heat capacity for ice is C =37.7 J/K-mole Assume all heat capacities are constant between 263K and 273K.

6 Solution: All steps in the three step cycle will be reversible HO 2 ( l,263 K) HO 2 ( s,263k) HO l,273 K HO s,273k 2 ( ) 2 ( ) Step 1: HO( l,263 K) HO( l,273k) 2 2 Tf 273 S1 = nc ln = ( 2moles)( 75.3 J / K) ln Ti 263 = 5.6 J / K 3 H1 = nc T = ( 2moles)( 75.3 J / Ki mole)( 1K) = J Step2: HO 2 ( l,273 ) HO 2 ( s,273) H freeze 61 J / mole S2 = n = ( 2moles) = 44 J / K Tfreeze 273K 4 H 2 = n H freeze = ( 2moles)( 61 J / mole) = J Step 3: HO s,273 K HO s,263k ( ) ( ) 2 2 Tf 263 S3 = nc ln = ( 2moles)( 37.7 J / K) ln = 2.8 J / K Ti 273 H3 = nc T = ( 2moles)( 37.7 J / Ki mole)( 1K) = 754J sys ( ) = ( ) = S = S1+ S2 + S3 = J / K = 41.2 J / K H J 11,254J sys Hsurr 11,J Hsurr = Hsys = 11, 254J Ssurr = = = 42.8 J / K T 263K Huniv = Hsys + Hsurr = S = S + S = 41.2 J / K J / K = 1.6 J / K univ sys surr b) Consider the formation of glucose from carbon dioxide and water, i.e. the reaction 6CO g + 6H O l C H O s + 6O g. of the photosynthetic process: ( ) ( ) ( ) ( ) surr Calculate the entropy and enthalpy changes for this chemical system at T=298K Solution: H = 6 H O g + H C H O s 6 H CO g 6 H H O l ( ( )) ( ) ( ) ( ) ( ) ( ) ( ) ( ( )) ( ( )) f 2 f f 2 f 2 = kJ kJ kJ = kJ

7 6 ( 2( )) ( ) ( ) ( ) ( ) ( ) ( ) ( ( )) S = S O g + S C H O s S CO S H O l = J / K J / K J / K = 259 J / K Calculate the free energy change at T=298K. Based on your answer is the production of glucose spontaneous at T=298? Explain. G = H T S = kJ ( 298 K)(.259 kj / K) = kJ G > so the reaction is not spontaneous. Repeat the entropy and enthalpy calculations for T=33K. Is the free energy change for glucose production via photosynthesis increasing or decreasing as the temperature increases? Explain. H ( T) = H ( 298K) + C ( T 298K) T S ( T) = S ( 298K) + C ln 298 K C = 6C O g + C C H O s 6C CO g 6C H O l ( ( )) ( ( )) ( ) ( J K) J K ( J K) ( J K) ( ) ( ()) = / / / / = J / K J / K J / K J / K = J / K Then H ( T) = H ( 298K) + C ( T 298K) = 281J + ( J / K)( 33K 298K) = kJ T 33K S ( T) = S ( 298K) + C ln = 259 J / K + ( J / K) ln = J / K 298K 298K G = H T S = 2792J 33K J / K = kJ ( )( ) The reaction at T=33K has a free energy change that is even more positive because the entropy change has become more negative. Helpful Information T=298K CO 2 (g) H 2 O C 6 H 12 O 6 O 2 H f kj/mole S f J/mole-K C p J/mole-K Assume all heat capacities are constant between T=298K and T=33K.

University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2008

University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 008 Midterm Examination Key July 5 008 Blue books are required. Answer only the number of questions requested. Chose wisely!