Chem 350 Problem Set 1 Thermodynamics Key

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Randolf McKenzie
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1 Chem 350 Problem Set 1 Thermodynamics 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 constant a nice day at 25 0 C. a) Calculate the work done in J. Analysis: The expansion is non ideal non reversible as it is done in a single step Needed equations: -p(v2-v1) (no ideal, constant temperature, constant pressure change in volume) Other data: since volume is in L and pressure at sea level is 1 atm conversion to J will be required. 1L*atm = J; Temperature is 298K although this value is not needed since it remains constant. If the temperature changed the analysis would be invalid. Solution: -p(v2-v1) -1 atm(1.6l-1.0l) = 0.6 Latm* J/L*atm = J The sign of the work is negative as appropriate for expansion. The system is doing work on the surroundings. b) Compare the work calculated in a) to the work done by a piston expanding a cylinder volume in a single step from 1.0 L to 1.6 L. The work is also accomplished at sea level. The work would be the same as both processes are no ideal and under the same conditions. Both processes are essentially single step processes c) Compare the work done in a) and b) to that of a computer-controlled piston that accomplishes the same expansion; however, the expansion is accomplished in millions of steps over 24 hours. Analysis: In this case we can assume the expansion accomplished is nearly ideal and a better approximation of the work would be obtained by using Needed Equations: w = -RT ln (V2/V1) Solution: w= J/molK * 298K ln (1.6L/1.0L) = J/mol In this case the ideal gas constant in J gives the answer directly. Even though the volume is in L the units cancel in the ln term. T must be used in K; 25 C +273= 298K d) Discuss different assumptions made in each part of the problem See Analysis above with each problem. Note that the work accomplished in c) is significantly more than in a) and b) as the maximum work possible is obtained in a reversible process. 2. Show that for reversible expansion of an ideal gas the pressure decreases. Analysis: The process is reversible and ideal. Therefore the ideal gas law applies. Equations: w = -RT ln (V2/V1) And P2V2=P1V1 (from the ideal gas law) Solution: V2/V1 = P1/P2 If V2>V1 as is the case for expansion then lnv2/v1 >0 and w<0

2 In order for w<0 P1 must also be greater than P2; therefore the pressure decreases when the gas expands. 3. A 1 meter tall daisy is drawing one 9 g of water from its root to its top. a) How much work is performed by the daisy in this process? Analysis: The work is work performed against gravity since the system (plant) is doing work the work will be positive. Needed equations: w = mgδh; gravitational force on earth is 9.8m/s 2 Solution: w = 9g * 9.8m/s 2 * 1m = 88.2gm 2 /s 2 = kgm 2 /s 2 = J b) How many calories (dietary) does the daisy consume or produce? Needed: 1cal=4.184J; 1000 cal = 1Cal = 1 kcal(dietary calorie) Solution: J * 1 cal/4.184 J = cal Dietary calories = cal * 1 Cal/1000cal = 2.1 E -5 Cal The positive sign indicates work done by the system work done by the system would result in a consumption of energy. c) What mass of glucose would be fixed or consumed to accomplish the work? Analysis: Sugars produce about 4 calories per gram (dietary calories) 2.1 E -5 Cal *1 g/4 Cal = 5.27 E-6 g or 5.27ug of glucose would need to be fixed by the plant to move the water if the process was 100% efficient. 4. What determines whether Δq and Δw are positive or negative? How are (or are) the two connected? Is temperature affected in these changes? Are they state functions? If Δq >0 then by convention this means that heat has been transferred into the system giving it an increase in internal energy ΔU>0 and the process is therefore termed endothermic. This often results in an increase in temperature of the system.. When q <0 het has been transferred to the surroundings, giving a decrease in the internal energy of the system ΔU<0 and the process is termed exothermic. When Δw >0 work has been don on the system giving and increase in internal energy. When Δw <0, work has been done on the surroundings resulting in a decrease in internal energy. The two functions are connected by ΔU = Δq + Δw. Therefore with a constant ΔU as heat increases work decreases and vice versa. 5. The changes in internal energy for the oxidation of glucose and stearic acid at 310K are -2.9E3 kj/mol and E3 kj/mol, respectively. a) Calculate the change in enthalpy for each reaction. b) Which substance is more likely to be useful for physiological energy storage? For each indicate any assumptions you make to complete the problem. a) Analysis: For glucose the complete combustion (oxidation) reaction is C6H12O6 + 6 O2 < == > 6 CO2 + 6 H2O In this case there is no volume change since 6 moles of gas goes to 6 moles of gas (condensed phases do not add appreciably to volume and at 310K water is a liquid) so ΔV = 0

3 Equations: ΔH = ΔU + PΔV (by definition) Since ΔV =0; ΔH = ΔU and the enthalpy is -2.9E3 kj/mol Analysis and equations: For stearic acid the reaction is C18H36O O2 < == > 18 CO H2O In this case the change in volume is not zero so the enthalpy and internal energy are not equal. To determine the change in volume, molar volumes must be assumed according to the ideal gas law at constant pressure and temperature PΔV = ΔnRT This reaction is a compression where 26 moles of gas are converted to 18. Again the water is in a condensed phase. So Δn = 8 moles ΔH = ΔU + PΔV or ΔH = ΔU + ΔnRT ΔH = E3 kj/mol + (-8mol)(8.314J/molK)(1kJ/1000J)(310K) = E3 kj/mol b) More heat is given out by the oxidation of one mol of stearic acid than 3 moles of glucose (same molecular formula). Clearly stearic acid (a fatty acid) is more energy rich than glucose (a simple sugar). The main reason for this is that in glucose carbons are predominately already oxidized (containing 5 OH groups) rather than the stearic saturated carbons (alkane). Thus fats are better energy storage than carbohydrates. 6. Using the oxidation of glucose as an example discuss the difference between the single step oxidation and the 8-step TCA oxidation of the molecule. Which thermodynamic functions are identical for each process and which differ. Explain. The overall or net equation for each process is C6H12O6 + 6 O2 < == > 6 CO2 + 6 H2O Therefore for the state functions, U, H and S the values calculated for the single step direct oxidation will be identical to the sum of those values across the 11 steps of the TCA cycle. State functions are path independent and depend only on physical states and bonds present. The net bonds broken and formed and identities of the overall starting materials and products are the same for both pathways. However, the total heat for the TCA cycle and the direct oxidation differ. The non-state functions are path dependent and the two different conversions result in different values. The same is true for work, w, which will change depending on the route taken. However, the sum of q and w will be invariant as the two together are related by ΔU = Δq + Δw 7. Compare the internal energies of two components of air H2O and O2 when the closed system is heated from 25 to 50 0 C at constant volume. Discuss any assumptions you have made. Analysis: To answer this question you need to assume ideality of the gas (a definite approximation!). Also you must know the geometry of both gasses in this case the Lewis structure and VSPER theory will show you both are linear. So we have one linear diatomic molecule and one non-linear triatomic molecule. Since the system is closed the number of moles is constant. Equations: dqv= du=cvdt, assuming Cv is constant over the change in temperature (you have no other data so you must assume constant behavior) ΔqV= ΔU=CvΔT for gasses Cv = df * nr/2 for a non linear triatomic molecule df=6 and for a linear diatomic df=5 ΔT = = 25K

4 for H2O ΔU = (6*1mol*8.314J/ mol K) *25K /2= 623 J (determined for 1 mol) for O2 ΔU = (5*1mol*8.314J/ mol K) *25K/2 = 520 J 8. The coefficient of thermal expansion of water at 25 0 C is E-6 1/K and the compressibility is 45.24E-6 bar -1. If the density of water is E3kg/m 3 calculate CP-CV. Equations needed CP-CV = α 2 VT/β Solution: Volume must be calculated from density * molecular weight (assuming we are doing this on a per mole basis) V = E-3 kg/mol* E3 kg/m 3 = kg 2 /mol*m 3 We should also convert pressure from bar to a more usable unit like atm. The conversion factor is 1 bar =0.987 atm E-6 bar * atm/bar = E-5 atm CP-CV = (257.05E-6 1/K * kg 2 /mol*m 3 * K) / E-5 atm -1 = 7.76 E -6 m 3 atm/molk a m 3 atm is a J = 7.76 E -6 J/molK for one mol 7.76 E-6 J/K 9. A dog with a normal temperature takes 400 J of heat (contained in chemical bonds) and converts it into 3 J of work. Assume that reversing the process using work to produce heat will convert the work back into heat with a 25% loss of energy. (Normal temp for a dog is F) a) How much of the original heat is not compensated for by reversing the process? b) What is the entropy change for this process at a temperature of 298K? c) Comment on the direction of entropy change does this make sense? a) uncompensated heat = original heat work back into heat(adjusted for loss of heat) 400J (3J(0.75) = J lost b) Entropy change by definition is ΔS = q/t this is a constant temperature process (it is in a body at constant temperature) so the equation can be used directly. The temperature of F corresponds to C which is 312 K. So ΔS = J/312K = 1.27J/K c) since ΔS >0 entropy is increasing. This makes since energy is being lost from the system to the surroundings. 10. A heat engine with 400 J of heat available transfers that heat quantitatively from a body at 300 K heating that body to 350 K what is the entropy change for the process? Is this process entropically spontaneous? Solution: As there is a temperature change the definition of entropy cannot be used. ΔS = q/t Since there is no pressure or volume changes and all of the heat is transferred the entropy can be calculated using the Clausius equation (alternatively each heat can be calculated separately) ΔS = q/ (1/T2-1/T1)

5 = 400J/ (1/350K - 1/300K) = J/K No the process is entropically spontaneous. ΔS <0 showing that heat does not spontaneously flow from a colder body to a hotter body. 11. When a particular protein binds directly to DNA the entropy of the system decreases by 2 kcal/k. At the same time 700kcal of heat is released (the binding is occurring at sea level, barometric pressure and temperature are not changing). a) What is the free energy change for the binding reaction at 310K? b) A protein mutation reduces the heat released to 600 kcal but does not alter the conformation of the protein nucleic complex or alter the number or residues involved in binding. What is the free energy of the mutated complex? c) Is the mutation thermodynamically beneficial? a) Analysis: In this problem the entropy is given and since it is a decrease, ΔS = -2kcal/K. Since the pressure is constant the heat released can be equated to enthalpy Δqp = ΔH = -700kcal (negative as heat is released) Equations needed: ΔG=ΔH-ΔTS Solution: ΔG= -700kcal (-2kcal/K *310K) = -80 kcal (spontaneous) b) the analysis is the same. The reduced heat results in the following: ΔG= -600kcal (-2kcal/K *310K) = 20 kcal (not spontaneous) c) The mutation is does not benefit direct binding of the protein to DNA. The wild type protein binding is thermodynamically spontaneous; however the mutant protein is not. Without coupling (change in mechanism of binding) the mutant will not bind DNA. 12. Calculate the heat of neutralization at 25 0 C when one mole of antacid (assume Ca(OH)2) encounters one mole of stomach acid. Analysis: This is a Hess Law problem. The reaction is H + (aq) + OH - (aq) === > H2O (l) Analysis can differ depending on the equation you write. If you include the calcium ion or express as hydronium for example. Since no other data are given this should be done using heats of formation data. This accounts for all of the heat in the bonds and treats the reaction like all bonds are broken in the reactants and all bonds are formed in the products. Recall the Hf for any element is 0. For any reaction: ΔH298 = ΔH298 products - ΔH298 reactants Looking up Hf data (your numbers may vary slightly depending on your source and your units) Hf H2O (l) = kcal/mol Hf H + (aq) = 0 Hf OH - (aq) = kcal/mol ΔH298 = [ (1)( )] [(1)(0) + (1)( ) ] = kcal (or kj)

6 13. The change in entropy for mixing can be obtained by summation of the individual components of the system. ΔS(system) = -R *sumi (nilnxi) where n is the moles of component i and x is the mole fraction of component i. a) What is the entropy change for preparing a 1 µmol aqueous solution of glucose at 25 0 C? b) Discuss your result in terms of the ideality of the solution. c) Explain why in an ideal solution heat of mixing (constant pressure) is 0? d) Calculate ΔG of mixing. Analysis: There are two components in the system glucose and water. At 25 0 C the molar volume of water is M (this is a standard value or you can use the density of water and the formula mass to calculate it) Solution: Components H2O assuming 1 L of solution the molar volume of water at mol/l the moles of water is mol Glucose C6H12O6 assuming 1 L of solution (the same L) 1 E-6mol/L * 1 L = 1 E-6mol Total moles = 1 E-6mol = The low concentration of glucose means that both its mole fraction and its number of moles make it essentially 0 compared to water. You can simplify the equation by considering the water only ΔS(system) = J/molK * mol * ln1 = 0 There is essentially 0 entropy change in mixing such a dilute solution. b) Since the solute has essentially no effect on the solvent the solution approaches ideality. c) By definition in an ideal solution there is no particle-particle interaction. If there is no interaction no heat can be exchanged. Under constant pressure qp = ΔH = 0. d) ΔG = ΔH- TΔS (as a summation for the system as expressed for entropy). Since the entropic contribution of the solute is negligible and the enthalpy is 0 for the solute the only factor for the free energy is the solvent and that is 0. Discuss the meaning of the value you obtain. ΔG = 0-298K (0) = 0; for an ideal solution the free energy of mixing is 0. That means the mixing is non spontaneous. In other words the solution has no interaction between solute and solvent molecules. Ideal solutions are not in fact solutions at all. 14. A ligation reaction has a Keq of 3.87 E-16 at 400 K and 1.67E-8 at 600K. a) Estimate ΔH o 500 b) Is this reaction favorable at any of the temperatures? a) Since temperature equilibrium data are provided and enthalpy is sought the applicable equation is the v anthoff equation: ln (KT2/KT1) = ΔH o /R (1/T1-1/T2) Rearranging for ΔH o provides ΔH o = R * [ln (KT2/KT1)]/ (1/T1-1/T2)

7 = J/molK * ln (1.67E-8/3.87 E-16)/(1/400-1/600)K = 175 kj This is the standard state enthalpy. The assumption is that the enthalpy is constant over the range of temperatures used to calculate the value. Since T1 and T2 are 400K and 600K and the problem asks for the enthalpy at 500K the value of ΔH o calculated should be valid at 500K. It is tempting to consider the ΔH o as the value at 298K. Remember however, that the 0 does NOT mean 298. Thermodynamic standard state does not specify temperature. b) Since the equilibrium constants are all very much less than 1 the concentration of products is small at equilibrium indicating an unfavorable reaction. This is borne out by the positive enthalpy component. 15. In terms of ΔS, ΔH ΔG and common sense discuss the spontaneity of the reaction H2O(s) < == > H2O(l) in an ice water bath maintained at 0 0 C. For the reaction as written (melting) the entropy change is positive because of the increase in randomness in going from a solid to a liquid. The liquid is less ordered than the solid (observation). Thus for a constant enthalpy system the spontaneous change would be for all of the ice to melt and the process be driven completely by entropy. However, the enthalpy change for the process is positive (because of overcoming intermolecular forces that hold the solid together), which would indicate a non spontaneous change (you have to heat ice to melt it put energy into the system). Thus if only enthalpy change is considered the spontaneous change would be for the water to freeze. If this system is held at 0 0 C the system will reach a state of equilibrium with ΔG = 0. In other words there will be ice and very cold water present. 16. Discuss the hydrophobic effect in terms of thermodynamic state functions G, H, S. The hydrophobic effect is the result of a lack of interaction of two phases often with a preference (or at least not an abhorrence) of self. When a hydrophobic molecule like a lipid is put in contact with water and forcibly dispersed by shaking or stirring, as soon as the agitation ceases the two substances form distinct layers. This process is observed and therefore thermodynamically spontaneous. With that reasoning ΔG<0. The question then becomes enthalpy and entropy. The hydrophobic effect supports organization of the layers rather than random distribution between the layers so on the surface it would appear that entropy is disfavored ΔS<0. Enthalpy of interaction between the substances is also absent so it would also appear that ΔH>0. Of course this can t be true since ΔG cannot be <0 with both enthalpy and entropy unfavorable. The answer comes in the nature of water. The hydrogen bonding in water is particularly strong. In fact hydrogen bonding is simply a particularly strong permanent dipole-dipole interaction. If water dissolves another compound interacting with it - it loses its highly energetic sphere of hydrogen bonds that it forms with other water molecules. So water maximizes its enthalpy by bonding with water; available interactions with other molecules like the non polar hydrocarbons in lipids are predominately less energetic. The enthalpy of the hydrophobic effect is actually <0 as water molecules reform hydrogen bonds after being separated by the lipid. So what about the entropy? In fact the entropy of the water increases as water hydrogen bonds with itself. Hydrogen bonds are not static and the fluctuating hydrogen bonds (from one pair to another) present a high level of entropy. In addition, if water solvates a lipid it does so at a high organizational cost to the water molecules. So despite the seaming appearance that enthalpy and entropy of the hydrophobic effect would be unfavorable the unique nature of water is the driving force that make both of these state functions promote spontaneity. 17. Calculate the maximum efficiency of an organism with a temperature of F that is trying to cool itself on a hot day (92 0 F as well as compared to a cool day, 70 0 F)

8 Analysis: The organism functions as a biological engine. In order to do any calculations the temperatures must be converted to K, they are 310, 306, 294 K, respectively. Equations: For the efficiency of an engine experiencing a temperature change: E = (Th-Tc)/Tc For the hot day: E = ( )/306 = = 1.31% For the cool day: E = ( )/306 = = 5.23% Your are almost 5x more efficient cooling yourself on a cool day than on a hot one. 18. Compare the entropy changes for the system surroundings and universe for the reversible isothermal expansion of one mole of an ideal gas from 0.010L to 0.1L at 298 K to the entropy changes for the same expansion performed irreversibly against a constant external pressure of 0.1 atm. Analysis: no change in temperature, reversible process for an ideal gas expansion should result in a positive entropy for the system (favorable) Equations: ΔS=nRln(V2/V1) ΔS(system) = (1.00mol)(8.314 J/molK) ln 0.100/0.01 = J/molK positive as expected For a reversible process ΔS of the system is compensated for by ΔS surroundings. ΔS= -nrln(v2/v1) = J/mol K For the universe (ideal reversible process) the total entropy change ΔSsystem + ΔSsurroundings is 0. For the irreversible process: Since a state function for the system (if the system has not changed) must be path independent ΔS(system) = (1.00mol)(8.314 J/molK) ln 0.100/0.01 = J/molK the reversible path is used to calculate both values. For the surroundings - in an isobaric isothermal expansion of an ideal gas the entropy must be in the opposite direction (as the gas expands it takes up more space at the expense of the surroundings) ΔU = ΔH =0 Therefore ΔU=q+w and q=-w = PΔV ΔS= q/t ΔS= PΔV/T = [0.1 atm *( )L ]/298K * J/Latm = J ΔS universe for real process = J/molK J/mol K = J/molK For real processes the entropy of the universe must expand. 19. The intracellular concentration of a protein is 1000 times the extracellular concentration. a)calculate the free energy change per mole required for transfer of the protein to the extracellular space at 37 o C. b) How does the free diffusion differ thermodynamically from protein channel mediated diffusion? The numerical solution to this problem depends on whether you are considering the movement in to out or out to in

9 [protein (in)] = 1000[protein out] G = G + RTln[protein] The transfer is from inside the cell to out, so G= Gout Gin = Gout + RTln out (Gin + RTln in) = (Gout Gin ) + RT(ln out ln in) = (Gout Gin ) + RTln(out/in). We assume that the standard state free energy of the protein does not depend on whether it is inside our outside the cell the structure of the protein and therefore its energy does not change. So we are left with G = RTln([protein]out/[protein]in) = 8.314J/molK * 335K *ln(1/100) = J/mol = J/mol this is a spontaneous process as would be expected since the intracellular concentration is much greater than extracellular and the natural direction of diffusion would be from the higher to the lower concentration. b) Free energy is a state function. The energy for the process does not change. The feasibility due to a membrane barrier may affect the rate of the diffusion but once it happens the energy is invariant.

reduction kj/mol

reduction kj/mol 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 -2870 kj/mol. a) How much energy is required to produce glucose from