ANSWER KEY. Chemistry 25 (Spring term 2015) Midterm Examination

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1 Name ANSWER KEY Chemistry 25 (Spring term 2015) Midterm Examination 1 (15 pts) Short answers 1A (3 pts) Consider the following system at two time points A and B. The system is divided by a moveable partition. The molecules (X) cannot move across the partition, but they can move between the grid boxes on the same side of the partition. i. Calculate the multiplicities of the two systems A and B. ii. At which time point (A or B) is the system closer to equilibrium? Briefly explain (one sentence). iii. Indicate the equilibrium position of the partition in time point C and calculate the corresponding multiplicity. i. time point A, W = (3!/(2!1!) x (7!/(2!5!)) = 3 x 21 = 63 time point B, W = (6!/(2!4!)) x (4!/(2!2!)) = 15 x 6 = 90 ii. point B has the greater multiplicity and is hence closer to equilibrium iii. the equilibrium position is when both sides have 5 grids, where the multiplicity is (5!/(2!3!)) x (5!/(2!3!)) = 100 1

2 1B (3 pts) Assuming an ideal solution, calculate the Gibbs free energy, entropy and enthalpy of mixing when 1.00 mol hexane is mixed with 1.00 mol heptane at 298 K. Report your answers in kj mol -1, J mol -1 K -1, and kj mol -1 for G, S and H, respectively. S mixing = nr x i ln x i = ln ln0.5 = J mol -1 K -1 G mixing = T S mixing = H mixing = 0 ( ) ( ) = 3.43 kj mol -1 2

3 1C (3 pts) In class, we calculated dlnk/dt and derived the van't Hoff equation relating lnk to H. lnk is also a function of P; following the same general approach as used for the van't Hoff equation, derive an expression for dlnk/dp of the form: " ln K % $ ' = xr γ V δ T β # P & T where x, γ, δ, β, γ are integers (and may be positive, negative or zero). ln K = ΔG! RT $ ln K ' & ) = 1 $ ΔG! ' & ) % P ( RT % P ( T T = ΔV RT x = 1; γ = 1; β =1; δ = 1 3

4 1D (6 pts) Here is a three-pack of rubber band questions to end this section on a snappy note. i. Which statement is valid for a rubber band? a. compact state + heat stretched state b. compact state stretched state + heat c. compact state stretched state + Higgs boson d. compact state stretched state ii. The temperature of an ideal elastic band obeying Hooke's law increases 1 C when the band is stretched in length by 10%. What is the temperature increase when the band is stretched 20%? (hint: the choices are 1 C, 2 C, 4 C or 8 C). as derived in the review Q = 1 2 kx2 by 4x, so that the temperature increase is 4 C, so if the length doubles, the heat released goes up iii. It is apparently well known in the slingshot community (ie - I saw this on the internet) that the longer one waits before releasing an extended slingshot, the less energy the projectile has. Based on your understanding of the rubber band mechanism, suggest in 1-2 sentences a reason why it is, or it is not, plausible that immediately releasing an extended slingshot would maximize the energy of an ejected projectile. When the elastic is stretched, heat is released and the temperature increases. The force constant is proportional to temperature, so that if the elastic is allowed to cool before firing, the force constant will be reduced and some of the stored energy released as heat, so that the energy of the ejected projectile will be diminished. 4

5 2 (NOT RELEVANT) 5

6 3 (20 pts) The pressure is on and temper(ature)s are boiling The boiling point of a liquid occurs when the vapor pressure equals the ambient pressure; by going to pressures above 1 atm, the boiling point of water can be raised above 100 C. This effect is utilized in the operation of an autoclave where water under pressure boils at 121 C and the elevated temperatures are used for sterilization. Let's assume that you need to use an autoclave; unfortunately, the temperature gauge is broken, but not the pressure gauge. Having taken Ch25, you realize that you can calculate the pressure where water boils at 121 C based on your knowledge of thermodynamics and the properties of water (liquid and steam) at 100 C. quantity value at 373 K S liq 24. J mol -1 K -1 S gas 132. J mol -1 K -1 C p,liq 75.3 J mol -1 K -1 C p,gas 37.5 J mol -1 K -1 V liq 18.9 cm 3 mol -1 Assume that these constants are independent of T and P, except for V gas given by the ideal gas law = RT/P. 3A (8 pts) From this data, calculate G( T) in kj mol -1 between T 1 and T 2 for liquid water and for water vapor (gas), ie G liq ( T) = G liq (T 2,P 2 ) - G liq (T 1,P 2 ) and G gas ( T) = G gas (T 2,P 1 ) - G gas (T 1,P 1 ). G( T 2 ) G( T 1 ) = S( T 1 ) T 2 T 1 ( ( " ( )+ C p T 2 1 ln T % * * $ '- T 1 ) ) # T 1 & -,, G liq ( T ) = G liq (394,P 2 ) - G liq (373,P 2 ) ( ( " = 24. ( ) ln 394 % + + * * # $ 373& ' ) ),, = 548 J mol 1 = kj mol 1 G gas ( T ) = G gas (394,P 1 ) - G gas (373,P 1 ) ( ( " = 132. ( ) ln 394 % + + * * # $ 373& ' ) ),, = 2790 J mol 1 = 2.79 kj mol 1 6

7 3B (8 pts) Determine the value of P 2 where water boils at 121 C from the appropriate thermodynamic cycle. For this calculation, assume: a) water vapor (gas) behaves as an ideal gas to determine the dependence of G gas ( P) = G gas (T 2,P 2 ) - G gas (T 2,P 1 ) on P 2. b) since liquid water is nearly incompressible and has a small volume relative to the gas, the contribution of G liq ( P) = G liq (T 1,P 2 ) - G lig (T 1,P 1 ) to the thermodynamic cycle can be neglected in this calculation. Report your answer for P 2 in both Pa and atm (hint, P 2 is between 1-10 atm). Also calculate G gas ( P) in kj mol -1 when water vapor goes from P 1 and P 2. P 2 = 1.98 atm = 2.01 x 10 5 Pa G liq ( P)+ G liq ( T ) = G gas ( T )+ G gas ( P)! = RT ln P $ 2 # & " % P 2 P 1 = Exp[ =1.98! G gas ( P) = RT ln P $ 2 # & " P 1 % P 1 ( ) ( )! = ( )ln 1.98 $ " # 1 % & = kj mol 1 7

8 3C (4 pts) Calculate G liq ( P) = G liq (T 1,P 2 ) - G liq (T 1,P 1 ) in kj mol -1 between P 1 and P 2 for liquid water to verify that this side of the thermodynamic cycle only makes a minor contribution to this calculation, relative to the magnitudes of the other 3 sides of the cycle. G liq ( P) = V P ( ) ( ) 10 6 m 3 =18.9cm 3 cm 3 =1.88 J mol 1 = kj mol 1 G liq ( P) is several orders of magnitude smaller than the other 3 sides of the thermodynamic cycle. 8

9 4 (25 pts) On the edge of stability 4A (6 pts) In the vicinity of the melting temperature of a wild type protein, the free energy of protein unfolding, G wt, is found to vary linearly with T by the equation G wt = T kj mol -1. What are the corresponding values of T m, H m and S m in K, kj mol -1 and kj mol -1 K -1, respectively? Assume C p =0. at T m, G wt = T m = 0 T m = = 332 K S = G T =1.12 kj mol 1 K 1 at T m, H = T S = 372 kj mol 1 9

10 4B (4 pts) A mutation is made in the protein characterized in 4A that alters G mut such that the native state of the mutant is stabilized by 6.00 kj mol -1 relative to the wild type at all temperatures. The stability of the denatured state is unaffected by this mutation. What is T m (in K) for this mutant? Since N for the mutant is stabilized relative to the wildtype by 6 kj mol -1, G mut is increased (more positive) by +6 kj mol -1 since these Gs are the free energy of unfolding.!! G mut = G wt T m = = T T m = = 337.5K 10

11 4C (6 pts) Under 1000 atm pressure, the melting temperature of the wild type protein equals that of the mutant protein at 1 atm. From this result, calculate V, the volume change associated with protein unfolding. Express your answer in units of m 3 mol -1 and Å 3 molecule -1. Assume V is independent of both P and T (that is, there is no thermal expansion and the protein is incompressible (not to mention incomprehensible)). the pressure dependence of contributes a V P term to the G of unfolding; for the native protein, application of 1000 atm pressure stabilizes the N state raising the Tm to that of the mutant. The magnitude of V (= V D - V N ) can be determined either by recognizing that V P = 6 kj mol -1 (the stabilization of the mutant given in Prob. 4B) or (equivalently) by solving G wt ( P) = T + V P = 0 at T = K. (note P = = 999 atm, but using 1000 atm is OK (I meant to use P = 1001 atm.)! or G wt V P = 6 kj mol 1 = 6000 J mol 1 V = 6000J mol 1 (1000 1) Pa = m 3 mol 1 = m 3 mol Å 3 1 m 3 N A molecules mol 1 = 98.4 Å 3 molecule 1 ( P) = T + V P = 0 at T = # & V = $ % P ' ( = m 3 mol 1 11

12 4D (9 pts) A more detailed experimental analysis suggests G has a quadratic dependence on temperature that can be parameterized as G = at 2 + bt + c Derive expressions for H, S and C p for protein unfolding as a function of temperature in terms of a, b, c and T. G S = T = 2aT b H = G +T S = at 2 + bt + c 2aT 2 bt = at 2 + c or since C p = S T H T = C p T = 2aT ; C # p = T % $ S T & ( = 2aT ' 12

13 5 (25 pts) Powering Transport The transport of molecules in and out of cells constitutes an important component of cellular metabolism. As an example, the transport of glucose from the outside to the inside of a cell may be described by the following equation: Eq. 5.1 Glucose out Glucose in Specialized integral membrane proteins known as transporters are responsible for glucose uptake; in mammals, glucose is imported by members of the GLUT family. The most widely distributed version is GLUT1 that is responsible for getting glucose into red blood cells and across the blood brain barrier, among many other roles. 5A (5 pts) An essential role for GLUT1 is keeping brain cells fueled with glucose. The brain operates at ~20 watts; assuming that oxidation of glucose releases 3000 kj mol -1 of energy, calculate the number of glucose molecules that must be oxidized per second to sustain this level of metabolic activity in the brain. the oxidation of one glucose per second will release 3 x 10 6 J mol -1 /(6.02 x mol -1 ) = 5 x J s -1 glucose -1 production of 20 W = 20 J s -1 will require oxidation of 20 J s -1 /(5 x J s -1 glucose -1 ) = 4 x glucose s -1 13

14 5B (3 pts) Consider a cell at 25 C, where the concentrations of glucose outside and inside the cell are 1.00 µm and 1.00 mm, respectively. What is G, the standard free energy change for this reaction as written in Eq. 5.1 (transport from the exterior to the interior)? Express your answer in kj mol -1. G = 0 ( G refers to the free energy change when reactants and products are both in standard state (= 1 M) and so the concentrations are the same on both sides on the membrane in the standard state) (or G = -RT ln K eq, with K eq = 1 in equilibrium (concentrations will be the same on both sides of the membrane at equilibrium) 5C (3 pts) What is G, the free energy change for Eq. 5.1 as written? Express your answer in kj mol -1. G = G +RT lnq ( ) ( ) = G +RT ln Gluc in = ( )ln = kj mol -1 14

15 5D (5 pts) Suppose a non-glut glucose transporter utilizes ATP hydrolysis to drive the uptake of glucose by cells according to the reaction: Eq. 5.2 natp + Glucose out nadp + npi + Glucose in G ATP = -28 kj mol -1 If the transporter is 100% efficient in utilizing the free energy of ATP hydrolysis, what is the maximum ratio of glucose uptake ((Glucose in )/(Glucose out )) that can be achieved when G = 0 for the coupling of n = 1 ATP per glucose transported? For this problem, use (ATP) = (ADP) = (Pi) = 1 M. G = G +RT lnq G = G +RT ln Gluc in ( )( ADP) n ( Pi) n ( )( ATP) n 0 = 28n + RT ln Gluc in 28n = RT ln Gluc in ( Gluc in )( ADP) n ( Pi) n = e 28n/RT ( )( ATP) n ( Gluc in )( 1) ( 1) ( )( 1) ( )( ADP) n ( Pi) n ( )( ATP) n ( )( ADP) n ( Pi) n ( )( ATP) n = e 28/RT = the rule of thumb is that coupling ATP hydrolysis shifts the equilibrium constant by ~ E (5 pts) Under the same conditions, what is the maximum ratio of glucose uptake ((Glucose in )/(Glucose out )) that can be achieved when G = 0 for the coupling of n = 2 ATP per glucose transported in Eq. 5.2? ( Gluc in )( ADP) 2 ( Pi) 2 = e 28 2/RT ( )( ATP) 2 ( Gluc in )( 1) ( 1) ( )( 1) = e 56/RT =

16 5F (4 pts) Under physiological conditions, the concentrations of ATP, ADP and Pi are not 1 M, but are typically found to be 10 mm, 10 µm and 10 mm, respectively. In this case, what is the maximum ratio of glucose uptake ((Glucose in )/(Glucose out )) that can be achieved when G = 0 and n = 1, in Eq. 5.2? G = G +RT lnq ( )( 10 2 ) ( ) ( Gluc in ) 10 5 ( ) 10 2 G = G +RT ln Gluc in = e 28/RT ( Gluc in ) ( ) = 105 e 28/RT = ( )( ADP) n ( Pi) n = 0 ( )( ATP) n 16

17 6 (NOT RELEVANT) 17

Chemistry 25 (Spring term 2015) Midterm Examination

Name Chemistry 25 (Spring term 2015) Midterm Examination Distributed Thursday, April 30, 2015 Due Thursday, May 7, 2015 by 1 pm in class or by 12:45 pm in 362 Broad Conditions (see 2016 midterm exam) 1