Chem/Biochem 471 Exam 3 12/18/08 Page 1 of 7 Name:

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1 Che/Bioche 47 Exa /8/08 Pae of 7 Please leave the exa paes stapled toether. The forulas are on a separate sheet. This exa has 5 questions. You ust answer at least 4 of the questions. You ay answer ore questions if you wish. Answerin 5 questions can be an advantae if you are unsure of soe of your answers (this will distribute the risk ). Answerin 4 questions is advantaeous if you are very sure of your answers. Each pae is worth 0 points. The total exa rade will be noralized so that the axiu nuber of course points for this exa will be 0. For exaple, ettin 80 points on 4 questions equals 00 points on 5 questions equals 0 points toward the final rade. Gettin 80 points on 5 questions would be worth 80% of the axiu rade. If you leave a pae blank, it will not be included in the radin. If you work on a pae and then decide that you do not want it to be raded, be sure to ark the box at the botto of the pae. If you work on the pae and fail to ark the box, the pae will be raded. Work at least 4 probles (of your choosin) or ore, as you prefer. Answers without explanations (where indicated) are not coplete.

2 Che/Bioche 47 Exa /8/08 Pae of 7. Two saples of as are at P.50 at and T 5.0 o C. One saple contains e of N and the other saple contains ½ e of O. The ar ass of N is 8.0 / and that of O is.0 /. You ay assue that the as ecules behave ideally. The collisional diaeter of N is.74 Å and that of O is.50 Å. (One Å 0-0.) a) For N, indicate the rank fro hihest to lowest of the rs ecular speed, averae speed and ost probable speed. (Nubers are not necessary, only the relative rank is needed.) Hihest speed: rs speed Interediate speed: averae speed Lowest speed: ost probable speed k T > 8k BT > π B k BT b) How does the averae kinetic enery of a N ecule in the first as saple copare to that of an O ecule in the second as saple? Averae KE of N is reater than that of O Averae KE of N is less than that of O Averae KE of N is the sae as that of O Averae KE per ecule is sae for both ases k B T and T is the c) Which as saple has ore collisions per second on averae, N or O? Explain. The total nuber of collisions per second for a as saple is N RT Z π σ. For both V M N and O the value of N/V is N N AP fro the ideal as law. Thus, dividin Z N by Z O V RT ives ( ) Z N σ.74 N M O A.. Thus, there are ore collisions for N. Z (.50 ) 8 O σ N M N A d) What is the averae distance an O ecule will ove between ecular collisions in the second as saple? ( )( ) We have that N N P at A Thus, the V RT ( L at )( 98K ) L K ean free path (the distance between ecular collisions) is N 5 ( 0 8 l π σ π.69 0 )(.5 0 ) Å. V

3 Che/Bioche 47 Exa /8/08 Pae of 7. You are studyin bushy stunt virus. You place 00 μ of virus into water at 0 C and find that the volue of the solution increases by 0. μl. ( L 000 c.) a) What is the partial specific volue of bushy stunt virus in water at 0 C? V v w P, T, K ΔV 0. 0 L Δw L c 0.74 b) You next put the solution of virus into a centrifue and spin it at 000 rp. You easure the position of the solvent-solution boundary (r ½ ) at several ties and plot lo r ½ versus tie. The slope of the plot is sec -. What is the value of the sedientation coefficient for bushy stunt virus?.0 d lo r We know that s. Thus, the slope of a lo r ½ versus tie plot is sω ω dt.0. Thus, s ( slope).0. The value of ω 000 rev π 05. Pluin this into the ω 60sec rev sec equation for s ives s ( slope) sec S ( 05. ω sec ) sec c) The density of water at 0 C is /c. What is the apparent ar ass of bushy stunt virus? Given the values of s and the partial specific volue, we can calculate the ar ass fro RTs M : D( vρ ) J 8.45 ( 9K )(. 0 sec) K M ,700, 000 c c sec c c d) If a uch saller virus (i.e saller ar ass) is ixed in with the bushy stunt virus in the centrifuation of part b and the partial ar volues of the two viruses are equal, will the saller virus irate faster, slower or at the sae speed as the bushy stunt virus? Explain. The saller virus will ove ore slowly. The saller virus has a ass that is less than that of the bushy stunt virus. It also has a radius a that is saller. Because the partial specific volues are the sae, the volues of these two viruses will be proportional to their asses, thus the radius of a virus is proportional to the cube root of its ass: a. Given that the frictional force on a sphere is 6πηa, we see that ( v ) the dependence of s on oes like ρ s. Therefore, the larer virus will have a larer ( 6 πη ) sedientation coefficient and thus a larer teral velocity. f sphere

4 Che/Bioche 47 Exa /8/08 Pae 4 of 7. The conversion of urea to aoniu follows the reaction k urea + H O NH CO -. k - A plot of the enery of the syste as a function of proress of the reaction is shown at riht. At 6.0 C the forward rate constant k equals At 7. C it equals a) What is the value of the activation enery barrier for the forward reaction? k We can apply the forula E A ln : k R T T 7. 0 ln E A 8.45 J 4.5K 44. 5K K b) You do several easureents of the activities of the reactants and products at equilibriu and find that the equilibriu constant equals 70 at 6 C. What is the value of the rate constant k -? urea E reaction NH CO -. Solvin for E A ives 7 kj/. For equilibriu, we see that the velocities of both forward and reverse reactions are equal, leadin us to k the relationship K eq. Thus, k ( 7 k / 7. 0 )/( 70).0 0 k K eq. c) What is the value of the Gibbs free enery chane for the conversion of urea to aoniu and carbonate at 6 C? We can do this in two ways. One is fro ΔG 0 -RT lnk. The other is to copute E A for the reverse rate constant and then use the E RT difference between it and that for the forward rate constant to et ΔG. To do the latter, we need to use k Ae A to et A: 5 ( J ) {( J )( 4.5K )} K 7. 0 Ae or A (.40 0 ). Fro this we et A {( J 7 4 E. Usin this to copute E A for the reverse rate constant, we et ( ) A, rev 8.45 )( 4.5K )} K e, or E A,rev 5 kj/. The difference between the two activation eneries equals ΔG: - kj/. d) If a catalyst is introduced to the reaction, the rate constant k is found to increase by a factor of Which of the followin is true? Because both forward and reverse rate constants depend on k - also increases by a factor of 5000 the sae enery barrier, they will both be affected in the sae way by chanes in it. Thus, k k - decreases by a factor of will increase. The enery barrier heiht ust the Gibbs free enery chane in the reaction becoes ore neative the activation enery barrier in the forward direction increases the activation enery barrier in the reverse direction decreases decrease for the forward rate constant to speed up, so the reverse barrier heiht will also decrease.

5 Che/Bioche 47 Exa /8/08 Pae 5 of 7 4. Heolobin can be considered to be a square with four O bindin sites rouhly at the four corners of the square (see illustration). The bindin affinity of O to a sinle site has a value of K. When a bindin site is occupied with O, the affinity of all sites that are next-neihbors of that site increases by a factor of τ. (For exaple, with an O bound to site, sites and 4 would have bindin affinities of τk while site would have affinity K.) Use the shorthand with 0 representin no liand bound to a site and representin O bound to a site. Then the state with one O bound to sites and would be 00. a) How any different ways can the syste be arraned when there are two O s bound to a heolobin? 4 The possible two-bound states are 00, 00, 00, 00, 00 and 00. There are 6. b) Let the reference state be heolobin with no O bound and abbreviate the partial pressure of O as p. What is the statistical weiht of the state 00? Any liand bound will ive a factor of Kp to the statistical weiht. Any next-neihbor pairs will ive a factor of τ. There are two liands bound and one neihbor pair, so ω τk. 00 p c) Consider the case when τ (i.e. the bindin affinity for O doubles when a neihbor is occupied) and the partial pressure of O equals (K) -, i.e. /K. Is it ore likely that an individual heolobin ecule will have one O bound or ore likely that it will have 4 O s bound? Explain. The statistical weiht for one O bound is ω 000 Kp K There are four such states (000, 000, K , 000). The statistical weiht for four O bound is ω τ ( Kc) K. The total K statistical weiht for sinle oxyen bound heolobin is and that for the 4-bound heolobin is, so it is ore likely that a heolobin will have O s bound. d) How would you obtain the bindin curve for O to heolobin fro the inforation you have been iven. By bindin curve I ean the plot of the nuber of O s bound, on averae, to a heolobin ecule as a function of the partial pressure of O. You don t need to do a calculation; just describe the procedure in words. First, the su over states Z N i ω needs to be coputed. Then takin the derivative with respect to the i concentration p, ultiplyin by p and dividin by Z will ive the ean nuber of liands bound per p Z ln Z heolobin: ν. One could also do a coplicated alebraic calculation of all possible Z p τ ln p τ siultaneous equilibriu reactions.

6 Che/Bioche 47 Exa /8/08 Pae 6 of 7 5. Quantu dots are sall collections of 00 to 00,000 atos. Electrons can ove within the quantu dots in a way that an electron can be considered to be a quantu echanical particle in a box. Let s odel a quantu dot as a cube with sides of lenth a. In such a -D box, the Schrödiner equation will have solutions alon the perpendicular diensions x, y and z. a) Consider the diension x only. Draw the shape of the wave function solution to the Schrödiner equation for an electron in a quantu dot for the quantu nuber n. The wave function is Ψ ± π sin x a. Either the + or solution, or both, are correct. a Ψ 0 a x b) What is the enery of the electron shown in the plot in part a for a quantu dot that has a side of lenth equal to 50 Å? ( Å 0-0.) h n The enery of a particle in a box is E n. When n, this enery is, for an electron, 8a 4 (.66 0 J sec) () 0 ( 0 k)( 50 0 ) 6 0 E.7 0 J c) What would be the iu enery photon that could be absorbed by a quantu dot of side lenth equal to 50Å? h n Because possible enery levels o as E n, the two closest spaced enery levels are for n 8a and n. The lowest enery photon absorption will be between these two closest spaced enery levels. The photon enery is the difference between the two enery levels: h E E E 7. 0 J. 8a 4 ( ) ( J sec) ( 4 ) 0 8( 9. 0 k)( 50 0 ) d) Would you want to ake the quantu dot larer or saller than 50 Å on a side if you wanted to decrease the wavelenth of the iu enery photon that you coputed for part c? Explain. Saller. The enery levels (and thus their spacin) are proportional to /a. Because the enery hc of a photon is E hν, we see that to have shorter wavelenth of the photon, its enery ust λ increase. To et the enery levels of the trapped electron to increase, the value of a (size of the box) ust decrease, thus the dot ust be ade saller.

7 Che/Bioche 47 Exa /8/08 Pae 7 of 7 Scratch pad: For radin purposes: question 4 5 Su Adjusted score

PHY 171. Lecture 14. (February 16, 2012)

PHY 171. Lecture 14. (February 16, 2012) PHY 171 Lecture 14 (February 16, 212) In the last lecture, we looked at a quantitative connection between acroscopic and icroscopic quantities by deriving an expression for pressure based on the assuptions