Chemistry 271, Section 22xx Your Name: Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Your Section #: Exam II (100 points total) November 7,2011 You have 53 minutes for this exam. Exams written in pencil or erasable ink will not be re-graded under any circumstances. Explanations should be concise and clear. I have given you more space than you should need. There is extra space on the last page if you need it. You will need a calculator for this exam. No other study aids or materials are permitted. Partial credit will be given, i.e., if you don't know, guess. Useful Equations: K" = [W][A-]/[HA] pH = -log([WD Kb = [HA][HO-]/[A-] Kw = [H+][HO-] pH = pKa + log [A-]/[HA] I'1GD = - RTlnKeq R = 0.08206 L·atm/mole K 0 DC = 273.15 K InKeq::: -M-J"/(RT) + I'1SD /R I'1S - qlT"? 0 R =8.314 J/mole K ::: 1.987 call mole K S::: kBlnW I'1G::: M1-TI'1S E::: 2: n; c; W =N!I(II n;!) n/no::: exp[-(c;-co)lkT] N::: 2: n; R ::: NAkB kB::: 1.38 X 10-23 J/K t' = t -vxlc2 Chemical standard state: 1 M solutes, pure liquids, 1 atm gases 2a Biochemical standard state: pH 7, all species in the ionic form found at pH 7 nano: 10-9 pico: 10-12 zepto: 10-21 Honor Pledge: At the end of the examination time, please write out the following sentence and sign it, or talk to me about it: "I pledge on my honor that I have not given or received any unauthorized assistance on this examination." +1 point extra credit for filling in this box and 23xx217 o Chemistry 27 1 sectioll 22xx Exam [( 1117I1! 1. (15 pts) Short Answer (2 pts each) Fill in the blanks: ( or Y' cv-h?kJ-(~+2.. ud.. -1'2. The Boltzmann distribution describes the distribution of etll..(r~ among PC{r b"t.le S; that is observed in each of the m (~~clJf-ck.f . h d' '"-I'~ " {~ of the e r~L/ "71/7~~'V-,...-\t at com h pnse t e pre omlll an t ---'---='-'C'\..!.77'1A.&.-'-.,..----"'... _ _ lG'CMI""",,.,,.,,,,,,,,,,-,IJ£JL..:~,,-_--,7,--lt'L"!..!....:W\.=--..:,---,-_, which is all of the microstates available at a given total __---=€.'-'.vt-'-e"",r.'--~7n?----(I pt each) r-, What is the sign of 6.G for any process that occurs spontaneously at constant P, T? ___-__--II--_C/This a special case of the (circle one) First Law €ond L!3 or Third Law of thermodynamics. +/ ~ The free energy has reached a minimum when a system has reached eat< .rf! 'h.l', .....'h 2. (15 pt~) van 't Hoff (a; 8 pts) Draw a van't Hoff plot for an endothermic disordering reaction. Label the axes, and show how you would determine 6,H0 and 6.50 from the plot. If there is a region where the reaction is spontaneous, label it. e~ttv",,~bk'~>O ~-: -1.°~O c(i'l~1-. bSO >0 Ll SO/((. 'It -IV.(~(->0 fclo-.. C4v~ S' ~ l"'-f~ c4ff / fj)6,. '1" [41 -:l.C-($ Score for the page, ____ _Chemistry 271 sect jon 22xx Exam II 1117111 B 6.1' ~ £. 3n L .)t-c, ""'-(b; 7 ptsJ'From the 'i!!n't Hoff equation, show that if you know the equilibrium constant at one temperature T" the equilibrium constant at temperature T2 is given by the following equation: 1Vl ct1-) -(~ ( 1(, J= (-~0) ( i~ -i,) lV\(~l.) := \~ (~'J -t-C:;J(t\ ~f~ ) -e¥~"l.1c-Wl,... ).~ cM0 ( j _.1 ) I.., {k:z,} ~ I~l.. ~ K, • e fl I. Tz. /" @ f:-f,vJ&. 1"vY [i Il Cp.... ~~~3. 120 pts) Boltzmann Distribution (a; 6 pts) We often speak of "kT" as the energy available from thermal motion. What is the population relative to the ground state (n / no) of a state with an energy of kT above the ground state? What about a state with an energy of 10 kT? t i. -0""" ~0. 004:>3}. Score for the page ____B6S C.£Chemistry 271 secti on 22" Exam II 1117111 Boltzmann Distributions at Three Temperatures, N = 10000 750 1-700 ~--~~~~ c::: .c 650 Ji ~ 0 0 400(')• .:.e 1Il .... o 300 ~ C1) c.-c::: o 200 '';::; m ::::I c. o 100 c.. o t-L.{OO~ @ , . o 2 4 6 8 10 12 Energy per particle (units of k . 300 K)B (b; 6 pts) The temperatures of the three curves are 400, 800, and 1200 K. Indicate which temperature is which on the graph above, and how do you know? (c; 2 pts) Identify the ground state populations for each distribution on the graph ~>/..1< (d; 3 pts) Sketch in the distribution for T = 1600 K. 0 r:r Q c.....,. vt/ffJr. r t'liJ.,..".J..-~ (e; 3 pts) Why are the areas under all four curves the same? ~ . I. I _, _ L. \:t:l;~ ,(-In tcN"1f ~T'-<-\~ eu....V(s <1.V( I-tW",", r.. kA ~{;.-51JrJ lf.... t.:\ ~/t,.c. .fv(~ JU,~;r-If' @ -\t..t arc..... ..... ".Jq. It... Cl,..{,,-..,.e -~f~. L-hW ",..~a-~(I\'£; -;: I i1 K~ tv.) Score for the page. ____Chemistry 271 seclion 12xx Exam II . I I i7i I I sn 4. (30 pts) Practical Thermodynamics (a; 25 pts) The water-gas shift reaction [CO (g) + H,0 (g) '" CO2 (g) + H2 (g)J is exothermic, which makes hydrogen production complicated because the syngas reaction that provides the feedstock is done at very hi gh temperature. From the data in the table for 25 °C, calculate~'-~;x> and€~SSUming that ~W'''' and ~o.," areSQnstant with T, calculate the free energy change G "0 t 200°C, the yquilibrium .. ,,- -' ~a~, and the temperature at whoIC t Tbt . constant IS. equaI h h e eqUl num to 1 fl...-P-f -R--' -(-393.5"-~)14/",,( ~ ( -Ito,S -l~I. fJ k.1/-< ~Wr in kJ/mol So in J/(mol K) ,~Wr (CO) =-110.5 So (CO) =197.7 ~~Wr (C02) =-393.5 So (C02 ) = 213.6 ~~W, (H2) =0 So (H2) =130.6 ~Mfor (HzO, g) =-241.8 So (HzO ,,R) =188.8 Mfo, (CH4) =-74.8 So (CH4) =1862 Mfo, (C, graphite) =0 So (C, graphite) =5.7 Mfo, (0,) =0 So (02) =2056nChemistry 271 seclion 22xx Exam II 11!7I11 (b; 5 pts) Why is choosing the temperature at which to carry out an exothermic reaction often a balancing act? Name another reaction that poses the same challenge. -()~ t.J~ {;((l h; re. ~ .ft-. few.,,~ Iv ~fee. ~ ~ ('-~~ t \ J ~L.(.f-~ ft...; ~,/,rh./'.'~ ~ -t~ {/ ~itS-I ~(/'u/lec, ~• [ . 1-/... .u_ .. Or S"f'V\I I~J O<.l r.Yu< T ,I'8l'fr1tMSn Oil@-~ ('{el,v-f" lci/l f'mtR-IV\ ~Clc\IV\(J lOw -(l> 1 1(~]J TIJ.J~ ~\\\~n &o\~C\\\C (\., (0 ~r~C,lC\l\\)\\ rJI \\J\\tt ~'" -+~+r..\ C\~,, <.\ct)\'\'\';/l\: ~~\A I )5. (20 pts)\tiemoglooin and Linkage ~~~1 ~t.il;." Consider the competitive binding of chloride (CI-) and oxygen (02 ) to hemoglobll1. We know that the R state of Hb binds O2 better than the T state. We are told that Ct binds better to the T state than to the R state. (6 pts) Draw the linkage cycle that shows that O2 binding favors the R state over the T state. @'-lV'-j l-ib ~l +-\b ~ -\\tI~ R.. + 6 2-~ (( . 01.. 1(\ 1(1. :;:. kJ I(~' ~~~ 1(\ > ILJ"'~~ @w-T"