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Unformatted text preview: Chemistry 271, Section lex
Prof. Jason Kahn University of Maryland, College Park Xem: SID {2; General Chemistry and Energetics Yen: Seeﬂen {1; Exam I (100 points total) March 7, 2011
N 2 l6“! You have 50 minutes for this exam. Exams written in pencil or erasable ink will not be regraded under any circumstances. Explanations should be concise and elear. 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: = [11+] [A‘1/ [HA] PH = 10g([H+]) Kb = [BH+][HO‘]/[B]
F=ma ‘e'”+1=0 PV=nRT Kw = [H*][HO‘] = 10‘14 pH = pKa + log([A‘]/[HA]) pH (ep) = (pKu/ + pKuz)/2
R = 0.08206 Latm/mole K 0 °C = 273.15 K pKu = —log(Ka) K p = KART)“ PZ/aj = 4n2/MG VXE = —aB/6t Honor Pledge: At the end ef the examination timeI please write out the following sentence and sign it, gr
talk to me abeut it: “I pledge on my honor that I have not given or received any unauthorized assistance on this examination +l point extra credit for ﬁlling in this box Chet tistrv 27! section Zixx Exam l 3/7/ 1 l, [20 pts) Fill in the Blanks 6 48 Read the whole sentence before you fill in the blanks. (a: 2 pts) KL. and KI. are equal if the “(A 1 does not
change during the reaction. r ‘ ‘1 L ,1 L/ f (b; 2 pts) The value of the reaction quotient Q approaches 411‘ \ muff C» as the reaction progresses. (c: 4 pts) At equilibrium, the rates of forward and reverse reactions are and are ‘_‘ 1—19 g g m 0 3+2; as each other. (ct,4 pts) The Cm ; “ﬂ @0456 of a weak acid is a strong base whose pKh can be predicted from the pKu of the weak acid using the  Kw \
equation p K2 7‘ 1 "' N )4, 1 . '4; Qty
(e; 2 pts) Kw is the equilibrium constant for the Q23 8/ 8(1 ’Céxs Gall/V” real/5““ Kill/1‘ (f; 2 pts) The concentration of a solid or a pure liquid is constant as long as some of it is still present, so it is .\ “N “A 92" the reaction quotient expression. —
~’+a
(g; 4 pts) Ideal gases are deﬁned by two properties: the particles have no {‘FCA ' w W “M .' ‘3
volume and they (11% V20 ! L2 attract each other. Score for the page '17 i 37 1
2 . \g/ (a; [8 pts) Explain the effect of each of th ollowin stresses on the position of the following equilibrium: W‘ “R
& Ki 9‘ )
equilibrium shifts and explain why. The five cases are independent of each other. \f' (i) N20 (g) is added to a mixture at equilibrium without changing the volume or the temperature. (ii) The volume of the equilibrium mixture is reduced at constant temperature. @ pNSSMﬂ IV‘CW ——) 67¢"lt'bnm INT}? {v m ——> 6R7”; h; turgid iii) The eq 'librium mixture ' ooled at constant volume. Fir Uhkwcék M1 a~ axe/r We.”
M7” ‘51 aw w T“ Questions (iv) and (v) are frequently answered incorrectly. Hint: Think about the Q and its ingredients. (iv) Gaseous argon (which does not react) is added to the equilibrium mixture and the total gas pressure and the temperature are maintained constant. (Hint: how would this be done?) ZE¥ K2 4')“ pry/(55cm: MTan Camila: U‘va—L
mu‘s‘f be. MGM}? wake hot/W MM {)Msmﬂ 6’ “MW?” 0%
(Larch/we SWH? l9 lire leﬂ'gg) Score for the page h .' 7 " 8 (v) Gaseous argon is added to the equilibrium mixture without changing the volume or temperature.
(Hint: what will the effect on Q be?) ’i'UM r/“Kwt a", «7 bar pa/hax fﬂffuny 4. “PM TMB
At equilibrium at 425.6 °C, a sample of ci‘slmethy—2ethycyclopropane is 73.6 % c0nverted to the trans
form (i.e. 73.6 % of the material is found in the trans form at equilibrium): I (
Cl 5 1 QQH 7.2
cis :2 trans A (b; 4 pts) Compge ‘ ibrium constant K for this reaction at 425 .6 °C. (c; 6 pts) Suppose that 0&5 moi of the cis compound is initially placed in a 15.0 L vessel and is then heated to 425.6 °C. Compute the equilibrium partial pressure of the cis and trans compounds. [Hintz you don’t 1
actually need the answer to part 12)“ to solve this] jh‘rhu z 525 M . Leta(MK " (Y2£‘+2?‘3lf_) 15.0 L.“ \3 4% 2,2.ol + on ,/
M “is @t
so KM Pﬁzoaséwm 21.48am @
, A
P4499 30.26‘( i‘2.Ol"=C9.$'3O aim @
(J » ~ Score for the page ,h 's 7 ‘ i ' l /7 i
AddBase Titration 3 22 '  ili i 14 Consider four separate titrations of a weak acid HA I
(pKa = 3.7) with four different weak bases B2, B4, ml
B6, and BS, with pr’s = 2,4,6, and 8 respectively. ; The family of titration curves is shown in the graph 5
at the right. The total concentration of acid is 0.1 M
and the ﬁnal concentration of base is 0.2 M. Ignore dilution. 2 o 0.025 0.05 0.075 0.1 0.125 0.15 0415' 0.2
Addodlaaso] (a; 6 pts) What is the dominant reaction occurring
in the solution as HA and B are mixed?
Explain why all the titration curves look the same on the left half of the graph. 69 HAfaHi “\Q ’4» [Ls/yui‘ie/v ban. 5o tacit/aid {u law; Mae mic: mwa
@oétﬁwu 5’ W5 4AMV (b; 6 pts) When the ﬁnal total concentrations of acid and base are 0.1 M and 0.2 M respectively, what are
the concentrations [B] and [EH], approximately (no ICE tablesl). Using the base dissociation
equilibrium,justify the circled pH values on the right hand side of the graph: show the calculation only once since it is the same for all of them. "fielzalm Elﬁntj=aiw W Wfﬂ‘
law Luz»; l’t/i {991W /4 1—;an +3: ‘
lie/w. POVU (7le pH 2 Nle Score for the page C . (c; 10 pts) The exact calculation of the titration curve of HA with B is complicated, especially at low
concentrations of HA and B. _We would like to solve for the six concentrations [HA], [A‘], [B], [BH+], [H], and [HO'] given the K” for HA, the K, for B, and the input concentrations CA and CB. Write
down the six equations that we would need to solve for the six variables. Q) [nappy] = a} @w LWJMJ®
@ [8W] 1 [K] 3 C5 @[MJ+ [3”] = [4] +[Hoj The molecule 2,3bisphosphoglycerate (2,3BPG) controls shortterm O
adaptation to high altitude. lt binds to Tstate hemoglobin about 250—fold _ 9 ﬂ)
more tightly than it does to Rstate hemoglobin. We know that the R and T
states also interconvert. (a; 8 pts) Given the three equilibria below, provide a linkage argument to show that the equilibrium constant for the interconversion of ROBPG and
TBPG is 25000.
" ROBPG "=1 R + BPG Kdiss,R = 3.0 x 10‘3 M R :3 T Km: 100
ToBl’G =5 T + BPG 101mm: 1.2 x 10‘5 M  RBPG ‘—‘ ToBPG KRTBPG = ?
Show that KRTBPG 25000: 3x‘o‘3 Score for the page is l ' 7/8 (b; 12 pts) The mathematics of linkage is complicated. We will consider simpliﬁed cases. First, consider
oxygen binding in the absence of BPG. The R state binds 02 about 100 time better than the T state:
R002: R + 02 Kan = 1.3 x 104 atm : Da] [(7,] not]
T002 .—‘ T + 02 Km = 1.3 x 10'2 atm
R : T Km: 100
lfwe ﬁnd in the end that [R] = 1.5 uM and [02] = 2.6 x 10'2 atm (assumed constant) use the equilibria
above to give the concentrations of R02,T, and TOz. The calculation is very straightforward:
since i give you the equilibrium concentrati ns ofR and 02 you do not need an ICE table. Calculate the fraction ofthe total hemoglobin that is bound to oxygen, i.e. calculate ([R002]+'[T002il)/(
[R002]+[T02J+[R]+[T]). ia'dw] + LT‘OL] (WMZ) Score for the page Chemit 7 se ' 2lxxEx I? /ll 8 (c; 10 pts) Next will consider the case when [BPG] is present. at high enough concentration so that all of the
hemoglobin is in the BPGbound form, and we will assess the effect of BPG on oxygen binding. We
assume that the dissociation constants for the R and T states are the same with and with0ut BPG. RBPGOzr: RBPG + 02 Km; = 1.3 x 104 atm 2 (ﬂ. 5P6J[ongp4. 0L]
TBPG'OZF‘ TBPG + 02 K0] = 1.3 X 10‘2 atm ROBPG :2 TBPG KRTBPG= 25000 Based on LeChatelier, circle whether more or \
versus in its absence. If [RBPG] = 10 nM and [02 = 2.6 x 10'2 atm [assumed constant) calculate the concentrations of RBPG02,TBPG, and TBPGOz. What fraction ofthe total hemoglobin is 0 IS bound to 02 in the presence of BPG bound to oxygen?
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