Chemistry 271, Section 21xx Your Name: Prof. Jason Kahn University of Maryland, College Park Your SID #: General Chemistry and Energetics Your Section #: Exam I (100 points total) March 7, 2011 You have 50 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: Ka = [H+][A–]/[HA] pH = –log([H+]) Kb = [BH+][HO–]/[B] F = ma ¸eiπ + 1 = 0 PV = nRT Kw = [H+][HO–] = 10–14 pH = pKa + log([A–]/[HA]) pH (e.p.) = (pKa1 + pKa2)/2 R = 0.08206 L·atm/mole K 0 °C = 273.15 K pKa = –log(Ka) Kp = Kc(RT)∆n P2/a3 = 4π2/MG ∇×E = –∂B/∂t 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 boxChemistry 271, section 21xx Exam I, 3/7/11 2/8 Score for the page 1. (20 pts) Fill in the Blanks Read the whole sentence before you fill in the blanks. (a; 2 pts) Kc and Kp are equal if the does not change during the reaction. (b; 2 pts) The value of the reaction quotient Q approaches as the reaction progresses. (c; 4 pts) At equilibrium, the rates of forward and reverse reactions are and are as each other. (d; 4 pts) The of a weak acid is a strong base whose pKb can be predicted from the pKa of the weak acid using the equation . (e; 2 pts) Kw is the equilibrium constant for the . (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 the reaction quotient expression. (g; 4 pts) Ideal gases are defined by two properties: the particles have no volume and they attract each other.Chemistry 271, section 21xx Exam I, 3/7/11 3/8 Score for the page 2. (28 pts) Chemical Equilibria [Adapted from Oxtoby]. (a; 18 pts) Explain the effect of each of the following stresses on the position of the following equilibrium: 3 NO (g) ⇌ N2O (g) + NO2 (g) The reaction as written is exothermic (releases heat). In each case, briefly explain which way the equilibrium shifts and explain why. The five cases are independent of each other. (i) N2O (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. (iii) The equilibrium mixture is cooled at constant volume. 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?)Chemistry 271, section 21xx Exam I, 3/7/11 4/8 Score for the page (v) Gaseous argon is added to the equilibrium mixture without changing the volume or temperature. (Hint: what will the effect on Q be?) At equilibrium at 425.6 °C, a sample of cis-1-methyl-2-ethylcyclopropane is 73.6 % converted to the trans form (i.e. 73.6 % of the material is found in the trans form at equilibrium): cis$⇌"trans$⇌"$(b; 4 pts) Compute the equilibrium constant K for this reaction at 425.6 °C. (c; 6 pts) Suppose that 0.525 mol 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. [Hint: you don’t actually need the answer to part (a) to solve this.]Chemistry 271, section 21xx Exam I, 3/7/11 5/8 Score for the page 3. (22 pts) Acid-base equilibria Consider four separate titrations of a weak acid HA (pKa = 3.7) with four different weak bases B2, B4, B6, and B8, with pKb’s = 2, 4, 6, and 8 respectively. The family of titration curves is shown in the graph at the right. The total concentration of acid is 0.1 M and the final concentration of base is 0.2 M. Ignore dilution. (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. (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 [BH+], approximately (no ICE tables!). 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.Chemistry 271, section 21xx Exam I, 3/7/11 6/8 Score for the page (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 Ka for HA, the Kb for B, and the input concentrations CA and CB. Write down the six equations that we would need to solve for the six variables. 3. (30 pts) Hemoglobin and Linkage The molecule 2,3-bisphosphoglycerate (2,3-BPG) controls short-term adaptation to high altitude. It binds to T-state hemoglobin about 250-fold more tightly than it does to R-state 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 R•BPG"and"T•BPG"is"25000. R•BPG"⇌"R"+"BPG" Kdiss,R"="3.0"×"10D3"M" " R"⇌"T" " KRT="100"T•BPG"⇌"T"+"BPG" Kdiss,T"="1.2"×"10D5"M" " R•BPG"⇌"T•BPG" KRTBPG = ? " Show"that"KRTBPG ="25000:Chemistry 271, section 21xx Exam I, 3/7/11 7/8 Score for the page (b; 12 pts) The mathematics of linkage is complicated. We will consider simplified cases and ignore the fact that Hb binds 4 (not 1) molecules of O2. First, consider oxygen binding in the absence of BPG. The R state binds O2 about 100 times better than the T state: R•O2"⇌"R"+"O2""KO,R"="1.3"×"10D4"atm" " (this"is"a"dissociation"equilibrium)"T•O2"⇌"T"+"O2""KO,T"="1.3"×"10D2"atm"