Chemistry 271, Section 21xx Your Name: University of Maryland, College Park Your SID #: General Chemistry and Energetics Prof. Jason Kahn Exam II (100 points total) April 13, 2009 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. Numbers to Live By: ∆S – q/T ≥ 0 pH = – log([H+]) E = mc2 S = klnW ∆G = ∆H – T∆S PV = nRT Ka = [H+][A–]/[HA] ∆G° = – RTlnKeq ¸eiπ + 1 = 0 W = N!/(∏ ni!) ni/n0 = exp[–(Ei-E0)/kT] TM = ∆H°/[∆S°+ R ln(CT/4)] °C = °K – 273.15 R = 8.314 J/mole K E = E° – 2.303(RT/nℱ)log10Q 2.303RT/ℱ = 0.0592 Volts at 25 °C ∆G = –nℱE ℱ = 96500 C(oulomb)/mole 1 V = 1 J/C a2 + b2 – 2abcosC = c2 Chemical standard state: 1M solutes, pure liquids, 1 atm gases Biochemical standard state: pH 7, all species in the ionic form found at pH 7 lnKeq = –∆H°/(RT) + ∆S°/R 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.”Chemistry 271, section 21xx Exam II, 4/13/09 2/7 Score for the page 1. van’t Hoff plots and DNA Thermodynamics (35 pts) (a; 18 pts) Sketch a single van’t Hoff plot comparing an exothermic-disordering reaction and an endothermic-disordering reaction, with the two having the same value of ∆S°. Label your axes. Which reaction is thermodynamically spontaneous (favorable) at some but not all temperatures? Indicate on the graph the temperature range over which that reaction is spontaneous. Give an example of each type of reaction.Chemistry 271, section 21xx Exam II, 4/13/09 3/7 Score for the page (b; 10 pts) The melting curve below shows the absorbance as a function of temperature for a short oligonucleotide. Recall (or believe) that the nearest-neighbor enthalpy change for base pair formation is always negative and that TM decreases as concentration decreases. Given these facts, we could make a longer oligonucleotide that has exactly the same melting temperature as the one on the curve shown. On the graph, draw the resulting melting curve. Below, briefly give your reasoning for its shape. (c; 7 pts) In DNA melting experiments, ∆G°37 can actually be determined with better precision (lower % error) than ∆H° or ∆S°, even though ∆G°37 = ∆H° + 310*∆S°. Why is this initially surprising, and how is it possible?Chemistry 271, section 21xx Exam II, 4/13/09 4/7 Score for the page 2. Free Energy and Equilibrium (35 pts) (a; 22 pts) For the reaction A ⇌ B + C, ∆H° = –75 kJ/mol and ∆S° = –120 J/mol K, and as usual we assume that ∆H° and ∆S° are constant. Calculate the free energy change and the equilibrium constant at 25 °C and at 150 °C. If we add in more A at constant temperature, what happens to the reaction mixture as it equilibrates, and how does the equilibrium constant change? ∆G° at 25 °C: Keq at 25 °C: ∆G° at 150 °C: Keq at 150 °C:Chemistry 271, section 21xx Exam II, 4/13/09 5/7 Score for the page (b; 5 pts) The equation ∆G = ∆G° + RTlnQ was derived from the idea that the molar free energy Gof a substance is given by G = G0+ RT ln P, where G° is the standard state free energy (at 1 atm) and P is the partial pressure of the gas. We will not reproduce the details of the derivation here, but we emphasized that the increase in Gwith partial pressure makes intuitive sense even if the functional form is not obvious. What reasoning did we give for the dependence (i.e., why does Gincrease with partial pressure)? (c; 8 pts) The graph below shows the total free energy of a reaction mixture as a function of the mole fraction X of products. Indicate the point at which equilibrium is reached. Sketch in the aspects of the graph that express the ideas that ∆G is negative for spontaneous changes in reactant and product concentrations and that ∆G = 0 for small changes at equilibrium.Chemistry 271, section 21xx Exam II, 4/13/09 6/7 Score for the page 3. Redox Reactions (30 pts) (a; 16 pts) Consider the reduction half-reactions below, combined to make a voltaic cell, operating in basic solution. Give a balanced equation for the overall spontaneous reaction and calculate its voltage under standard conditions (1 M everything). Which is the anode reaction and which is the cathode reaction? ZnO(s) +H2O(ℓ) + 2 e– → Zn(s) + 2 HO–(aq) E° = –1.25 V Ag2O(s) → Ag(s) [You balance this one] E° = 0.34 VChemistry 271, section 21xx Exam II, 4/13/09 7/7 Score for the page (b; 11 pts) Practical Electrochemistry. Circle whether a car battery is acting as a voltaic cell or an electrolytic cell under the following four conditions: 1) You are starting the car as usual. Voltaic Electrolytic 2) You are driving normally and the alternator is working. Voltaic Electrolytic 3) The engine is off and the battery is powering your monster car stereo and portable refrigerator at the beach, much to everyone’s delight. Voltaic Electrolytic 4) The friendly AAA mechanic is jump-starting your car later that evening after the music has wound down and the soda has warmed up. Voltaic Electrolytic Give one reason that we can’t just put eight AA batteries in series to provide the same 12V that a car battery provides: (c; 3 pts) Why does oxidation happen at the anode of an electrochemical cell? Page Score 1 2 3 4 5 6 7