Download Chemistry Exam Review: Equations, Reactions, and Electrochemistry - Prof. Jason D. Kahn and more Exams Chemistry in PDF only on Docsity! Chemistry 271, Section 23xx University of Maryland, College Park General Chemistry and Energetics Prof. Jason Kahn Final Exam (200 points total) December 14, 2010 You have 120 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. 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.” Your Name: Your SID #: Viewing: Friday, December 17, 11-2 Chemistry 1110-1111 Useful Equations: ∆S – q/T ≥ 0 pH = – log([H+]) S = klnW ∆G° = – RTlnKeq ∆G = ∆H – T∆S PV = nRT Ka = [H+][A–]/[HA] ni/n0 = exp[–(ε i-‐ ε 0)/kT] ∆G = –nℱE °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 ℱ = 96500 C(oulomb)/mole Standard hydrogen electrode: 2 H+ (aq, 1 M) + 2 e– → H2 (g) E° = 0.000 V lnKeq = –∆H°/RT + ∆S°/R ln k = (–Ea/RT) + ln A 1 Volt = 1 Joule/Coulomb [A] = [A]0 – kt ln[A] = ln[A]0 – kt 1/[A] = 1/[A]0 + 2kt P(v)dv = Cv2exp(–mv2/2kT) dS = (cp/T) dT cp = qp/dT = dH/dT Chemistry 271, section 23xx Final Exam, 12/14/10 2/10 Score for the page 1. Multiple Choice (24 pts) (i, 4 pts) The Third Law of Thermodynamics gives us a reference state of zero entropy. What else is necessary for measuring absolute entropies of pure substances at non-‐zero temperature? (a) The First Law of Thermodynamics. (b) Measurements of heat capacity as a function of temperature. (c) Calculation of the free energy of formation from tabulated data. (d) Enumeration of microstates. (e) We cannot measure absolute entropy, only changes in entropy. (ii; 4 pts) A second-‐order rate constant (a) has units of time–1. (b) changes during the course of the reaction as the reactant is depleted. (c) can almost never be faster than diffusion-‐controlled. (d) has units of M–1s–1. (e) (c) and (d). (iii, 4 pts) The Steady State Approximation (a) is not useful in enzyme kinetics because substrate concentration is constantly changing. (b) can be applied only to reactions of the type A → B → C. (c) holds only at equilibrium, when the state of the system is steady. (d) includes rapid pre-‐equilibrium as a special case. (e) is applicable to the product of any rapid reaction. (iv; 4 pts) The half life t1/2 of a second order reaction A+A → B (a) = (ln2)/k (b) = 1/{k([A]0) 2} (c) = [A]0/k (d) is always shorter than the half life of a first-‐order reaction. (e) none of the above (v, 4 pts) The function of Photosystem I in plants is to (a) use light energy to reduce NADP+ to NADPH. (b) collect light energy and deliver it to Photosystem II. (c) oxidize water to provide light. (d) re-‐reduce the reaction center of Photosystem II. (e) reduce water to maintain a basic pH. (vi, 4 pts) Assimilation is defined as (a) using a substance as a terminal electron acceptor and making no further use of the product. (b) converting (typically through reduction) inorganic material into biomass. (c) nitrogen fixation and all other processes that split diatomic molecules. (d) absorption of enemies by an oppressive regime. (e) none of the above. Chemistry 271, section 23xx Final Exam, 12/14/10 5/10 Score for the page (c; 4 pts) Two possible failure pathways for photosynthesis are _____________________________________, if the electron is not transferred away from the excited state fast enough, or __________________________________ ___________________________________, if the electron is not removed from the immediate vicinity fast enough. (d; 9 pts) How does increased atmospheric CO2 cause global warming? 4. Kinetics and Thermodynamics (45 pts) (a; 7 pts) The Maxwell-‐Boltzmann distribution and Arrhenius equation state that the rate of an elementary reaction always increases with increasing temperature. So how is it possible for the equilibrium constant of a reaction A ⇌ B to decrease as temperature increases? Would the reaction be (circle one) exothermic or endothermic? Chemistry 271, section 23xx Final Exam, 12/14/10 6/10 Score for the page (b; 24 pts) Fill in the blanks. A catalyst can change the ________________________ and/or the _____________________________of a reaction but not its ___________________________ constant. The underlined “v2” factor in the Maxwell-‐Boltzmann distribution Cv2(exp(–mv2/2kT)) arises because of the ω(ε) factor in the Boltzmann distribution, the __________________________________________, which is the number of ways in which a particle can have _______________________________________. The underlined v2 comes from the _____________________________________ of a sphere of radius v. A rate law can be determined by inspection only for a(n) ________________________________________reaction. The “2” in SN2 comes from the fact that_______________________________________________________________________. (c; 8 pts) Sketch the Arrhenius plot used to measure activation energy Ea and the preexponential factor A. Label the axes. What is the molecular meaning of Ea according to collision theory for gas phase reactions? (d; 6 pts) In discussing gas-‐phase reactions we frequently discussed the need for a collision gas M to catalyze either dissociation or recombination reactions. It is easy to understand that the collision gas can deliver energy to break apart molecules (think of any parking lot). Why is the collision gas needed to allow atoms/molecules to recombine? Chemistry 271, section 23xx Final Exam, 12/14/10 7/10 Score for the page 5. Kinetics and the Steady State Approximation (35 pts) Leucine zipper proteins (Z) are monomeric in solution by themselves but they bind to DNA (D) as dimers, with the overall reaction being Z + Z + D Kbind Z2D There has been debate about the reaction mechanism. One possibility is that a monomer Z must dimerize to form an unstable intermediate Z2 that either readily falls apart or else binds to DNA. The proposed mechanism is as follows, where we ignore the reverse of the second step (this is reasonable if the complex is very stable): Z + Z k1 k−1 Z2 Z2 + D k2⎯ →⎯ Z2D (a; 5 pts) Write down the differential rate law for the appearance of Z2D. It includes the unknown [Z2]. (b; 12 pts) Apply the Steady State Approximation to the unstable Z2 intermediate to determine its steady state concentration during the binding reaction.