Exam2_example_key - Exam 2 (PChem for the Biosciences –...

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Unformatted text preview: Exam 2 (PChem for the Biosciences – CHM3400), Nov. 6, 2009 Name:___________ KEY________________________ UFID:_____________________ Instructions: Use a pen, NOT a pencil to answer the questions. Fill in your name and UFID here above. Give your answers in the blank spaces following each question/sub‐question. If you need more space, use the back side, clearly indicating this on the front page where the question is given. On the back side clearly refer to the question that you are answering (e.g. Q. 1). The total credit for this exam is 100 points, but there are 106 available. The credit of each sub‐question is indicated in bold (credit). Numeric values should be given with a sufficient number of digits and the units must be indicated. Useful tables, physical constants and formulae for this test are given at the end of the exam as an Appendix – make sure that you check the content of the Appendix before reading the questions. Be concise with your answers, but remember that you are graded on the information that you put down on paper – if intermediate steps in a calculation are missing and the result is incorrect, no credit can be given. Always show the equations and units that you use. Good luck! Conceptual questions: 1. BONUS QUESTION: When mixing two gases A and B, what are the plots for ΔmixH, ΔmixS and ΔmixG, as a function of mole fraction of A (x1). (6 pts) 2. Shown below is the phase diagram for water. The dashed lines represent the phase diagram upon adding a salt to the solution. Explain why the phase diagram has shifted lines, and extrapolate what this means in terms of the boiling and melting points compared to pure water. (8 pts) 1 Upon mixing, the free energy (=chemical potential) of the solution is lowered relative to the other phases (vapor, solid). This means that the phase diagram for the liquid phase is expanded. This results in an increase in the boiling point and a lowering of the melting point compared to the pure liquid. 3. Shown below are some dielectric constant for liquids. State the equation you would use to calculate the attractive force between a cation and an anion, which makes use of the dielectric constant. When comparing H2O and C6H6, where would this attractive force be larger, if everything else remains the same? (7 pts) F= q Na + qCl − 4πε 0εr 2 The attractive force would be larger in a C6H6 solution. 4. The activity coefficient relates the thermodynamic solubility product to the apparent solubility product. It is stated that γ=1 for infinitely dilute solutions. What process occurs in the solution when γ<1. (4 pts) Ion pair formation 2 5. For the equilibrium reaction shown below, state the rate of change of A in terms of appearance and depletion (4). Derive the equilibrium constant in terms of k1 and k‐1 from this (4). (8 pts) d [ A] = −k1[ A] + k −1[ B] dt At equilibrium: k1[ A] = k −1[ B ] [ B ] k1 = =K [ A] k −1 6. Below is shown a Van’t Hoff plot. In terms of enthalpy, rationalize what type of a reaction this is (4). How would you determine the entropy from this plot? Again show the mathematical value (4). (8 pts) The slope is positive. Since the slope represents ‐ΔrH/R, this means that the enthalpy change is negative ⇒ exothermic reaction. The entropy can be determined from the intercept. lnK (for 1/T =0): ΔrS/R. 7. Below is shown an Arrhenius plot. In this case, what does the slope represent? (4) Why could a positive slope never be observed in this case? (4) (8 pts) 3 Slope = ‐Ea/R. A positive slope would indicate a negative activation barrier (Ea) to a reaction. 8. For the two metals Ag and Mn, which way would you set up the galvanic cell below to obtain a flow of electrons, explaining what occurs at the anode and cathode. (6) What electromotive force (emf) would you measure for 1 M concentration solutions? (3) (see appendix) (9 pts) Cathode: Ag+ + e‐ → Ag Cathode: Mn → Mn2+ + 2 e‐ E0 = 0.8‐(‐1.180)=1.980 V 9. For ligand L binding to a protein you are told that the fractional concentration of L bound to P is Y= n[ L] [ L] + K Plot Y vs. [L], and explain on the graph how you find n and K. (8 pts) 4 Numeric problems: 10. For: A + B → C. Write the kinetics expression for the appearance of C as a function of [A] and [B] (4). Rationalize the units for k (4). (8 pts) Rate / M s‐1 [A] / M [B] / M 1.2x10‐5 0.02 0.01 0.3x10‐5 0.02 0.005 0.6x10‐5 0.01 0.01 d[C]/dt = k[A][B]2 Units for k: M s‐1 = k M3 ⇒ k (M2 s‐1) 11. Radioactive decay The half‐life for 14C is 5,720 years. Calculate the rate constant for this reaction in years‐1 (6). Let’s say that the natural background radiation due to 14C is 500 cts/s. An archeological excavation exhibits a radiation level of merely 10 cts/s. How old is this sample? (6) (Hint: concentrations are proportional to radioactivity levels). (12 points) ln 2 ln 2 0.693 ⇒k = = = 1.21 ⋅10 −4 ( years −1 ) k t1/ 2 5,720 [ A]0 500 ln ln [ A]t [ A]t ln = 10 = 32,330( years) [ A] = − kt ⇒ t = k 1.21 ⋅10 − 4 0 t1/ 2 = 5 12. Nitric oxide formation Nitric oxide is formed in car engines by the reaction of N2 with O2: N2(g) + O2(g) → 2 NO(g) Calculate the equilibrium constants for this reaction at 25oC (10) and at 1700 oC (10) using Table 4.3 in the Appendix, and knowing that ΔrH0 = 180.8 (kJ mol‐1). (20 points) o o Δ r G o = ΣvΔ f G ( products ) − ΣvΔ f G (reac tan ts ) = 2xΔGf(NO) ‐ [ΔGf(N2)+ΔGf(O2)] = 2x(86.7) – [0+0] = 173.4 (kJ mol‐1) ΔrGo = ‐RT lnK ΔrG0 173.4 ⋅103 ( Jmol −1 ) ln =− = −69.9 K =− 8.314( JK −1mol −1 ) ⋅ 298.2( K ) RT K = 4.22 x 10‐31 Apply the Van’t Hoff equation to determine K2 at 1700 oC: 0 ⎞ ⎛ ln K 2 = Δ r H ⎜ 1 − 1 ⎟ ⎟ ⎜ K1 ln R K2 4.22 ⋅10 −31 ⎝ T1 T2 ⎠ 180.8 ⋅103 ⎛ 1 1⎞ = − ⎜ ⎟ 8.314 ⎝ 298.2 1973.2 ⎠ = 3.2 x 10‐4 Appendix: Boltzmann’s constant: kB = 1.38 x 10‐23 J K‐1 (= kg m2 s‐2 K‐1) Avogadro’s number: NA= 6.02214 x 1023 mol‐1 Ideal gas constant: R = 8.314 J K‐1 mol‐1 Speed of light in vacuum c = 3 x 108 m s‐1 eV = 1.602 x 10‐19 J 6 7 ...
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This note was uploaded on 05/29/2011 for the course CHM 3400 taught by Professor Seabra during the Spring '08 term at University of Florida.

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