課程名稱︰普通化學二 課程性質︰系內必修 課程教師︰鄭原忠 開課學院：理學院 開課系所︰化學系 考試日期（年月日）︰2012/5/4 考試時限（分鐘）：120分鐘 是否需發放獎勵金：是 （如未明確表示，則不予發放） 試題 : General Chemistry (II) Mid-term Exam #2 Date: 5/4/2012 ***Refer to the last page for thermodynamic data and physical constants*** (108 points as a total score) 1. (25%) Acetic acid is a weak acid with pKa=4.85. (a) Give the chemical equation for the dissociation of acetic acid in water. Clearly point out conjugated acid-base pairs. Calculate ΔG° of this reaction. What does the value of ΔG° tell you about the diresctionali- ty of the reaction? (b) Calculate the pH value of a 0.1 M acetic acid solution. How many percen- tage of the acetic acid molecures are in their dissociated form? (c) 50.00 mL of acetic acid of unknown concentration is titrated with a 0.1000 M solution of NaOH. The equivalence point is reached after 39.30 mL of NaOH solution has been added. Calculate the original concentration of the acid and the pH at the half-equivalence point (19.65 mL NaOH so- lution added). (d) Sketch the titration curve in (c), be as specific as you can. (e) Give how to make a 50 mL buffer with pH=5.0 using a 0.1 M acetic acid solution and a 0.1 M NaOH solution. (HINT: skip to the next question and come back later if you do not recall the Handerson-Hasselbalch equation) 2.(12%) Consider the galvanic cell: Pt(s) | Cr(2+)(aq), Cr(3+)(aq) || Cu(2+)(aq) | Cu(s) (a) Diagram the cell, indicating the direction of electron flow and the mo- tion of ions in the salt bridge. (b) Write a balanced equation for the overall reaction in this cell. (c) Calculate the standard potential of the cell using the data in the last page. (d) Calculate the free energy change at 25°C when 0.1 mole of Cu plates out if all concentrations remain at the standard value of 1 M throughout the process. In this experiment, what is the maximum electrical work done by the cell on its surroundings? 3.(6%) An electrolytic cell consists of a pair of inert electrodes in a buffered solu- tion of pH=5.0 containing NiSO4 at a concentration of 1.0 M. (a) What product is formed at the cathode? (b) If the pH is changed to pH=1.0, what product will form at the cathode? 4.(10%) Given the following standard reduction potentials: Tl(3+) + e(-) ---> Tl(2+) E°= -0.37 V Tl(3+) + 2e(-) ---> Tl(+) E°= 1.25 V (a) Calculate the half-cell potential for the half-reaction Tl(2+) + e(-) ---> Tl(+) (b) Determine of the following disproportionate reaction will occur sponta- neously in aqueous solution in the standard condition. 2Tl(2+)(aq) ---> Tl(3+)(aq) + Tl(+)(aq) 5. A student decides to measure the solubility of lead sulfate in water and sets up the electrochemical cell Pb(s) | PbSO4 | SO4(2-)(aq, 0.0500 M) || Cl(-)(aq, 1.0 M) | AgCl | Ag(s) At 25°C the student finds the cell voltage to be 0.546 V. Gicen that the half- cell potential AgCl(s) + e(-) ---> Ag(s) + Cl(-)(aq) E°=0.222 V (a)(2%)Give the half-reaction and half-cell potential of the other half-cell. (b)(5%)Calculate the concentration of Pb(2+) in the solution. (c)(3%)Calculate the Ksp of PbSO4. 6.(15%) The temperature dependence of solubility for several salts are depicted on the right. [說明：本題的圖太難畫出來，所以用中文大略說明圖表重點。 (1) 橫軸：溫度。縱軸：溶解度(g per 100g water) (2) 共畫了5種鹽類，包含(c)中所述之Na2SO4 (3) 其圖形有：隨溫度上升溶解度上升者;隨溫度上升溶解度下降者;以及低溫時 隨溫度上升溶解度上升、高溫時隨溫度上升溶解度下降（即c中敘述）者 ] (a) The solubility of most salts increases as temperature increases. Does this indicate that most ionic dissolution processes are endothermic or exothermic? Give your rationale and explain why this is the case based on the microscopic picture of salt dissolution. (b) Derive an equation that relates the slope of the curve (solubility vs. temperature) to the enthalpy of dissolution. Clearly specify all the va- riables. (c) Na2SO4 shows increasing solubility with temperature below 32°C, but a decreasing solubility at higher temperature. Give an explanation to this phenomenon and sketch the dissolution enthalpy as a function of tempera- ture for Na2SO4. 7.(15%) The rate for the oxidation of iron(II) by cerium(IV) Ce(4+)(aq) + Fe(2+)(aq) ---> Ce(3+)(aq) + Fe(3+)(aq) is measured at the several different initial concentrations of the two rxns: [Ce(4+)] (mol/L) [Fe(2+)] (mol/L) Rate (mol/L-s) ------------------------------------------------------------------ 1.1 * 10^(-5) 1.8 * 10^(-5) 2.0 * 10^(-7) 1.1 * 10^(-5) 2.8 * 10^(-5) 3.1 * 10^(-7) 3.4 * 10^(-5) 2.8 * 10^(-5) 9.5 * 10^(-7) (a) Write the rate expression for this reaction. (b) Calculate the rate constant k, and give its units. (c) Consider a solution in which [Ce(4+)] is 3*10^(-5) M and [Fe(2+)] is 0.1 M, calculate the concentration of cerium(IV) ion after 60 seconds. (HINT: You can assume [Fe(2+)] is constant here). 8.(15%) Consider Michaelis-Menten mechanism for enzyme catalysis of a single subtract/ product. (a) Describe the reaction mechanism and five the chemical equations for the elementary steps. Clearly define all the chemical species in your for- mula. (b) Derive a equation that relates the rate of product formation to the con- centration of subtract and total enzyme concentration (i.e., the Michaelis-Menten equation). (c) Plot the rate as a function of substrate concentration. What kinds of information can be obtained from this plot? Specify them on the plot clearly. [註：本題中出現兩次 subtract 應為 substrate 的誤植] *** Thermodynamic functions: H=U+PV A=U-TS G=H-TS Thermodynamic equations: ΔG=ΔG°+RTlnQ ΔG°= -nFE° Standard reduction potentials at 25°C (volts): Cu(+) + e(-) ---> Cu(s) 0.522 O2(g) + 2H2O + 4e(-) ---> 4OH(-) 0.401 Cu(2+) + 2e(-) ---> Cu(s) 0.3402 PbO2(s) + H2O + 2e(-) ---> PbO(s) + 2OH(-) 0.28 Hg2Cl2(s) + 2e(-) ---> 2Hg(l) + 2Cl(-) 0.2682 AgCl(s) + e(-) ---> Ag(s) + Cl(-) 0.2223 Pb(2+) + 2e(-) ---> Pb(s) -0.1263 Sn(2+) + 2e(-) ---> Sn(s) -0.1364 Ni(2+) + 2e(-) ---> Ni(s) -0.23 Co(2+) + 2e(-) ---> Co(s) -0.28 Mn(OH)3(s) + e(-) ---> Mn(OH)2(s) + OH(-) -0.40 Cd(2+) +2e(-) ---> Cd(s) -0.4026 Fe(2+) + 2e(-) ---> Fe(s) -0.409 Cr(3+) + e(-) ---> Cr(2+) -0.424 Fe(OH)3(s) + e(-) ---> Fe(OH)2(s) + OH(-) -0.56 Physical constants: R=0.082 L-atm/K-mol =8.3 J/mol-K *** --

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