Chemistry 4A, Fall 2024 Practice Midterm 4 PDF

Potentially Useful Information (Thermodynamics) Kinetic energy: KE = mv2/2; Gravitational potential energy: Eg = mgh 1 3RT Result of the kinetic theory of gases: PV = Nmv 2 ; Root-mean-square velocity: vR...

Potentially Useful Information (Thermodynamics) Kinetic energy: KE = mv2/2; Gravitational potential energy: Eg = mgh 1 3RT Result of the kinetic theory of gases: PV = Nmv 2 ; Root-mean-square velocity: vRMS = v2 = 3 M 1 Mean free path of gas molecules: λ = 2nV πd 2 3/ 2  M   1 Mv 2  Maxwell–Boltzmann distribution: f (v) = 4πv 2  exp −   2πRT   2 RT   n  2 PV Van der Waals equation:  P + a   (V − nb) = nRT ; “Compressibility Factor”: z =   V   nRT V∞ P Work in reversible isothermal processes: w = −nRT ln = −nRT ln 1 V1 P∞ Permutation of n different items: n! = n(n−1)(n−2)…1 N! Combination: choosing L items from N items: N C L = (N − L )! L! Number of microstates for an ideal gas: Ω = gV NU 3N / 2 P Difference between actual G and G◦ for ideal gas: ∆PPref G = nRT ln Pref K2 ∆H   1 1  P2 ∆H   1 1  Van’t Hoff equation: ln =−  − ; ln =−  −  K1 R  T2 T1  P1 R  T2 T1  nB nB Molality: bB = = mA nA M A g = 9.8 m/s2; 1 atm ~101325 Pa ~760 mm Hg; 1 bar = 105 Pa R = 8.314 J mol−1 K−1 = 0.08206 atm L mol−1 K−1 kB = 1.381 × 10-23 m2 kg s-2 K-1 T (0 ◦C) =273.15 K; 1 cal = 4.18 J Density of liquids: ρ(H2O) = 1.00 g/mL; ρ(Hg) = 13.6 g/mL  Chemistry 4A, Fall 2024 Name, Section # ___________________________ ID # ___________________________ Practice Midterm 4 (100 points) 1. Multiple choices (Please solid-fill the bubbles as. 6 pts each question; only one correct answer) 1.1. A system went through a series of irreversible (spontaneous) and reversible processes and was eventually brought back to its initial state. We thus expect that for the system: 〇 ΔS = 0 〇 ΔS ≥ 0 〇 ΔS > 0 〇 ΔS < 0 1.2. Given that the molar specific heat capacity of Au is higher than Cu for the entire temperature range of 0−300K, we conclude that at 300 K: 〇 Au has higher absolute entropy per mole 〇 Cu has higher absolute entropy per mole 〇 Au and Cu have the same absolute entropy per mole 〇 No conclusion can be drawn regarding the entropy difference between Au and Cu 1.3. Given ∆ f H  (diamond) = 1.9 kJ/mol , S  (diamond) = 2.38 J ⋅ mol-1 ⋅ K -1 , and S  (graphite) = 5.74 J ⋅ mol-1 ⋅ K -1 : 〇 Diamond is thermodynamically stable at all temperatures 〇 Diamond is thermodynamically stable at low temperatures only 〇 Diamond is thermodynamically stable at high temperatures only 〇 Diamond is thermodynamically unstable at all temperatures 1.4. Which of the following is a common weak Brønsted-Lowry base? 〇 NaOH 〇 HCN 〇 NH4+ 〇 HCO3− 1.5. Which one of the solutions prepared below is a buffer? 〇 Mixing 1 L of 0.1 M NaCl and 1 L of 0.1 M HCl 〇 Mixing 1 L of 0.1 M NaOH and 1 L of 0.1 M acetic acid 〇 Mixing 1 L of 0.1 M NaOH and 1 L of 0.2 M acetic acid 〇 Mixing 1 L of 0.2 M NaOH and 1 L of 0.1 M acetic acid 1.6. At room temperature: Recall from our lecture that the osmotic pressure of a 1 M urea solution is 25 atm. If we consider the seawater as a 0.6 M NaCl solution, approximately how much pressure needs to be exerted to desalinate seawater by pushing it through a semipermeable (water-selective) membrane: 〇 8 atm 〇 15 atm 〇 30 atm 〇 60 atm Page 1 of 4  Chemistry 4A, Fall 2024 Name, Section # ___________________________ ID # ___________________________ 1.7. Which one of the below phase diagrams qualitatively matches that of CO2? P P P P Solid Liquid Solid Liquid Solid Liquid Solid Liquid 1 atm 1 atm Gas Gas Gas Gas 1 atm 1 atm T T T T 〇 〇 〇 〇 𝑚𝑚𝑚𝑚𝑚𝑚 2. (6 pts) Henry’s Law constant of CO2 in seawater is 𝑘𝑘𝐻𝐻 = 3.3 × 10−2 𝐿𝐿 ×𝑎𝑎𝑎𝑎𝑎𝑎. The atmospheric concentration of CO2 (as mole fraction xCO2) has risen from ~280 ppm (parts per million) in pre-industrial times to 410 ppm today. What are the expected molarities of CO2 in seawater, pre-industrial and current? 3. (12 pts) In one of the demos in our lectures, we showed that adding water to the below system shifted the equilibrium to the left (reactants): [Co(H 2 O)6 ]2+ (aq) + 4 Cl− (aq)  [Co(Cl) 4 ]2− (aq) + 6 H 2 O (a) (6 pts) Why is it incorrect to say that this was because adding water increased the concentration of H2O? Page 2 of 4  Chemistry 4A, Fall 2024 Name, Section # ___________________________ ID # ___________________________ (b) (6 pts) What is the correct explanation? 4. (16 pts) Given that at 800 K (1) CO2 (g) + H2 (g) ⇌ CO (g) + H2O (g) K1 = 0.15 (2) CuO (s) + H2 (g) ⇌ Cu (s) + H2O (g) K2 = 60 (a) (8 pts) Calculate ΔG° of reaction (1) (b) (8 pts) Calculate the equilibrium constant K3 for reaction: CuO (s) + CO (g) ⇌ Cu (s) + CO2 (g) at 800 K Page 3 of 4  Chemistry 4A, Fall 2024 Name, Section # ___________________________ ID # ___________________________ 5. (25 pts) A 1 M solution of acetic acid (CH3COOH; shorthanded as HAc) is titrated by 1 M NaOH. Given that for the acetic acid, K a = 1.8 × 10 −5 (pKa = 4.74), calculate: (a) (5 pts) pH of the starting HAc solution (b) (6 pts) pH of the solution as it is exactly half-titrated, i.e., half of the initial HAc was neutralized. (c) (6 pts) pH of the solution as it is titrated to the point that the concentration of sodium acetate (NaAc) is 10 times the concentration of acetic acid. (d) (8 pts) pH of the solution at the equivalence point, when equal amounts of NaOH and HAc are mixed. Page 4 of 4