Chem 1B Fall 2024 Exam 4 Notes

Exam Information

  • Duration: 90 minutes

  • Type: Closed book, one 8.5 x 11 sheet of notes (one side).

  • Materials Allowed: Scratch paper (provided), sheets of constants, useless data, formulae, and a periodic table.

  • Prohibited: White-out.

Thermodynamics

  • Work: w=−PΔVw = -P\Delta V

  • Heat: q=smΔTq = sm\Delta T

  • Enthalpy: ΔH=qp\Delta H = q_p

  • Enthalpy Definition: H=U+PVH = U + PV

  • Gibbs Free Energy: G=H−TSG = H - TS

  • Internal Energy: ΔU=q+w\Delta U = q + w

  • Enthalpy of Reaction: ΔH<em>rxn=∑ν</em>iΔH<em>of(products)−∑ν</em>iΔHof(reactants)\Delta H<em>{rxn} = \sum \nu</em>i\Delta H<em>{of}(products) - \sum \nu</em>i\Delta H_{of}(reactants)

  • Entropy of Reaction: ΔS<em>rxn=∑ν</em>iS<em>o(products)−∑ν</em>iSo(reactants)\Delta S<em>{rxn} = \sum \nu</em>iS<em>o(products) - \sum \nu</em>iS_o(reactants)

  • Gibbs Free Energy of Reaction: ΔG<em>rxn=∑ν</em>iΔG<em>of(products)−∑ν</em>iΔGof(reactants)\Delta G<em>{rxn} = \sum \nu</em>i\Delta G<em>{of}(products) - \sum \nu</em>i\Delta G_{of}(reactants)

  • Entropy and Microstates: S=kln⁡WS = k\ln W

  • Gibbs Free Energy Change: ΔG=ΔGo+RTln⁡Q\Delta G = \Delta G_o + RT\ln Q

Equilibrium

  • Equilibrium Constant: K<em>eq=exp(−ΔG</em>oRT)K<em>{eq} = exp(-\frac{\Delta G</em>o}{RT})

Electrochemistry

  • Cell Potential: E<em>cello=E</em>redo−EoxoE<em>{cell}^o = E</em>{red}^o - E_{ox}^o

  • Equilibrium Constant and Cell Potential: ln⁡K<em>eq=nFEo</em>cellRT\ln K<em>{eq} = \frac{nFE^o</em>{cell}}{RT}

Constants and Conversions

  • 1 Ampere: 1A=1JC1 A = 1 \frac{J}{C}

  • 1 Volt: 1V=1JC1 V = 1 \frac{J}{C}

  • 1 Coulomb: 1C=1A⋅s1 C = 1 A \cdot s

  • 1 Faraday: 1F=96,485Coulombs1 F = 96,485 Coulombs

Safety Questions

  • General lab safety procedures are emphasized for future science courses and application at home and work.

Multiple Choice Questions

  1. In a galvanic cell, reduction occurs at the cathode.

  2. Determining the number of electrons transferred in the reaction.

    • 3I2(s)+2Fe(s)→2Fe3+(aq)+6I−(aq)3 I_2(s) + 2 Fe(s) \rightarrow 2 Fe^{3+}(aq) + 6 I^-(aq)

    • 6 electrons are transferred.

  3. Cell Diagram for the reaction:

    • Mg(s)+2AgNO<em>3(aq)→2Ag(s)+Mg(NO</em>3)2(aq)Mg(s) + 2AgNO<em>3(aq) \rightarrow 2Ag(s) + Mg(NO</em>3)_2(aq)

    • The cell diagram is: Mg(s)∣Mg2+(aq)∣∣Ag+(aq)∣Ag(s)Mg(s) | Mg^{2+}(aq) || Ag^+(aq) | Ag(s)

  4. Determine EcelloE^o_{cell} for the reaction:

    • 2Ag++Mg→2Ag+Mg2+2 Ag^+ + Mg \rightarrow 2 Ag + Mg^{2+}

    • Half-reactions:

      • Mg2+(aq)+2e−→Mg(s)Eo=−2.356VMg^{2+}(aq) + 2 e^- \rightarrow Mg(s) \quad E^o = -2.356 V

      • Ag+(aq)+e−→Ag(s)Eo=0.800VAg^+(aq) + e^- \rightarrow Ag(s) \quad E^o = 0.800 V

    • Ecello=0.800−(−2.356)=3.156VE^o_{cell} = 0.800 - (-2.356) = 3.156 V

  5. Consider the cell:

    • Ni∣Ni2+(?M)∣∣Cu2+(0.136M)∣CuNi | Ni^{2+} (?M) || Cu^{2+} (0.136 M) | Cu

    • Half-reactions:

      • Ni2++2e−→NiEo=−0.257VNi^{2+} + 2e^- \rightarrow Ni \quad E^o = -0.257 V

      • Cu2++2e−→CuEo=0.340VCu^{2+} + 2e^- \rightarrow Cu \quad E^o = 0.340 V

    • Nernst Equation:

      • E=Eo−0.0592nlog⁡QE = E^o - \frac{0.0592}{n} \log Q

      • 0.621=(0.340−(−0.257))−0.05922log⁡[Ni2+][Cu2+]0.621 = (0.340 - (-0.257)) - \frac{0.0592}{2} \log \frac{[Ni^{2+}]}{[Cu^{2+}]}

      • 0.621=0.597−0.05922log⁡[Ni2+]0.1360.621 = 0.597 - \frac{0.0592}{2} \log \frac{[Ni^{2+}]}{0.136}

      • Solving for [Ni2+][Ni^{2+}] gives approximately 0.4M0.4 M

  6. Given the reduction potentials:

    • Mg+2+2e−→Mg−2.37VMg^{+2} + 2e^- \rightarrow Mg \quad -2.37V

    • V+2+2e−→V−1.18VV^{+2} + 2e^- \rightarrow V \quad -1.18V

    • Cu+2+e−→Cu+1+0.15VCu^{+2} + e^- \rightarrow Cu^{+1} \quad +0.15V

    • The reaction that will proceed spontaneously from left to right is:

      • V+2Cu+2→V+2+2Cu+1V + 2 Cu^{+2} \rightarrow V^{+2} + 2 Cu^{+1}

  7. Determine KeqK_{eq} at 25°C for the reaction:

    • Cu2+(aq)+Pb(s)⇌Pb2+(aq)+Cu(s)Cu^{2+}(aq) + Pb(s) \rightleftharpoons Pb^{2+}(aq) + Cu(s)

    • Half-reactions:

      • Pb2+(aq)+2e−⇌Pb(s)Eo=−0.125VPb^{2+}(aq) + 2 e^- \rightleftharpoons Pb(s) \quad E^o = -0.125 V

      • Cu2+(aq)+2e−⇌Cu(s)Eo=+0.337VCu^{2+}(aq) + 2 e^- \rightleftharpoons Cu(s) \quad E^o = +0.337 V

    • Ecello=0.337−(−0.125)=0.462VE_{cell}^o = 0.337 - (-0.125) = 0.462 V

    • ln⁡K=nFEoRT=2<br>96485<br>0.4628.314<br>298≈35.7\ln K = \frac{nFE^o}{RT} = \frac{2 <br>96485 <br>0.462}{8.314 <br>298} \approx 35.7

    • K=e35.7≈2×1016K = e^{35.7} \approx 2 \times 10^{16}

Polymer Chemistry

  1. Addition polymer formed from the monomer CH2=CHClCH_2=CHCl

  2. Stick/angle notation structure for CH<em>3CH(OH)CH(CH</em>3)C<em>2H</em>5CH<em>3CH(OH)CH(CH</em>3)C<em>2H</em>5

  3. Correct name for the structure given.

Isomers

  1. Two structures with a double bond are classified as geometric isomers.

  2. Two structures, one with flourine and one with chlorine, are classified as geometric isomers.

  3. Two structures, again with flourine and chlorine, are classified as geometric isomers.

  4. Classifying polymers as atactic, isotactic or syndiotactic and amorphous/soft or crystalline/hard

    • Polymer A is atactic & soft, polymer B is syndiotactic & hard

Redox Reaction Balancing

B. Balance the following redox reaction using the ½ reaction method.

  • H<em>3AsO</em>4+Fe→AsH3+Fe3+H<em>3AsO</em>4 + Fe \rightarrow AsH_3 + Fe^{3+} (ACIDIC solution)

  • Balancing steps:

    1. Write two ½ reactions.

    2. Balance atoms other than H or O in each ½ reaction.

    3. Balance O by adding H2OH_2O’s in each ½ reaction.

    4. Balance H by adding H+H^+ in each ½ reaction.

    5. Add e−e^- to balance charge in each ½ reaction.

    6. Multiply each ½ reaction by a factor so that e−e^- will cancel when the ½ reactions are added.

    7. Add the two ½ reactions together. Cancel out things as needed.

  • The balanced reaction is:

    • Reduction: H<em>3AsO</em>4+8H++8e−→AsH<em>3+4H</em>2OH<em>3AsO</em>4 + 8H^+ + 8e^- \rightarrow AsH<em>3 + 4H</em>2O

    • Oxidation: Fe→Fe3++3e−Fe \rightarrow Fe^{3+} + 3e^-

    • Overall: 3H<em>3AsO</em>4+8Fe+24H+→3AsH<em>3+8Fe3++12H</em>2O3H<em>3AsO</em>4 + 8Fe + 24H^+ \rightarrow 3AsH<em>3 + 8Fe^{3+} + 12H</em>2O

Electrolysis

C. A current of 10.0 amp is passed for 1.00 hour through an electrolysis cell containing a molten gold salt, AuCl3AuCl_3. How many grams of gold are deposited at the cathode?

  • [1F=96495c/mole−1F = 96495c/mol e^-, 1amp=1c/s1amp=1c/s]

  • Calculations:

    • 10Cs<br>3600s=36000C10 \frac{C}{s} <br>3600 s = 36000 C

    • 36000C96495C/mol=0.373mole−\frac{36000 C}{96495 C/mol} = 0.373 mol e^-

    • 0.373mole−3mole−molAu=0.124molAu\frac{0.373 mole^-}{3 \frac{mol e^-}{mol Au}} = 0.124 mol Au

    • 0.124molAu<br>196.97gmol=24.4gAu0.124 mol Au <br>196.97 \frac{g}{mol} = 24.4 g Au

Reduction Potential

D. Determine the standard reduction potential (EoE^o) at 298K for the half reaction

  • AgI(s)+e−→Ag(s)+I−(aq)AgI(s) + e^- \rightarrow Ag(s) + I^-(aq)

  • Use the following, KspK_{sp} for AgI is 8.51×10−178.51 \times 10^{-17} and Eo=+0.800VE^o = +0.800V for Ag+1+e−→Ag(s)Ag^{+1} + e^- \rightarrow Ag(s)

  • Calculations:

    • AgI(s)⇌Ag+(aq)+I−(aq)Ksp=8.51<br>10−17AgI(s) \rightleftharpoons Ag^+(aq) + I^-(aq) \quad K_{sp} = 8.51 <br>10^{-17}

    • Ag+(aq)+e−→Ag(s)Eo=+0.800VAg^+(aq) + e^- \rightarrow Ag(s) \quad E^o = +0.800 V

    • AgI(s)+e−→Ag(s)+I−(aq)AgI(s) + e^- \rightarrow Ag(s) + I^-(aq)

    • ΔGo=−RTln⁡K=−8.314<br>298ln⁡(8.51<br>10−17)=92600J/mol\Delta G^o = -RT \ln K = -8.314 <br>298 \ln (8.51 <br>10^{-17}) = 92600 J/mol

    • Eo=−ΔGonF=−9260096485=−0.96VE^o = \frac{-\Delta G^o}{nF} = \frac{-92600}{96485} = -0.96 V

    • The overall EoE^o is 0.800−0.96=−0.16V0.800 - 0.96 = -0.16 V

Organic Products

E. Give the major organic products in stick-angle notation.
F. Circle the chiral carbons in the two structures. Points will be deducted both for chiral carbons missed as well as non-chiral carbons circled. (8pts) Be neat in circling single carbons