Question 1 (e)

assume that х << 0.200 М (0.200- х) ' х = \[Нзо+\] = З.5ХШ-3 М рн --1од\[НзО+\] = -log(3.5x10-3) - 2.45

Question 3 (b)

  • ΔH° = enthalpy of bonds broken - enthalpy of bonds formed

    Average Bond Energies Reactants Covalent Bonds = AH(reactlon) Atomization State + 21-420 Products Covalent Bonds C02) 11 g kcal/ mol 111 lg2 -CE(Product Bond Energies) -E(Reactant Bond Energies)(444 + 384) - + 23B) = -B2B - 634 = -1 g4 kcal/mol Correcting for the heat of condensation of 2 H20 product molecules 2(-10.5) -215 kcal/mol AHO = -194 -21=

Question 3 (d)

Reaction Zero order reaction First order reaction Second order reaction Order 2 Units of rate mol mol s mol S constant = mol la-Is- (mol ) o (mol 1--1 (mol 1--1 ) = mol-Il,s 2

Question 5 (c)

(c) 12(s) and Br2(l) can react to form the compound IBr(I). Predict which would have the greater molar enthalpy of vaporization, IBr(l) or Br2(l). Justify your prediction. IBr(I). Two reasons may be given. First, IBr is polar, and dipole-dipole forces would tend to increase the enthalpy of vaporization. Second, IBr should have stronger London dispersion forces because of the greater number of electrons in the larger IBr molecule. 1 point is earned for the correct choice with either or both of the acceptable reasons.

Question 5 (d)

(d) Explain why the hexane layer is light purple while the water layer is virtually colorless. Your explanation should reference the relative strengths of interactions between molecules of 12 and the solvents H20 and C H and the reasons for the differences. The hexane layer is purple because most of the 12 is dissolved in it. The entrance of the 12 into water requires disruption of the hydrogen bonds in water, which are much stronger than the London dispersion forces in hexane. Meanwhile, the London dispersion forces between 12 and hexane would be stronger than the London dispersion forces between I and water. (Water and 12 can also interact through a dipole-induced dipole force, but this attraction is insufficient to overcome the other differences noted above.) 1 point is earned for recognizing from the experimental observations that the iodine dissolved in the hexane. 1 point is earned for a correct explanation referencing the differences between water and hexane in their interactions with 12.

Question 5 (e)

3 Formal Charge Valence Electrons 7 7 NonBonding Val Electrons 6 6 Triiodide Bondinq Electrons 2 4/2 =

(ii) In which layer, water or hexane, would the concentration of 13- be higher? Explain. 13- would be more soluble in water because of the ion-dipole 1 point is earned for the correct interactions that would occur between the ions and the polar water choice and explanation. molecules. No such interactions are possible in the nonpolar hexane.

Question 6 (c)

cathode anode cathode = + ran ode cell

Question 6 (e)

(i) A student bumps the cell setup, resulting in the salt bridge losing contact with the solution in the cathode compartment. Is V equal to 0.47 or is V equal to 0? Justify your choice. V 0 V. The transfer of ions through the salt bridge will stop. A charge imbalance between the half-cells will prevent electrons from flowing through the wire. 1 point is earned for the correct choice with an appropriate explanation. (ii) A student spills a small amount of 0.5 M Na2S04(aq) into the compartment with the Pb electrode, resulting in the formation of a precipitate. Is V less than 0.47 or is V greater than 0.47 ? Justify your choice. V > 0.47 V. The sulfate ion will react with the Pb2+ ion to form a precipitate. This results in a thermodynamically favored anode half-cell reaction and hence a larger potential difference. The choice may also be justified using the Nernst equation. 1 point is earned for the correct choice with an appropriate explanation. cell RT \[Pb2+\] cell — nF \[Cu2+\] Decreasing the \[Pb2+\] will increase the cell voltage.

(iii) After the laboratory session is over, a student leaves the switch closed. The next day, the student opens the switch and reads the voltmeter. Is V less than 0.47 or is V equal to 0.47? Justify your choice. V < 0.47 V. Over time, \[Pb2+\] increases and \[Cu2+\] decreases, making both half-cell reactions less thermodvnamicallv favorable. The choice may also be justified using the Nernst equation. Increasing \[Pb2+\] and decreasing \[Cu2+\] decreases the cell voltage. The choice may also be justified by stating that the voltage is zero as a result of the establishment of equilibrium. 1 point is earned for the correct choice with an appropriate explanation.

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