AP Chemistry Practice Exam 2 ❮ 365(c) As the water evaporates, the concentration of sulfuric acid will increase, which leads to an increase in
the hydrogen ion concentration. An increase in the hydrogen ion concentration will promote the PbO 2
half-reaction more than it will change the Pb half-reaction. The result is a shift toward the left, which
will increase the voltage.
Give yourself 1 point for this answer.
(d) It is necessary to have a conductor present for the transfer of electrons. Lead(IV) oxide is an ionic
compound, which does not conduct electricity. Lead is a metal, which does conduct electricity.
Give yourself 1 point for this answer. Subtract 1 point if any of the answers has the wrong number
of significant figures.
Total your points for the various parts. There are 10 possible points.Question 2
(a) ΔHrxn° = [2(50)] - [2(90) + 1(0)] = - 80 kJ
The answer with appropriate units is worth 1 point. You do not need to get the exact answer, but your
answer should round to this one.
(b) The answer from part (a) equals the bonds broken minus the bonds formed.
[(2 NO) + (Cl–Cl)] - 2 [(N=O) + (NCl)] = - 80 kJ
[(2 NO) + (240)] - 2 [(610) + (200)] = - 80 kJ
NO = 650 kJ
The setup (broken - formed) is worth 1 point, and the answer is worth 1 point. You do not need to get
the exact answer, but your answer should round to this one. If your answer from part (a) was wrong,
but you use it correctly in this part, you still get your setup and answer points.
(c) The value is higher than the value given for an N=O bond. The NO molecule has an odd number of
electrons and this will alter its bonding.
You get 1 point for the correct observation, and 1 point for any explanation involving an odd
number of electrons. If you got the wrong answer for part (b), you can still get 1 or 2 points if you
used the answer correctly on this part.
(d) ΔSrxn° = [2(260)] - [2(210) + 1(220)] = - 120 J/K
The setup (products - reactants) is worth 1 point, and the answer is worth 1 point. You do not need
to get the exact answer, but your answer should round to this one.
(e) It is necessary to calculate the free-energy change.
ΔGrxn° = ΔHrxn° - TΔSrxn° = - 80 kJ - (298 K)(1 kJ/1000 J)(-120 J/K) = - 44 kJ
The negative value means the reaction is spontaneous.
You get 1 point for the prediction that the reaction is spontaneous. The setup (plugging into the
equation) is worth 1 point if you remember to change the temperature to kelvin and the joule-to-
kilojoule conversion (or vice versa for an answer in J). An additional 1 point comes from the answer.
If you got the wrong value in either part (a) or (b), but used it correctly, you will still get the point
for the answer. The free-energy equation is part of the material supplied in the exam booklet.
Total your points for the different parts. There is a maximum of 10 points possible. Subtract 1 point
if all your answers do not have the correct number of significant figures.21-Moore_PE02_p341-370.indd 365 31/05/18 1:54 pm