5 Steps to a 5 AP Chemistry

AP Chemistry Practice Exam 2  341

Comparing Experiments 1 and 3: The concentration of ClO–is doubled, and the rate is doubled.

This leads to ClO–having an order of 1. You get 1 point for this reasoning.

Comparing Experiments 1 and 4: The concentration of OH–is doubled, and the rate is halved. This

leads to OH–having an order of –1. You get 1 point for this reasoning.

(b) Rate =k[I–][ClO–][OH–]–1This answer is worth 1 point. You will still get 1 point if you use incorrect

orders from part (a).

(c) Rearrange the rate law to: k=Rate/[I–][ClO–][OH–]–1The values from any of the experiments may

be plugged into this formula. Using Experiment 1:

You get 1 point for entering the values into the rearranged equation from part (b). You get 1 point for

the units.

You get 1 point for showing the products being lower than the reactants. You get 1 point for labeling

the activation energy. You get 1 point for correctly labeling ΔH.

Total your points for the different parts. There are 10 possible points. Subtract one point if you did not

report the correct number of significant figures in part (c).

Question 3.

(a) ΔH°rxn=[4(–393.5) +2(–241.83) +2(0)] – [4(135) +5(0)] =–2598 kJ

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 you should be able to round to this one.

(b) ΔS°rxn=[4(213.7) +2(188.72) +2(191.5)] – [4(201.7) +5(205.0)] =–216.6 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 yours should round to this one.

(c) ΔG°rxn=ΔH°rxn– TΔS°rxn=–2598 kJ–(298 K)(l kJ/1000J)(–216.6 J/K) =–2533 kJ

The setup (putting values into the equation) is worth 1 point if you remember to change the temper-

ature to kelvin and convert joules to kilojoules. 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.

(d) This requires a repeat of parts (a)–(c) using the values for H 2 O(l).

ΔH°rxn=[4(–393.5) +2(–285.84) +2(0)] – [4(135) +5(0)] =–2686 kJ

ΔS°rxn=[4(213.7) +2(69.94) +2(191.5)] – [4(201.7) +5(205.0)] =–454.1 J/K

k==1 22 0 200 0 200 0 200. /[. ][. ][. ]−−^11 6 10. s

activated complex

reactants

products

reaction progress

Ea

ΔHrxn

(d)