AP® CHEMISTRY 2001 SCORING GUIDELINES Question 61.0 M NaOH from a stock solution of 3.0 M NaOH...

AP® CHEMISTRY 2001 SCORING GUIDELINES

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Question 6

(10 points) (a) The reaction is an oxidation-reduction (redox) reaction. 1 point To go from I − to I3

−, the iodine has to be oxidized, so credit is earned for the 1 point formula of any reasonable oxidizing agent (e.g., F2 , Co 3+, O2 , MnO4

− ) (b) (i) One point is earned for an “X” drawn on the curve at a point corresponding 1 point

to approximately 37-42 minutes (see graph below). This is where the curve levels off.

(ii) The instantaneous rate of the reaction at 20 minutes is the slope of the line 1 point tangent to the curve at 20 minutes. (See the graph below.)

• No credit is earned if the student confuses “rate” with “rate constant”. (b) Keep the initial concentration of S2O8

2− constant, change the initial concentration 2 points of I − by a known amount, and see what effect this change has on the measured rate of the reaction. This procedure should be repeated, but this time keeping the concentration of I − constant and changing the concentration of S2O8

2−.

Note: Points are earned for a method that involves controlling one reactant concentration and varying the other while comparing the reaction rates.

• No credit is earned if student confuses product and reactant concentrations.


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Question 6 (cont.)

(d) The general rate law expression is: rate = k [ I −]x [ S2O82−] y 1 point

The values of x and y can be determined by the experiments described 1 point in (c) above. These values can then be put into the equation with the rate and concentrations at a particular time in a particular trial, and the equation can then be solved for k.

• No credit is earned if student confuses rate law with equilibrium expression. • No credit is earned if student uses reverse reaction expression.

(e) On the graph, a curve is drawn that starts at the origin, rises more quickly 2 points than the original curve, and levels off at [I3

− ] = 0.0020 M. (See example drawn on the graph below.)

Note: One point is earned if the curve starts at the origin and rises more quickly than the original curve; the other point is earned if the curve levels off at the same [ I3

− ] as the original curve.


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12

Question 5

Total Score 10 Points

H+(aq) + OH−(aq) H2O(l)

5. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut , for the reaction represented above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q = mc∆T .

Assume the following. • Both solutions are at the same temperature before they are combined.

• The densities of all the solutions are the same as that of water.

• Any heat lost to the calorimeter or to the air is negligible.

• The specific heat capacity of the combined solutions is the same as that of water. (a) Give appropriate units for each of the terms in the equation q = mc∆T .

q has units of joules (or kilojoules or calories or kilocalories)

m has units of grams or kilograms

c has units of J g−1 °C−1 or J g−1 K−1 (calories or kilograms acceptable alternatives)

T has units of °C or K

1 point earned for any two units

2 points earned for all four units

(b) List the measurements that must be made in order to obtain the value of q .

• volume or mass of the HCl or NaOH solutions

• initial temperature of HCl or NaOH before mixing

• final (highest) temperature of solution after mixing

1 point earned for any volume (mass of reactant)

1 point earned for initial and final (highest) temperature

(∆T is not a measurement)



13

Question 5 (cont�’d.)

(c) Explain how to calculate each of the following. (i) The number of moles of water formed during the experiment

Since there is mixing of equal volumes of the same concentration and the reaction has 1:1 stoichiometry, moles of H

2O = moles of

HCl = moles NaOH. To determine the number of moles of HCl:

(volume HCl) !"

#$%

&

L 1HCl mol

!"

#$%

&

HCl mol1OH mol1 2

= mol H2O

OR

(volume NaOH) !!"

#$$%

&

L1NaOHmol0.1

!!"

#$$%

&

NaOHmol1OHmol1 2 = mol H2O

OR

OH2n = nHCl = nNaOH = VHCl × 1 M = VNaOH × 1 M

1 point earned for the number of moles of H2O using the stoichiometric relationship between HCl (or NaOH) and H2O

(ii) The value of the molar enthalpy of neutralization, ∆Hneut , for the reaction between HCl(aq) and NaOH(aq)

Determine the quantity of the heat produced, q, from q = mc∆T, where m = total mass of solution; divide q by mol H2O determined in part (c) (i) to determine ∆Hneut :

∆H neut = OH mol 2

q− OR

OHmol 2

q

(mol reactant can substitute for mol H2O)

1 point earned for q 1 point earned for ∆H

neut



14

Question 5 (cont�’d.)

(d) The student repeats the experiment with the same equal volumes as before, but this time uses

2.0 M HCl and 2.0 M NaOH. (i) Indicate whether the value of q increases, decreases, or stays the same when compared to the

first experiment. Justify your prediction.

The ∆T will be greater, so q increases. There are more moles of HCl and NaOH reacting so the final temperature of the mixture will be higher.

1 point earned for direction and explanation

Note: Arguments about increased mass are not acceptable because the total mass increase is negligible (the solutions have virtually the same density) and is not the driving force for increases in q.

(ii) Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut , increases, decreases, or stays

the same when compared to the first experiment. Justify your prediction.

Both q and mol H2O increase proportionately. However,

when the quotient is determined, there is no change in ∆Hneut

Molar enthalpy is defined as per mole of reaction, therefore it will not change when the number of moles is doubled.

1 point earned for correct direction and explanation

(e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect

would this have on the calculated value of the molar enthalpy of neutralization, ∆Hneut ? Justify your answer.

Heat lost to the air will produce a smaller ∆T. In the equation q = mc∆T a smaller ∆T will produce a smaller value for q (heat released) than it should. In the equation

∆Hneut = OHmol 2

q−

the smaller magnitude of q and the constant mol H2O means that

∆Hneut will be less negative (more positive).

1 point earned for correct direction and explanation

Notes: ∆H decreases because q decreases earns 1 point ∆T decreases because ∆H decreases earns 1 point No points earned for ∆T decreases therefore q decreases


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9

Question 5

5. Three pure, solid compounds labeled X , Y , and Z are placed on a lab bench with the objective of identifying each one. It is known that the compounds (listed in random order) are KCl , Na2CO3 , and MgSO4 . A student performs several tests on the compounds; the results are summarized in the table below.

Compound pH of an Aqueous

Solution of the Compound

Result of Adding 1.0 M NaOH to a

Solution of the Compound

Result of Adding 1.0 M HCl Dropwise to

the Solid Compound

X > 7 No observed reaction Evolution of a gas

Y 7 No observed reaction No observed reaction

Z 7 Formation of a white precipitate

No observed reaction

(a) Identify each compound based on the observations recorded in the table.

Compound X ______ Na2CO3 _______

Compound Y ______ KCl _________

Compound Z _____ MgSO4 _________

One point is earned for one correct identification, and a second point is earned for a second correct identification.

(No points are earned if all three identifications are the same compound; no second point is earned if two identifications are the same compound.)

(b) Write the chemical formula for the precipitate produced when 1.0 M NaOH is added to a solution of compound Z .

Mg(OH)2 One point is earned for the correct formula.

(c) Explain why an aqueous solution of compound X has a pH value greater than 7. Write an equation as part of your explanation.

CO32− reacts with water to form OH−.

CO3

2−(aq) + H2O(l) → OH −(aq) + HCO3−(aq)

One point is earned for identifying CO32−

as a base.

One point is earned for a correct equation.



10

Question 5 (continued)

(d) One of the testing solutions used was 1.0 M NaOH . Describe the steps for preparing 100. mL of 1.0 M NaOH from a stock solution of 3.0 M NaOH using a 50 mL buret, a 100 mL volumetric flask, distilled water, and a small dropper.

1,000 mL of 1.0 M NaOH contains 1.0 mol NaOH; therefore, 100. mL of 1.0 M NaOH contains 0.10 mol NaOH (i.e., 0.10 mol NaOH is needed for the solution)

volume of 3.0 M NaOH needed = 0.10 mol NaOH × 1,000 mL

3.0 mol NaOH

= 33 mL

Step 1: Use the buret to deliver 33 mL of the 3.0 M NaOH stock solution into the clean 100 mL volumetric flask.

Step 2: Add distilled water to the flask until the liquid level is just below the calibration line in the neck of the flask; swirl gently to mix.

Step 3: Use the small dropper to add the last amount of distilled water, drop by drop, until the bottom of the meniscus in the flask neck is level with the calibration line. Insert the stopper, and invert the flask to mix.

One point is earned for

using the buret to dispense 33 mL of

NaOH(aq).

One point is earned for

adding distilled water to the calibration

mark.

(e) Describe a simple laboratory test that you could use to distinguish between Na2CO3(s) and CaCO3(s).

In your description, specify how the results of the test would enable you to determine which compound was Na2CO3(s) and which compound was CaCO3(s) .

A water solubility test would work. Put a small amount of one substance in a beaker of distilled water. If the substance dissolves readily when stirred, then it is Na2CO3 ; if it does not dissolve, it is CaCO3 .

OR

A flame test would work. Dip a moistened wire into a sample of one of the substances and place the wire in the flame of a bunsen burner. If a bright orange-yellow color is observed, then the sample is Na2CO3 ; if a brick red color is observed, it is CaCO3 .

Note: The student does NOT have to perform a confirmatory test on

the other substance if one has already been identified with a test.

One point is earned for any reasonable test.

One point is earned for interpreting the results that

will identify one compound.



11

Question 6

6. Answer each of the following in terms of principles of molecular behavior and chemical concepts.

(a) The structures for glucose, C6H12O6 , and cyclohexane, C6H12 , are shown below.

Identify the type(s) of intermolecular attractive forces in

(i) pure glucose

Hydrogen bonding OR dipole-dipole interactions OR van der Waals interactions (London dispersion forces may also be mentioned.)

One point is earned for a correct answer.

(ii) pure cyclohexane

London dispersion forces

One point is earned for London dispersion forces.

(b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain.

The hydroxyl groups in glucose molecules can form strong hydrogen bonds with the solvent (water) molecules, so glucose is soluble in water. In contrast, cyclohexane is not capable of forming strong intermolecular attractions with water (no hydrogen bonding), so the water-cyclohexane interactions are not as energetically favorable as the interactions that already exist among polar water molecules.

OR • Glucose is polar and cyclohexane is nonpolar. • Polar solutes (such as glucose) are generally soluble in

polar solvents such as water. • Nonpolar solutes (such as cyclohexane) are not soluble in

the polar solvent.

One point is earned for explaining the solubility of glucose in terms of hydrogen bonding or dipole-dipole

interactions with water.

One point is earned for explaining the difference in the polarity of

cyclohexane and water.

OR

One point is earned for any one of the three concepts; two points are earned

for any two of the three concepts.



12


(c) Consider the two processes represented below.

Process 1: H2O(l) → H2O(g) ∆H ° = + 44.0 kJ mol−1

Process 2: H2O(l) → H2(g) + 21 O ( )2

g ∆H ° = + 286 kJ mol−1

(i) For each of the two processes, identify the type(s) of intermolecular or intramolecular attractive

forces that must be overcome for the process to occur.

In process 1, hydrogen bonds (or dipole-dipole interactions) in liquid water are overcome to produce distinct water molecules in the vapor phase.

In process 2, covalent bonds (or sigma bonds, or electron-pair bonds) within water molecules must be broken to allow the atoms to recombine into molecular hydrogen and oxygen.

One point is earned for identifying the type of intermolecular force involved

in process 1.

One point is earned for identifying the type of intramolecular bonding

involved in process 2.

(ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation.

When water boils, H2O molecules break apart to form hydrogen molecules and oxygen molecules.

I disagree with the statement. Boiling is simply Process 1, in which only intermolecular forces are broken and the water molecules stay intact. No intramolecular or covalent bonds break in this process.

One point is earned for disagreeing with the statement and providing a

correct explanation.



13


(d) Consider the four reaction-energy profile diagrams shown below.

(i) Identify the two diagrams that could represent a catalyzed and an uncatalyzed reaction pathway for the same reaction. Indicate which of the two diagrams represents the catalyzed reaction pathway for the reaction.

Diagram 1 represents a catalyzed pathway and diagram 2 represents an uncatalyzed pathway for the same reaction.

One point is earned for identifying the correct graphs and indicating which

represents which pathway.

(ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation.

Adding a catalyst to a reaction mixture adds energy that causes the reaction to proceed more quickly.

I disagree with the statement. A catalyst does not add energy, but provides an alternate reaction pathway with a lower activation energy.

One point is earned for disagreeing with the statement and providing

an explanation.

AP® CHEMISTRY 2008 SCORING GUIDELINES (Form B)

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Question 2

A(g) + B(g) → C(g) + D(g)

For the gas-phase reaction represented above, the following experimental data were obtained.

Experiment Initial [A] (mol L−1)

Initial [B] (mol L−1)

Initial Reaction Rate (mol L−1 s−1)

1 0.033 0.034 6.67 × 10−4

2 0.034 0.137 1.08 × 10−2

3 0.136 0.136 1.07 × 10−2 4 0.202 0.233 ?

(a) Determine the order of the reaction with respect to reactant A. Justify your answer.

Between experiments 2 and 3, [B] stays the same and [A] is quadrupled, but the initial reaction rate stays the same. This means that the initial reaction rate is not dependent on [A], so the reaction is zero order with respect to A. (May also justify using mathematics as shown in part (b).)

One point is earned for the correct order and for the justification.

(b) Determine the order of the reaction with respect to reactant B. Justify your answer.

2

1

raterate

= 2 2

1 1

[A] [B]

[A] [B]

yx

yx

k

k

2

41.08 106.67 10

−

−××

= (0.034) (0.137) (0.033) (0.034)

x y

x yk

k where x = 0

16.2 = (4.03) y

y = 2, so the reaction is second order with respect to B

OR

Between experiments 1 and 2, [A] stays the same, [B] is multiplied by 4, and the initial reaction rate is multiplied by 16. This means that the reaction is second order with respect to B.

One point is earned for the correct order and for the justification.

(c) Write the rate law for the overall reaction.

rate = k [B]2 One point is earned for the correct rate law (or a rate law consistent with the answers in part (a) and part (b)).

AP® CHEMISTRY 2008 SCORING GUIDELINES (Form B)

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(d) Determine the value of the rate constant, k , for the reaction. Include units with your answer.

Using experiment 2:

rate = k [B]2

k = 2

rate[B]

= 4 1 1

1 26.67 10 mol L sec

(0.034 mol L )

− − −

−× = 0.577 M −1 sec−1

One point is earned for the correct numerical value of the rate constant.

One point is earned for the correct units.

(e) Calculate the initial reaction rate for experiment 4.

rate = k [B]2

rate = (0.577 M −1 sec−1) × (0.233 mol L−1)2

= 3.13 × 10−2 mol L−1 sec−1

One point is earned for the correct answer, including units.

(f) The following mechanism has been proposed for the reaction.

Step 1: B + B → E + D slow

Step 2: E + A →← B + C fast equilibrium

Provide two reasons why the mechanism is acceptable.

(1) When steps 1 and 2 are added together, the overall reaction is A + B → C + D. This is the stoichiometry that was given for the overall reaction.

(2) The rate-determining step (slow step) is consistent with the rate law because only reactant B occurs in the rate law and it occurs to the power of 2, which is the number of B molecules colliding in the rate-determining step.

(3) The rate-determining step is consistent with the rate law because A is absent from the rate-determining step and the reaction is zero order—i.e., reactant A does not appear in the rate law.

One point is earned for each correct reason, with a maximum of 2 points.

(g) In the mechanism in part (f), is species E a catalyst, or is it an intermediate? Justify your answer.

Species E is an intermediate; it is formed in step 1 and consumed in step 2.

AND/OR

Species E is not a catalyst because a catalyst occurs as a reactant in an earlier step and is then reproduced as a product in a later step.

One point is earned for the correct answer with justification.

AP® CHEMISTRY 2001 SCORING GUIDELINES Question 61.0 M NaOH from a stock solution of 3.0 M NaOH...

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