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ap05_sg_chemistry_form_b

AP? Chemistry

2005 Scoring Guidelines

Form B

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Question 1

K a =

[H O][OCl]

[HOCl]

= 3.2 × 10?8

Hypochlorous acid, HOCl, is a weak acid in water. The K a expression for HOCl is shown above.

(a) Write a chemical equation showing how HOCl behaves as an acid in water.

HOCl(aq) + H2O(l) → OCl?(aq) + H3O+(aq)One point is earned for the correct chemical equation.

(b) Calculate the pH of a 0.175 M solution of HOCl.

NaOH(aq).

HOCl(aq) + OH?(aq) → OCl?(aq) + H2O(l)One point is earned for both of the correct reactants.

One point is earned for both of the correct products.

(d) In an experiment, 20.00 mL of 0.175 M HOCl(aq) is placed in a flask and titrated with 6.55 mL of 0.435 M

NaOH(aq).

(i) Calculate the number of moles of NaOH(aq) added.

Question 1 (continued) Calculate

[H3O+] in the flask after the NaOH(aq) has been added. (ii)

(iii) Calculate [OH?] in the flask after the NaOH(aq) has been added.

Question 2

Water was electrolyzed, as shown in the diagram above, for 5.61 minutes using a constant current of

0.513 ampere. A small amount of nonreactive electrolyte was added to the container before the electrolysis began. The temperature was 298 K and the atmospheric pressure was 1.00 atm.

(a) Write the balanced equation for the half reaction that took place at the anode.

2 H2O(l) → O2(g) + 4 H+(aq) + 4 e?One point is earned for the correct half reaction.

(b) Calculate the amount of electric charge, in coulombs, that passed through the solution.

diagram?

When water decomposes according to the balanced chemical equation 2 H2O(l) → O2(g) + 2 H2(g), twice as many moles of hydrogen are produced than moles of oxygen.One point is earned for the correct explanation based on the

stoichiometry of the

decomposition reaction.

Question 2 (continued)

(d) Calculate the number of moles of H 2(g ) produced during the electrolysis.

(e) Calculate the volume, in liters, at 298 K and 1.00 atm of dry H 2(g ) produced during the electrolysis. (f) After the hydrolysis reaction was over, the vertical position of the tube containing the collected H 2(g ) was

adjusted until the water levels inside and outside the tube were the same, as shown in the diagram below. The volume of gas in the tube was measured under these conditions of 298 K and 1.00 atm, and its volume was greater than the volume calculated in part (e). Explain.

Because the electrolysis of water occurs in water, there is some water vapor in the tube of H 2(g ) that was collected. The volume calculated in part (e) was the volume of only the H 2(g ) in the tube at the given temperature and pressure. The presence of another gas (water vapor) results in a greater volume at the given temperature and pressure.

One point is earned for recognizing that there is some water vapor in the sample of

hydrogen gas.

Question 3

X → 2 Y + Z

The decomposition of gas X to produce gases Y and Z is represented by the equation above. In a certain

experiment, the reaction took place in a 5.00 L flask at 428 K. Data from this experiment were used to produce the information in the table below, which is plotted in the graphs that follow.

Time

(minutes) [X] (mol L ?1) ln [X]

[X] ?1 (L mol ?1 )

0 0.00633 ?5.062 158 10. 0.00520 ?5.259

192 20. 0.00427 ?5.456 234 30. 0.00349 ?5.658 287 50. 0.00236 ?6.049 424 70. 0.00160 ?6.438 625 100. 0.000900 ?7.013

1,110

(a) How many moles of X were initially in the flask?

(b) How many molecules of Y were produced in the first 20. minutes of the reaction?

The reaction is first order with respect to X because a plot of ln [X] versus time produces a straight line with a negative slope.

One point is earned for the correct order and an explanation.

(d) Write the rate law for this reaction.

rate = k [X]1

One point is earned for the rate law consistent with part (c).

(e) Calculate the specific rate constant for this reaction. Specify units.

(f) Calculate the concentration of X in the flask after a total of 150. minutes of reaction.

Question 4

Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You need not balance the equations.

General Scoring: Three points are earned for each: 1 point for correct reactant(s) and 2 points for correct product(s).

(a) A solution of potassium carbonate is added a solution of strontium nitrate.

CO32? + Sr2+→ SrCO3

(b) Propene is burned in air.

C3H6 + O2→ CO2 + H2O

NH3 + Zn2+→ Zn(NH3)62+

(d) Ethanoic acid (acetic acid) is added to a solution of barium hydroxide.

OH? + HC2H3O2→ C2H3O2? + H2O

(e) A small piece of potassium is added to water.

K + H2O → OH? + H2 + K+

(f) Powdered iron metal is strongly heated with powdered sulfur.

Fe + S → FeS

Note: Fe2S3 also acceptable as a product

(g) A solution of sodium fluoride is added to a solution of hydrochloric acid.

H+ + F?→ HF

(h) A strip of lead metal is added to a solution of silver nitrate.

Pb + Ag+→ Pb2+ + Ag

Question 5

2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 22 H2O(l) → 2 KAl(SO4)2·12H2O(s) +

3 H2(g)

In an experiment, a student synthesizes alum, KAl(SO4)2·12H2O(s), by reacting aluminum metal with potassium hydroxide and sulfuric acid, as represented in the balanced equation above.

(a) In order to synthesize alum, the student must prepare a 5.0 M solution of sulfuric acid. Describe the

procedure for preparing 50.0 mL of 5.0 M H2SO4using any of the chemicals and equipment listed below.

Indicate specific amounts and equipment where appropriate.

10.0 M H2SO450.0 mL volumetric flask

Distilled water 50.0 mL buret

100 mL graduated cylinder 25.0 mL pipet

100 mL beaker 50 mL beaker

(b) Calculate the minimum volume of 5.0 M H2SO4that the student must use to react completely with

2.7 g aluminum metal.

Question 5 (continued)

crystals, then measures their mass.

(i) If the student weighs the crystals before they are completely dry, would the calculated percent yield be

greater than, less than, or equal to the actual percent yield? Explain.

If the KAl(SO4)2·12H2O(s) crystals have not been properly dried, there will be excess water present, making the mass of the product greater than it should be and the calculated percent yield too high. Therefore, the calculated percent yield will be greater than the actual percent yield.One point is earned for the prediction and a correct

explanation.

(ii) Cooling the reaction solution in an ice bath improves the percent yield obtained. Explain.

If the solubility of KAl(SO4)2·12H2O(s) decreases with decreasing temperature, cooling the reaction solution would result in the precipitation of more KAl(SO4)2·12H2O(s).One point is earned for the correct explanation.

(d) The student heats crystals of pure alum, KAl(SO4)2·12H2O(s), in an open crucible to a constant mass. The

mass of the sample after heating is less than the mass before heating. Explain.

KAl(SO4)2·12H2O(s) is a hydrate. For the mass of the sample to be less

after heating, the water of hydration must be lost. Heating the sample of KAl(SO4)2·12H2O(s) crystals will drive off the water first, decreasing the mass of the sample.One point is earned for the correct explanation.

Question 6

Consider two containers of volume 1.0 L at 298 K, as shown above. One container holds 0.10 mol N2(g) and the other holds 0.10 mol H2(g). The average kinetic energy of the N2(g) molecules is 6.2 × 10?21 J. Assume that the N2(g) and the H2(g) exhibit ideal behavior.

(a) Is the pressure in the container holding the H2(g) less than, greater than, or equal to the pressure in the

container holding the N2(g) ? Justify your answer.

(b) What is the average kinetic energy of the H2(g) molecules?

The average kinetic energy of the H2(g) molecules is

6.2 × 10?21 J because both gases are at the same temperature.One point is earned for the correct energy.

Question 6 (continued)

(d) What change could be made that would decrease the average kinetic energy of the molecules in the container?

The average kinetic energy of a gas particle depends on the temperature of the gas sample. To decrease the average kinetic energy of the gas particles in a gas sample, the temperature of the N2(g) would have to be lowered.One point is earned for the correct answer with an explanation.

(e) If the volume of the container holding the H2(g) was decreased to 0.50 L at 298 K, what would be the

change in each of the following variables? In each case, justify your answer.

(i) The pressure within the container

(ii) The average speed of the H2(g) molecules

The average speed is unchanged when the volume of the gas sample is halved. Average speed depends on changes in temperature, not changes in volume.One point is earned for the correct answer with an explanation.

Question 7

Answer the following questions about thermodynamics.

(a) In the empty box in the table above, write a balanced chemical equation for the complete combustion of

one mole of CH3OH(l). Assume products are in their standard states at 298 K. Coefficients do not need to be in whole numbers.

(b) On the basis of your answer to part (a) and the information in the table, determine the enthalpy change for

the reaction C(s) + H2(g) + H2O(l) → CH3OH(l).

formation for one mole of CH3OH(l).

Question 7 (continued)

(d) Predict the sign of ?S°for the combustion of H2(g). Explain your reasoning.

?S°for the combustion of H2(g) is negative. Both reactants are

in the gas phase and the product is in the liquid phase. The liquid phase is much more ordered than the gas phase, so the product is more ordered compared to the reactants, meaning that ?S° is negative.

(Note: There are fewer moles of products than reactants, which also favors a more ordered condition in the products, but the difference in phases is the more important factor.)One point is earned for the correct

sign of ?S°.

One point is earned for a correct

explanation.

(e) On the basis of bond energies, explain why the combustion of H2(g) is exothermic.

The combustion of H2(g) is exothermic ( ?H° < 0) because

more energy is released during the formation of two moles of O–H bonds than is required to break one mole of H–H bonds and one-half of a mole of O–O bonds.One point is earned for the correct

explanation.

Question 8

Use principles of atomic structure, bonding, and intermolecular forces to answer the following questions. Your responses must include specific information about all substances referred to in each part.

(a) Draw a complete Lewis electron-dot structure for the CS2molecule. Include all valence electrons in your

structure.

One point is earned for the correct structure. (b) The carbon-to-sulfur bond length in CS2is 160 picometers. Is the carbon-to-selenium bond length in CSe2

expected to be greater than, less than, or equal to this value? Justify your answer.

The carbon-to-selenium bond length in CSe2 is greater

than the carbon-to-sulfur bond length in CS2. Because the valence electrons in Se are in a higher shell (n = 4) than the valence electrons in S (n = 3), Se has a larger atomic radius than S has. Therefore, the carbon- to-selenium bond length is greater than the carbon-to- sulfur bond length.

One point is earned for indicating that the C-Se bond length is greater than the C-S

bond length.

One point is earned for indicating that Se is

larger than S.

selenium bond in CSe2expected to be greater than, less than, or equal to this value? Justify your answer.

The carbon-to-selenium bond energy in CSe2is less than the carbon-to-sulfur bond energy in CS2 because Se has a larger atomic radius than S. Because Se is a larger atom, the orbital overlap between the Se and C will be smaller than the orbital overlap between S and C.One point is earned for indicating that the C-Se bond energy is less than the C-S

bond energy.

One point is earned for the explanation.

Question 8 (continued)

(d) The complete structural formulas of propane, C3H8, and methanoic acid, HCOOH, are shown above. In the

table below, write the type(s) of intermolecular attractive forces(s) that occur in each substance.

Substance Boiling Point Intermolecular Attractive Force(s)

Propane229 K

Methanoic acid374 K

Propane has dispersion forces.

Methanoic acid has dispersion forces and hydrogen bonding forces.One point is earned for IMFs in

propane.

One point is earned for IMFs in

methanoic acid.

(e) Use principles of intermolecular attractive forces to explain why methanoic acid has a higher boiling point

than propane.

Hydrogen bonding IMFs among methanoic acid molecules are

much stronger than dispersion forces among propane molecules. The stronger the IMFs, the more energy it takes to overcome them. Therefore, methanoic acid has a higher boiling point than propane.One point is earned for comparing the strengths of the IMFs.

ap05_sg_chemistry_form_b

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