AP CHEMISTRY 2009 SCORING GUIDELINES - Glimme5-6-Review-ans.pdf · AP® CHEMISTRY 2009 SCORING...

AP® CHEMISTRY 2009 SCORING GUIDELINES

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Question 1 (10 points) Answer the following questions that relate to the chemistry of halogen oxoacids.

(a) Use the information in the table below to answer part (a)(i).

Acid Ka at 298 K

HOCl 2.9 10 8

HOBr 2.4 10 9

(i) Which of the two acids is stronger, HOCl or HOBr ? Justify your answer in terms of Ka .

HOCl is the stronger acid because its Ka value is greater than the Ka value of HOBr.

One point is earned for the correct answer with justification.

(ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part (a)(i).

One point is earned for a correct diagram.

(iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker acid than the acid that you identified in part (a)(i). Justify your prediction in terms of chemical bonding.

HOI is a weaker acid than HOCl because the O–H bond in HOI is stronger than the O–H bond in HOCl. The lower electronegativity (electron-drawing ability) of I compared with that of Cl results in an electron density that is higher (hence a bond that is stronger) between the H and O atoms in HOI compared with the electron density between the H and O atoms in HOCl.

OR

The conjugate base OCl is more stable than OIbecause Cl, being more electronegative, is better able to accommodate the negative charge.

One point is earned for predicting that HOI is a weaker acid than HOCl and stating that iodine has a lower electronegativity than chlorine and EITHER

stating that this results in a stronger O–Hbond in HOI

OR

stating that this decreases the stability of theOI ion in solution.



Question 1 (continued)

(b) Write the equation for the reaction that occurs between hypochlorous acid and water.

HOCl + H2O OCl + H3O+

OR

HOCl OCl + H+

One point is earned for the correct equation.

(c) A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K. The hydrolysis reaction OCl (aq) + H2O(l) HOCl(aq) + OH (aq) occurs.

(i) Write the equilibrium-constant expression for the hydrolysis reaction that occurs between OCl (aq) and H2O(l) .

Kb = [HOCl][OH ][OCl ]

One point is earned for the correct expression.

(ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction.

Kb = w

a

KK =

14

81.0 102.9 10

= 3.4 × 10 7 One point is earned for the correct value with supporting work.

(iii) Calculate the value of [OH ] in the 1.2 M NaOCl solution at 298 K .

Khyd = 3.4 × 10 7 = [OH ][HOCl][OCl ]

= ( )( )(1.2 )x x

x 2

1.2x

(1.2)(3.4 × 10 7) = x2

x = [OH ] = 6.4 × 10 4 M

[OCl ] [HOCl] [OH ]

initial value 1.2 0 0

change x x x

equilibrium value 1.2 x x x

One point is earned for the correct setup.

One point is earned for the correct answer with supporting

calculations.




(d) A buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCl at 298 K. The pH of the buffer solution is determined to be 6.48.

(i) Calculate the value of [H3O+] in the buffer solution.

[H+] = 10 6.48 = 3.3 × 10 7 M One point is earned for the correct value.

(ii) Indicate which of HOCl(aq) or OCl (aq) is present at the higher concentration in the buffer solution. Support your answer with a calculation.

[H+] = 3.3 × 10 7 M and Ka for HOCl = 2.9 × 10 8

Ka = [H ][OCl ][HOCl]

2.9 × 10 8 = 7(3.3 10 )[OCl ]

[HOCl]

[OCl ][HOCl] =

8

72.9 103.3 10

= 0.088 [HOCl] > [OCl ]

One point is earned for the correct answer with supporting buffer

calculations.



Question 5 (8 points)

Reaction Equation 298H 298S 298G

X

C(s) + H2O(g) CO(g) + H2(g) +131 kJ mol 1 +134 J mol 1 K 1 +91 kJ mol 1

Y

CO2(g) + H2(g) CO(g) + H2O(g) +41 kJ mol 1 +42 J mol 1 K 1 +29 kJ mol 1

Z

2 CO(g) C(s) + CO2(g) ? ? ?

Answer the following questions using the information related to reactions X, Y, and Z in the table above.

(a) For reaction X, write the expression for the equilibrium constant, Kp .

Kp = 2

2

CO H

H O

p pp

One point is earned for the correct expression.

(b) For reaction X, will the equilibrium constant, Kp , increase, decrease, or remain the same if the temperature rises above 298 K ? Justify your answer.

Kp will increase. If the temperature is increased for an endothermic

reaction ( 298H = +131 kJ mol 1), then by Le Chatelier’s principle the reaction will shift toward products, thereby absorbing energy. With greater concentrations of products at equilibrium, the value of Kp will increase.

OR Because G = RT ln Kp = 298H T 298S ,

then ln Kp = 298HRT + 298S

R .

An increase in T for a positive 298H results in an increase in ln Kp and thus an increase in Kp .

One point is earned for the correct answer with appropriate justification.



Question 5 (continued) (c) For reaction Y at 298 K, is the value of Kp greater than 1, less than 1, or equal to 1? Justify your answer.

Kp for reaction Y is less than 1.

For reaction Y, 298G = +29 kJ mol 1, a positive number.

Because G = RT ln K and G is positive, then ln Kpmust be negative. This is true when Kp is less than 1.

OR A positive G results in a nonspontaneous reaction under

standard conditions. This favors reactants over products as equilibrium is approached starting from standard conditions, resulting in a Kp less than 1.


(d) For reaction Y at 298 K, which is larger: the total bond energy of the reactants or the total bond energy of the products? Explain.

The total bond energy of the reactants is larger. Reaction Y is endothermic ( 298H = + 41 kJ mol 1 > 0), so

there is a net input of energy as the reaction occurs. Thus, the total energy required to break the bonds in the reactants must be greater than the total energy released when the bonds are formed in the products.

One point is earned for the correct answer with appropriate

explanation.

(e) Is the following statement true or false? Justify your answer.

“On the basis of the data in the table, it can be predicted that reaction Y will occur more rapidly than reaction X will occur.”

The statement is false. Thermodynamic data for an overall reaction have no

bearing on how slowly or rapidly the reaction occurs.





(f) Consider reaction Z at 298 K.

(i) Is S for the reaction positive, negative, or zero? Justify your answer.

S for reaction Z is negative.

In reaction Z, two moles of gas with relatively high entropy are converted into one mole of solid and one mole of gas, a net loss of one mole of gas and thus a net loss in entropy.

OR Reaction Z can be obtained by reversing reactions X and Y

and adding them together. Thus S for reaction Z is the sum of two negative numbers and must itself be negative.

One point is earned for the correct answer with an

appropriate justification.

(ii) Determine the value of H for the reaction.

Add the values of the negatives of 298H for reactions X and Y :

131 kJ mol 1 + ( 41 kJ mol 1) = 172 kJ mol

One point is earned for the correct answer.

(iii) A sealed glass reaction vessel contains only CO(g) and a small amount of C(s). If a reaction occurs and the temperature is held constant at 298 K, will the pressure in the reaction vessel increase, decrease, or remain the same over time? Explain.

The pressure in the flask decreases.

The reaction would proceed to the right, forming more C(s) and

CO2(g). Because two moles of CO(g) would be consumed for every mole of CO2(g) that is produced, the total number of moles of gas in the flask would decrease, thereby causing the pressure in the flask to decrease.

One point is earned for the correct answer with an

appropriate explanation.

AP® CHEMISTRY 2009 SCORING GUIDELINES (Form B)


Question 5 (9 points) Answer the following questions about nitrogen, hydrogen, and ammonia.

(a) In the boxes below, draw the complete Lewis electron-dot diagrams for N2 and NH3.

The correct structures are shown in the boxes above.

Two points are earned for the correct Lewis electron-dot diagrams (1 point each).

(b) Calculate the standard free-energy change, G , that occurs when 12.0 g of H2(g) reacts with excess N2(g) at 298 K according to the reaction represented below.

N2(g) + 3 H2(g) 2 NH3(g) 298G = 34 kJ mol 1

12.0 g H2 × 2

2

1 mol H2.0 g H ×

2

1 mol reaction3 mol H × 34 kJ

1 mol reaction = – 68 kJ

One point is earned for the correct stoichiometry.

One point is earned for

the correct answer.

(c) Given that 298H for the reaction is 92.2 kJ mol 1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain.

298H = (bond energy of the reactants) (bond energy of the products)

Based on the equation above, for 298H to be negative, the total bond

energy of the products must be larger than the total bond energy of the reactants.

OR More energy is released as product bonds are formed than is absorbed as

reactant bonds are broken.

One point is earned for the correct answer

with the correct equation and explanation.



Question 5 (continued) (d) The value of the standard entropy change, 298S , for the reaction is 199 J mol 1K 1. Explain why the value

of 298S is negative.

All of the reactants and products in the reaction are in the gas phase, so the sign of the entropy change will depend on the number of moles of particles in the reactants and products. There are more moles of reactants (four) compared with moles of products (two), so there is a greater number of microstates in the reactants than in the products. Therefore the entropy decreases as the reaction proceeds (fewer possible microstates), and the sign of the entropy change is negative.

One point is earned for the correct explanation.

(e) Assume that H and S for the reaction are independent of temperature.

(i) Explain why there is a temperature above 298 K at which the algebraic sign of the value of G changes.

G = H T S As the temperature increases |T S | will at some point exceed | H | . Because both H and S are negative, the sign of G will then change from negative to positive.


(ii) Theoretically, the best yields of ammonia should be achieved at low temperatures and high pressures. Explain.

Low temperatures: The reaction is exothermic. By Le Chatelier’s principle, decreasing the temperature drives the reaction to the right to produce more heat energy, and thus more ammonia is produced.

High pressures: For this reaction, higher pressure is achieved by decreasing the volume of the container. As pressure increases, the reaction equilibrium shifts in the direction that reduces the total number of particles (by Le Chatelier’s principle). In this case, the product has fewer moles of particles than the reactants; thus product would be favored. Higher pressure therefore results in an increase in the amount of ammonia.

One point is earned for explaining increased yield at low temperatures.

One point is earned for explaining increased yield at high pressures.



Question 6 (9 points) Answer the following questions about electrochemical cells.

It is observed that when silver metal is placed in aqueous thallium(I) fluoride, TlF, no reaction occurs. When the switch is closed in the cell represented above, the voltage reading is +1.14 V.

(a) Write the reduction half-reaction that occurs in the cell.

Ag+ + e Ag One point is earned for the correct equation.

(b) Write the equation for the overall reaction that occurs in the cell.

Tl + Ag+ Tl+ + Ag One point is earned for the correct equation.

(c) Identify the anode in the cell. Justify your answer.

The anode is where oxidation occurs. In the overall reaction Tl is oxidized to Tl+, so the anode is the Tl electrode in the left cell.


(d) On the diagram above, use an arrow to clearly indicate the direction of electron flow as the cell operates.

The arrow should show electron flow in the direction from the Tl electrode through the wire to the Ag electrode.

One point is earned for a correct arrow.



Question 6 (continued) (e) Calculate the value of the standard reduction potential for the Tl+/Tl half-reaction.

cellE = redE oxE

+1.14 V = +0.80V oxE

oxE = 0.34 V



The standard reduction potential, E , of the reaction Pt2+ + 2 e Pt is 1.20 V.

(f) Assume that electrodes of pure Pt , Ag, and Ni are available as well as 1.00 M solutions of their salts. Three different electrochemical cells can be constructed using these materials. Identify the two metals that when used to make an electrochemical cell would produce the cell with the largest voltage. Explain how you arrived at your answer.

E (V)

Ni2+ + 2 e Ni 0.25

Ag+ + e Ag 0.80

Pt2+ + 2 e Pt 1.20

cellE = redE oxE

The two metals that yield the largest cellE

are those with the biggest difference in E , namely, Pt and Ni (see cellE calculation below).

cellE = +1.20 ( 0.25) = +1.45 V


(g) Predict whether Pt metal will react when it is placed in 1.00 M AgNO3(aq). Justify your answer.

When Pt metal is added to 1.00 M AgNO3 , the only redox reaction that could occur would be for Pt to become oxidized as Ag+ is reduced.

cellE = redE oxE = +0.80 V (+1.20 V) = 0.40 V Because cellE for that reaction is negative, no reaction will

occur.

One point is earned for comparing E° values.

One point is earned for

the correct interpretation.



Question 1

C(s) + CO2(g) !" 2 CO(g)

Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred. As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container. Results are recorded in the table below.

Time (hours)

Total Pressure of Gases in Container at 1,160 K

(atm)

0.0 5.00

2.0 6.26

4.0 7.09

6.0 7.75

8.0 8.37

10.0 8.37

(a) Write the expression for the equilibrium constant, Kp , for the reaction.

Kp = 2

2CO

CO

( )P

P One point is earned for the correct expression.

(b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the volume of

the solid carbon is negligible.)

(5.00 atm)(2.00 L) 0.105 molL atm(0.0821 )(1,160 K)mol K

PVnRT



(c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO2(g) is 1.63 atm. Calculate

(i) the partial pressure of CO(g) , and

2CO CO totalP P P+ =

2CO COtotalP P P= # = 8.37 atm # 1.63 atm = 6.74 atm

One point is earned for the correct answer supported by a correct method.



Question 1 (continued) (ii) the value of the equilibrium constant, Kp .

Kp = 2

2CO

CO

( )PP

= 2(6.74 atm)

1.63 atm = 27.9

One point is earned for a correct setup that is consistent with part (a).

One point is earned for the correct answer according to the setup.

(d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst is negligible.)

The total pressure of the gases at equilibrium with a catalyst present would be equal to the total pressure of the gases without a catalyst. Although a catalyst would cause the system to reach the same equilibrium state more quickly, it would not affect the extent of the reaction, which is determined by the value of the equilibrium constant, Kp .


In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s) , plus CO(g) and CO2(g) , each at a partial pressure of 2.00 atm at 1,160 K.

(e) Predict whether the partial pressure of CO2(g) will increase, decrease, or remain the same as this system approaches equilibrium. Justify your prediction with a calculation.

Q =

2

2CO

CO

( )PP

= 2(2.00 atm)

2.00 atm = 2.00 < Kp ( = 27.9),

therefore 2COP will decrease as the system approaches

equilibrium.

One point is earned for a correct calculation involving Q or ICE

calculation.

One point is earned for a correct conclusion based on the calculation.



Question 6

(a) Structures of the pyridine molecule and the benzene molecule are shown below. Pyridine is soluble in water, whereas benzene is not soluble in water. Account for the difference in solubility. You must discuss both of the substances in your answer.

Pyridine is polar (and capable of forming hydrogen bonds with water), while the nonpolar benzene is not capable of forming hydrogen bonds. Pyridine will dissolve in water because of the strong hydrogen bonds (or dipole-dipole intermolecular interactions) that exist between the lone pair of electrons on pyridine’s nitrogen atom and the solvent water molecules. No such strong intermolecular interaction can exist between benzene and water, so benzene is insoluble in water.

One point is earned for identifying a relevant structural difference between pyridine and benzene.

One point is earned for indicating that pyridine is soluble in water because pyridine can form strong dipole-dipole interactions (or hydrogen bonds) with water, while benzene cannot.

(b) Structures of the dimethyl ether molecule and the ethanol molecule are shown below. The normal boiling point of dimethyl ether is 250 K, whereas the normal boiling point of ethanol is 351 K. Account for the difference in boiling points. You must discuss both of the substances in your answer.




The intermolecular forces of attraction among molecules of dimethyl ether consist of London (dispersion) forces and weak dipole-dipole interactions. In addition to London forces and dipole-dipole interactions that are comparable in strength to those in dimethyl ether, ethanol can form hydrogen bonds between the H of one molecule and the O of a nearby ethanol molecule. Hydrogen bonds are particularly strong intermolecular forces, so they require more energy to overcome during the boiling process. As a result, a higher temperature is needed to boil ethanol than is needed to boil dimethyl ether.

One point is earned for recognizing that ethanol molecules can form intermolecular hydrogen bonds, whereas dimethyl ether molecules do not form intermolecular hydrogen bonds.

One point is earned for recognizing that, compared to the energy required to overcome the weaker intermolecular forces in liquid dimethyl ether, more energy is required to overcome the stronger hydrogen bonds in liquid ethanol, leading to a higher boiling point.

(c) SO2 melts at 201 K, whereas SiO2 melts at 1,883 K. Account for the difference in melting points. You must discuss both of the substances in your answer.

In the solid phase, SO2 consists of discrete molecules

with dipole-dipole and London (dispersion) forcesamong the molecules. These forces are relatively weak and are easily overcome at a relatively low temperature, consistent with the low melting point of SO2 .

In solid SiO2 , a network of Si and O atoms, linked by

strong covalent bonds, exists. These covalent bonds are much stronger than typical intermolecular interactions, so very high temperatures are needed to overcome the covalent bonds in SiO2. This is consistent with the

very high melting point for SiO2.

One point is earned for recognizing that SO2 is a molecular solid with only

weak dipole-dipole and London forces among SO2 molecules.

One point is earned for recognizing that SiO2 is a covalent network solid, and that strong covalent bonds must be broken for SiO2 to melt.



Question 6 (continued) (d) The normal boiling point of Cl2(l) (238 K) is higher than the normal boiling point of HCl(l) (188 K).

Account for the difference in normal boiling points based on the types of intermolecular forces in the substances. You must discuss both of the substances in your answer.

The intermolecular forces in liquid Cl2 are

London (dispersion) forces, whereas the intermolecular forces in liquid HCl consist of London forces and dipole-dipole interactions. Since the boiling point of Cl2 is

higher than the boiling point of HCl, the London forces among Cl2 molecules must be

greater than the London and dipole-dipole forces among HCl molecules. The greater strength of the London forces between Cl2molecules occurs because Cl2 has more

electrons than HCl, and the strength of the London interaction is proportional to the total number of electrons.

One point is earned for recognizing that the London forces among Cl2 molecules must be

larger than the intermolecular forces (London and dipole-dipole) among HCl molecules.

One point is earned for recognizing that the strength of the London forces among molecules is proportional to the total number of electrons in each molecule.



Question 1 Answer the following questions regarding the decomposition of arsenic pentafluoride, AsF5(g) .

(a) A 55.8 g sample of AsF5(g) is introduced into an evacuated 10.5 L container at 105°C.

(i) What is the initial molar concentration of AsF5(g) in the container?

mol AsF5 = 55.8 g AsF5� ! 5

5

1 mol AsF169.9 g AsF

= 0.328 mol

[AsF5]i = 50.328 mol AsF10.5 L

= 0.0313 M

One point is earned for the correct molar mass.

One point is earned for the correct concentration.

(ii) What is the initial pressure, in atmospheres, of the AsF5(g) in the container?

PV = nRT

P = 1 10.328 mol 0.0821 L atm mol K 378 K

10.5 L

" "! ! = 0.969 atm

One point is earned for the correct substitution.

One point is earned for the correct pressure.

At 105°C, AsF5(g) decomposes into AsF3(g) and F2(g) according to the following chemical equation.

AsF5(g) #$ AsF3(g) + F2(g)

(b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF5(g).

K = 3 2

5

[AsF ] [F ][AsF ]

One point is earned for the correct equation.

(c) When equilibrium is established, 27.7 percent of the original number of moles of AsF5(g) has decomposed.

(i) Calculate the molar concentration of AsF5(g) at equilibrium.

100.0% " 27.7% = 72.3%

[AsF5] = 0.723 ! 0.0313 M = 0.0226 M One point is earned for the correct concentration.




(ii) Using molar concentrations, calculate the value of the equilibrium constant, Keq , at 105°C.

[AsF3] = [F2] = 0.277 ! [AsF5]i

= 0.277 ! 0.0313 M = 0.00867 M

Keq = 3 2

5

[AsF ] [F ][AsF ]

= [0.00867] [0.00867][0.0226]

= 0.00333

One point is earned for setting [AsF3] = [F2] .

Note: the point is not earned if the student indicates that [AsF3] = [F2] = [AsF5] .

One point is earned for the correct calculation of [AsF3] and [F2] .

One point is earned for the correct calculation of Keq .

(d) Calculate the mole fraction of F2(g) in the container at equilibrium.

mol AsF5 = 0.0226 M ! 10.5 L = 0.237 mol

mol F2 = mol AsF3 = 0.00867 M ! 10.5 L = 0.0910 mol

mol fraction F2 = 2

2 3 5

mol Fmol F + mol AsF + mol AsF

= 0.09100.0910 + 0.0910 + 0.237

= 0.217

OR

mol fraction F2 = 0.008640.00864 + 0.00864 + 0.0226

= 0.217

One point is earned for the correct calculation of the mole

fraction of F2(g).



Question 5 The identity of an unknown solid is to be determined. The compound is one of the seven salts in the following table.

Al(NO3)3. 9H2O BaCl2. 2H2O CaCO3 CuSO4

. 5H2O

NaCl BaSO4 Ni(NO3)2. 6H2O

Use the results of the following observations or laboratory tests to explain how each compound in the table may be eliminated or confirmed. The tests are done in sequence from (a) through (e).

(a) The unknown compound is white. In the table below, cross out the two compounds that can be eliminated using this observation. Be sure to cross out these same two compounds in the tables in parts (b), (c), and (d).


. 5H2O


One point is earned for each correctly crossed-out compound.

(b) When the unknown compound is added to water, it dissolves readily. In the table below, cross out the two

compounds that can be eliminated using this test. Be sure to cross out these same two compounds in the tables in parts (c) and (d).


. 5H2O


One point is earned for each additional correctly crossed-out compound.

(c) When AgNO3(aq) is added to an aqueous solution of the unknown compound, a white precipitate forms.

In the table below, cross out each compound that can be eliminated using this test. Be sure to cross out the same compound(s) in the table in part (d).


. 5H2O


One point is earned for crossing out Al(NO3)3. 9H2O or for crossing out Ni(NO3)2

. 6H2O if it had not been crossed out earlier.




(d) When the unknown compound is carefully heated, it loses mass. In the table below, cross out each compound that can be eliminated using this test.


. 5H2O NaCl BaSO4 Ni(NO3)2

. 6H2O

One point is earned for crossing out NaCl or for crossing out either CaCO3 or BaSO4 if they had not been crossed out earlier.

(e) Describe a test that can be used to confirm the identity of the unknown compound identified in part (d).

Limit your confirmation test to a reaction between an aqueous solution of the unknown compound and an aqueous solution of one of the other soluble salts listed in the tables above. Describe the expected results of the test; include the formula(s) of any product(s).

Mix an aqueous solution of BaCl2. 2H2O

with an aqueous solution of CuSO4. 5H2O.

The BaSO4 will precipitate.

One point is earned for describing a precipitation reaction between the compound left in part (d) and another compound given in the problem.

One point is earned for a correct identification of a precipitate that would form upon the mixing of the chosen solutions.



Question 6 Use principles of thermodynamics to answer the following questions.

(a) The gas N2O4 decomposes to form the gas NO2 according to the equation below.

(i) Predict the sign of !H° for the reaction. Justify your answer.

Bonds are broken when NO2 molecules form from N2O4

molecules. Energy must be absorbed to break bonds, so the reaction is endothermic and the sign of !H° is positive.

One point is earned for the correct sign and a correct explanation.

(ii) Predict the sign of !S° for the reaction. Justify your answer.

There are two gaseous product molecules for each gaseous reactant molecule, so the product has more entropy than the reactant. The entropy increases as the reaction proceeds, so the sign of !S° is positive.

One point is earned for the correct sign and a correct

explanation.

(b) One of the diagrams below best represents the relationship between !G° and temperature for the reaction given in part (a). Assume that !H° and !S° are independent of temperature.

Draw a circle around the correct graph. Explain why you chose that graph in terms of the relationship !G° = !H° – T!S° .

The leftmost graph should be circled.

!S° is positive, so as T increases, T!S° becomes a larger positive number. At higher temperatures, you are subtracting larger positive numbers from !H° to get !G°, so !G° decreases with increasing temperature.

One point is earned for the correct graph selection.

One point is earned for the explanation.



Question 6 (continued) (c) A reaction mixture of N2O4 and NO2 is at equilibrium. Heat is added to the mixture while the mixture is

maintained at constant pressure.

(i) Explain why the concentration of N2O4 decreases.

The reaction is endothermic. For endothermic reactions, increasing the temperature drives the reaction to the right. This increases the equilibrium concentration of NO2 and decreases the equilibrium concentration of N2O4.


(ii) The value of Keq at 25°C is 5.0 ! 10"3. Will the value of Keq at 100°C be greater than, less than, or equal to this value?

Because the reaction is endothermic, at higher temperatures the reaction goes further to the right. This means that the value of Keq at 100°C will be greater than the value of Keq at 25°C.

One point is earned for the correct choice. (No explanation required.)

(d) Using the value of Keq at 25°C given in part (c)(ii), predict whether the value of #H° is expected to be greater than, less than, or equal to the value of T#S° . Explain.

Keq at 25°C is less than 1, hence #G° must be positive. And in order for #G° to be positive, #H° must be greater than T#S°.

One point is earned for the correct prediction.

One point is earned for the explanation.

AP CHEMISTRY 2009 SCORING GUIDELINES - Glimme5-6-Review-ans.pdf · AP® CHEMISTRY 2009 SCORING...

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