unit f fr practice general equilibrium (pg 1 of 26) h o

1. Answer the following questions about glucose, C6H12O6, an important biochemical energy source.

(a) Write the empirical formula of glucose.

In many organisms, glucose is oxidized to carbon dioxide and water, as represented by the following equation. C6H12O6(s) + 6 O2(g) → 6CO2(g) + 6H2O(L)

A 2.50 g sample of glucose and an excess of O2(g) were placed in a calorimeter. After the reaction was initiated and proceeded to completion, the total heat released by the reaction was calculated to be 39.0 kJ.

(b) Calculate the value of ∆Hº, in kJ mol−1, for the combustion of glucose.

(c) When oxygen is not available, glucose can be oxidized by fermentation. In that process, ethanol and carbon dioxide are produced, as represented by the following equation.

C6H12O6(s) → 2C2H5OH(L) + 2CO2(g) ∆Hº = −68.0 kJ mol−1 at 298 K

The value of the equilibrium constant, Kp, for the reaction at 298 K is 8.9 × 1039

(i) Calculate the value of the standard free-energy change, ∆Gº, for the reaction at 298 K. Include units with your answer.

(ii) Calculate the value of the standard entropy change, ∆Sº, in J K−1 mol−1, for the reaction at 298 K

(iii) Indicate whether the equilibrium constant for the fermentation reaction increases, decreases, or remains the same if the temperature is increased. Justify your answer.

(d) Using your answer for part (b) and the information provided in part (c), calculate the value of ∆Hº following reaction.

C2H5OH(L) + 3O2(g) → 2CO2(g) + 3H2O(L)

Unit F FR Practice General Equilibrium (pg 1 of 26)

ANSWER # 1Much of this problem is a review of unit C, with only part (c iii) specific to unit F.

(a) Remember that empirical formulas are the lowest whole number ratio: CH2O

(b) The problem is asking you for kJ per mole. In the information at the top you are told the kJ that is released (exothermic, thus −∆H) when 2.5 g of glucose is burned. Take note that the molar mass of glucose is 180 g/mol.

−39.0kJ2.50g

× 180g1mol

= −2, 810 kJ mol−1

(c i) To do this problem, remember the relationship between ∆G and K that you learned in unit C.

AP® CHEMISTRY 2011 SCORING GUIDELINES (Form B)

© 2011 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 3 (9 points)

Answer the following questions about glucose, C6H12O6 , an important biochemical energy source.

(a) Write the empirical formula of glucose.

CH2O 1 point is earned for the correct formula.

In many organisms, glucose is oxidized to carbon dioxide and water, as represented by the following equation.

C6H12O6(s) + 6 O2(g) ! 6 CO2(g) + 6 H2O(l)

A 2.50 g sample of glucose and an excess of O2(g) were placed in a calorimeter. After the reaction was initiated and proceeded to completion, the total heat released by the reaction was calculated to be 39.0 kJ.

(b) Calculate the value of "H#, in kJ mol$1, for the combustion of glucose.

2.50 g % 6 12 61 mol C H O

180.16 g C H O6 12 6 = 0.0139 mol C6H12O6

39.0 kJ0.0139 mol! = $2,810 kJ mol$1

1 point is earned for the correct answer.

(c) When oxygen is not available, glucose can be oxidized by fermentation. In that process, ethanol and carbon dioxide are produced, as represented by the following equation.

C6H12O6(s) ! 2 C2H5OH(l) + 2 CO2(g) "H# = $68.0 kJ mol$1 at 298 K

The value of the equilibrium constant, Kp , for the reaction at 298 K is 8.9 ! 1039.

(i) Calculate the value of the standard free-energy change, "G#, for the reaction at 298 K. Include units with your answer.

"G# = $RT lnK

= $(8.31 J mol$1 K$1)(298 K)(ln 8.9 ! 1039)

= $228,000 J mol$1 = $228 kJ mol$1

1 point is earned for correct setup.

1 point is earned for correct answer.

(c ii) Now that you know ∆G, you can use the Gibbs free energy equation to calculate ∆S. Be alert that they want you to report the answer in J/Kmol.



Question 3 (continued)

(ii) Calculate the value of the standard entropy change, !S", in J K#1 mol#1, for the reaction at 298 K.

!G" = !H" # T!S"

!S" = H GT

! !!! !

= 1( 68.0 kJ mol ) ( 228 kJ mol )298 K

! !! ! !

1

= 0.537 kJ K#1 mol#1 = 537 J K#1 mol#1

1 point is earned for the correct setup.



!H" is negative, so when the temperature increases, the equilibrium for the reaction is shifted to the left (according to Le Châtelier’s principle). This means that the equilibrium constant decreases.

1 point is earned for the correct answer with justification.

(d) Using your answer for part (b) and the information provided in part (c), calculate the value of !H" for the following reaction.

C2H5OH(l) + 3 O2(g) $ 2 CO2(g) + 3 H2O(l)

C6H12O6(s) + 6 O2(g) $ 6 CO2(g) + 6 H2O(l) !H" = #2,810 kJ mol#1

2 C2H5OH(l) + 2 CO2(g) $ C6H12O6(s) !H" = 68.0 kJ mol#1

______________________________________________________________

2 C2H5OH(l) + 6 O2(g) $ 4 CO2(g) + 6 H2O(l) !H" = #2,740 kJ mol#1,

thus !H" for the reaction as written in (d) is #1,370 kJ mol#1.


1 point is earned for the correct answer.!

(c iii) ∆Hº is negative, so when the temperature increases, the equilibrium position will shift to the left, according to LeChatelier’s principle, which means that the equilibrium constant will decrease.

(d) This is a Hess’ law problem. Using the reaction at the top as written, and flipping the fermentation reaction given in part c.




(ii) Calculate the value of the standard entropy change, !S", in J K#1 mol#1, for the reaction at 298 K.

!G" = !H" # T!S"

!S" = H GT

! !!! !

= 1( 68.0 kJ mol ) ( 228 kJ mol )298 K

! !! ! !

1

= 0.537 kJ K#1 mol#1 = 537 J K#1 mol#1




!H" is negative, so when the temperature increases, the equilibrium for the reaction is shifted to the left (according to Le Châtelier’s principle). This means that the equilibrium constant decreases.

1 point is earned for the correct answer with justification.

(d) Using your answer for part (b) and the information provided in part (c), calculate the value of !H" for the following reaction.

C2H5OH(l) + 3 O2(g) $ 2 CO2(g) + 3 H2O(l)

C6H12O6(s) + 6 O2(g) $ 6 CO2(g) + 6 H2O(l) !H" = #2,810 kJ mol#1

2 C2H5OH(l) + 2 CO2(g) $ C6H12O6(s) !H" = 68.0 kJ mol#1

______________________________________________________________

2 C2H5OH(l) + 6 O2(g) $ 4 CO2(g) + 6 H2O(l) !H" = #2,740 kJ mol#1,

thus !H" for the reaction as written in (d) is #1,370 kJ mol#1.


1 point is earned for the correct answer.!


2. The compound butane, C4H10, occurs in two isomeric forms, n-butane and isobutane (2-methyl propane). Both compounds exist as gases at 25°C and 1.0 atm.

save parts a and b for unit J(a) Draw the structural formula of each of the isomers (include all atoms). Clearly label each structure.

(b) On the basis of molecular structure, identify the isomer that has the higher boiling point. Justify your answer.

The two isomers exist in equilibrium as represented by the equation below.

n-butane(g) ⇋ isobutane(g) Kc = 2.5 at 25°C

Suppose that a 0.010 mol sample of pure n-butane is placed in an evacuated 1.0 L rigid container at 25 °C.

(c) Write the expression for the equilibrium constant, Kc, for the reaction.(aka: Write the equation that defines the equilibrium condition.)

(d) Calculate the initial pressure in the container when the n-butane is first introduced (before the reaction starts).

(e) The n-butane reacts until equilibrium has been established at 25°C. (i) Calculate the total pressure in the container at equilibrium. Justify your answer.

(ii) Calculate the molar concentration of each species at equilibrium.

(iii) If the volume of the system is reduced to half of its original volume, what will be the new concentration of n-butane after equilibrium has been reestablished at 25°C? Justify your answer.

Suppose that in another experiment a 0.010 mol sample of pure isobutane is placed in an evacuated 1.0 L rigid container and allowed to come to equilibrium at 25°C. (f) Calculate the molar concentration of each species after equilibrium has been established.


ANSWER # 2

(a)



The compound butane, C4H10 , occurs in two isomeric forms, n-butane and isobutane (2-methyl propane).

Both compounds exist as gases at 25!C and 1.0 atm.

(a) Draw the structural formula of each of the isomers (include all atoms). Clearly label each structure.

Two points are earned for two

correct structures with correct labels.

(Note: 1 point can be earned for either two correct

structures that are mislabeled or one correct

structure with or without correct label.)

OR

1 point can be earned for two skeletal structures

(hydrogen atoms not shown) with proper labels.

(b) On the basis of molecular structure, identify the isomer that has the higher boiling point. Justify your

answer.

The isomer n-butane has the higher boiling point. London (dispersion)

forces are greater among molecules of n-butane than they are among

molecules of isobutane because molecules of n-butane, with its linear

structure, can approach one another more closely and can form a greater

number of induced temporary dipoles than molecules of isobutane, with

its more compact structure, can form.

One point is earned

for the correct choice of

isomer with justification.


n-butane(g) !"

isobutane(g) Kc = 2.5 at 25!C

Suppose that a 0.010 mol sample of pure n-butane is placed in an evacuated 1.0 L rigid container at 25!C.

(c) Write the expression for the equilibrium constant, Kc , for the reaction.

Kc = [isobutane]

[ -butane]n

One point is earned for the correct equation.

© 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com











OR




answer.







One point is earned




n-butane(g) !"




Kc = [isobutane]

[ -butane]n



(b)











OR




answer.







One point is earned




n-butane(g) !"




Kc = [isobutane]

[ -butane]n



(c) For this problem, AP asks for Kc, please be sure and use square brackets.











OR




answer.







One point is earned




n-butane(g) !"




Kc = [isobutane]

[ -butane]n



(d) A simple PV = nRT will solve for the pressure from the initial amount of moles of gas in the container.



(d) Calculate the initial pressure in the container when the n-butane is first introduced (before the reaction

starts).

L atm(0.010 mol)(0.0821 )(298 K )

mol KnRTP

1.0 LV

!

!" "

= 0.24 atm

One point is earned for the correct

substitution and numerical answer.

(e) The n-butane reacts until equilibrium has been established at 25#C.

(i) Calculate the total pressure in the container at equilibrium. Justify your answer.

The total pressure in the container remains the same, 0.24 atm.

As the reaction proceeds, the number of molecules in the

container remains constant; one molecule of isobutane is

produced for each molecule of n-butane consumed.

One point is earned for the

correct answer with justification.

(ii) Calculate the molar concentration of each species at equilibrium.

[isobutane]=

[ -butane]cK

n =

(0.010 )

x

x$ = 2.5

2.5 (0.010 ) 0.025 2.5x x" $ " $ x

% x = 0.0071 M isobutane 3.5 = 0.025x

(0.010 M $ 0.0071 M) = 0.003 M n-butane

One point is earned for the correct setup.

One point is earned for both correct numerical

answers.

(iii) If the volume of the system is reduced to half of its original volume, what will be the new

concentration of n-butane after equilibrium has been reestablished at 25#C ? Justify your answer.

Halving the volume of the container at equilibrium

doubles the pressure of both isobutane and n-butane,

which has no effect on the equilibrium because the

stoichiometry of the reaction is one mole of product

produced for each mole of reactant consumed. Since

the number of moles of each isomer is unchanged but

the volume is reduced by half, concentrations of both

isomers are doubled and the concentration of n-butane

will be 2 ! 0.003 M = 0.006 M .

One point is earned for the correct answer

with justification.


(e i) This may seem like a complicated calculation, however you may notice from the reaction, that every time one gaseous molecule reacts, one gaseous molecule is formed, thus the total pressure will remain the same, 0.24 atm, even as the reaction proceeds. Pressure at equilibrium = 0.24 atm.

(e ii) Kc =isobutane[ ]n − butane[ ] =

x0.010 − x( ) = 2.5 ➸ x = 2.5 0.010 − x( ) ➸ x = 0.025 − 2.5x ➸ 3.5x = 0.025

x = 0.0071 M isobutane, thus (0.010 M − 0.0071 M) = 0.003 M n−butane

(e iii) Halving the volume of the container at equilibrium doubles the pressure of both isobutane and n−butane, which has no effect on the equilibrium position because the stoichiometry of the reaction is the same for reactants and products. Although the number of moles of each isomer is unchanged, the volume is reduced by half, the concentration of the gases would be doubled. Thus n−butane would become (2 × 0.003 M) = 0.006 M.

(f) The concentrations of isobutane and n−butane would be exactly the same as they were calculated in part (e ii) since the stoichiometry is the same on both sides, the problem would solve to exactly the same results,this time 0.003 M for n−butane and 0.0071 M for isobutane.

Kc =isobutane[ ]n − butane[ ] =

0.010 − x( )x

= 2.5 ➸ 2.5x = 0.010 − x( ) ➸ 3.5x = 0.01 ➸ x = 0.003


Save for unit J

Save for unit J

2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS

© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

-11- GO ON TO THE NEXT PAGE.

Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each. Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.

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) ? ? ?

5. 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 .

(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.

(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.

(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.

(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.”

(f) Consider reaction Z at 298 K.

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

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

(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.

1. 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.

(aka: Write the equation that defines the equilibrium condition.)

(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.

(c) For reaction Y at 298 K, calculate the value of Kp.

(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.

(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.”

(f) Consider reaction Z at 298 K.

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

(ii) Determine the value of ∆Hº for the reaction.

(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.


ANSWER #3(a) Since we are asked for Kp, please do NOT use square brackets, and it would be best if you included P to indicate pressure.

Kp =PCO( ) PH2( )PH2O( )

(b) Kp will increase. If the temperature is increased for an endothermic reaction, as indicated by the positive ∆H, then by LeChatelier’s Principle the reaction will shift toward products, thereby absorbing energy. With greater concentrations of products, and lower concentration of reactants at equilibrium, the value of Kp will increase.

OR you can invoke a more mathematical explanation. Because ∆Gº= −RT lnKp and ∆Gº=∆ H 298o −T ∆ S298

o

We can set ∆Gº=∆Gº thus − RT lnKp =∆ H 298o −T ∆ S298

o and lnKp = −∆ H 298o

RT+ ∆ S298

o

R Thus an increase in T for a positive ∆H results in a more positive lnKp, and thus an increase in Kp (We can use this set of formulas because ∆H and ∆S are not very temperature dependent.)

However, you can not simply use just ∆Gº= −RT lnKp to determine the effect of a different T on K, because ∆G is VERY

temperature dependent and this equation is only valid when you know the T for which you know the ∆H. Thus the effect of temperature on K is a function of the sign of ∆H, not the sign of ∆G.

(c) ∆Gº= −RT lnKp thus Kp = e∆G º−RT substitute and solve Kp = e

+29kJ /mol−0.00831kJ /molK( ) 298K( )

Thus Kp = 8.2 × 10−6

(d) The total bond energy of the reactants is larger. Reaction Y is endothermic, since ∆Hº is positive, so there is a net input of energy as the reaction occurs. This means that the total energy required to break the bonds must be greater than the total energy released when the bonds are formed in the products.

(e) This statement is false. Thermodynamic data for an overall reaction have no bearing on how slowly or how rapidly a reaction occurs.

(f i) ∆Sº 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 you can invoke a more mathematical approach using Hess’ Law. Reaction Z can be obtained by reversing reactions X and Y and then adding them together. Thus the ∆Sº values for X and Y would be negative, and adding two negative values together must cause ∆Sº for X to be a negative number.

(f ii) Use Hess’ Law to see that if reaction X and Y are reversed, they will add together to produce reaction Z. Thus adding the negatives of ∆Hº298 for reactions X and Y yields:

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

(f iii) The pressure in the flask decreases. Since only CO and C were put in the reaction vessel, the reaction can only proceed in the forward direction to reach

equilibrium, since some CO2 must be formed in order to achieve equilibrium. The stoichiometry of the reaction shows that for every two moles of CO that reacts, only one mole of will be produced. Thus as the reaction proceeds in the forward direction to reach equilibrium the number of moles of gas in the flask would decrease.


1. Use the principles of thermodynamics to answer the following questions.(a) The gas N2O4 decomposes to form the gas NO2 according to the following equation below.

(i) Predict the sign of ∆Hº for the reaction. Justify your answer

(ii) Predict the sign of ∆Sº for the reaction. Justify your answer.

(b) One of the diagrams below best represents the relationship between ∆Gº and temperature for the reaction given in part (a). Assume that ∆Hº and ∆Sº are independent of temperature.

Draw a circle around the correct graph and explain why you chose that graph in terms of the relationship ∆Gº = ∆Hº − T∆Sº.

(c) A reaction mixture of N2O4 and NO2 is at equilibrium. Heat is added to the mixture while maintained at a constant pressure.(i) Explain why the concentration of N2O4 decreases.

(ii) The value of Keq at 25ºC is 5.0 × 10−3. Will the value of Keq at 100ºC be greater than, less than, or equal to this value?

(d) Using the value of Keq at 25ºC given in part (c)(ii), predict whether the value of ∆Hº is expected to be greater than, less than, or equal to the value of T∆Sº. Explain.

(e) The following diagram shows a representation for an equilibrium mixture of N2O4 and NO2 at a certain temperature. The following container is enlarged, keeping the temperature the same. Sketch a representation for the equilibrium mixture that is achieved with the larger volume.


N NO

O

O

ON

O

O•N

O

O•

∆Gº

T

∆Gº

T

∆Gº

T

N2O4

NO2

ANSWER # 4

(a i) During this decomposition reaction, the N−N bond is broken. Energy is always absorbed when bonds are broken, so the reaction is endothermic and the sign of ∆Hº is positive.

(a ii) 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 ∆Sº is positive.

(b) The leftmost graph should be circled. Since ∆Sº is positive, as T increases, T∆Sº becomes a larger positive number. Thus at higher temperatures, you would be subtracting higher positive numbers from ∆Hº to get the ∆Gº value, thus ∆Gº becomes more negative (decreases) with increasing temperature.

(c i) The reaction is endothermic. For endothermic reactions, increasing temperature will always drive the reaction to the right. This increases the concentration of NO2 and decreases the concentration of N2O4.

(c ii) Again, because the reaction is endothermic, higher temperatures will drive the reaction further to the right, and changes the equilibrium constant. Thus the Keq at 100ºC will be greater than the value of the Keq at 25ºC.

(d) Keq at 25ºC is less than 1, which tells us that ∆Gº must be positive. We know from the prediction in (a i) that the sign of ∆Hº is positive,and from (a ii) that ∆Sº is also positive. Thus considering these signs in the Gibbs equation: ∆Gº = ∆Hº − T∆Sº, we know that ∆Hº must be greater than T∆Sº

(e) When the container is expanded, the equilibrium position will shift to favor more molecules. The new diagram must indicate that 1, 2, or 3 of the N2O4 had decomposed and thus contains at least one molecule of N2O4 remaining. The new diagram must also shows the stoichiometrically correct amount of NO2 formed for every N2O4 decomposed, will be acceptable. In the diagram shown, two N2O4 have decomposed into four NO2.


N2O4

NO2

2. 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?

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

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). (aka: Write the equation that defines the equilibrium condition.)

(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.

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

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


ANSWER #5(a i) 55.8g × 1mol

169.9g= 0.328mol 0.328mol

10.5L= 0.0313M

(a ii) PV = nRT P =0.328mol( ) 0.08206Latmmol−1K −1( ) 378Κ( )

10.5L= 0.969atm

(b) Keq =AsF3[ ] F2[ ]AsF5[ ]

(c i) We always start with 100%, thus if 27.7% decomposed, there must be 72.3% remaining.

Thus [AsF5] = 0.723 × 0.0313 M = 0.0226 M remains at equilibrium.

(c ii) The stoichiometry of the reaction tells us that [AsF3] will equal [F2]. From part (c i) we know that 27% of the AsF5 decomposed, we know that 27% of the AsF5 original concentration will be the amount of AsF3 and F2 that form.

[AsF3] = [F2] = 0.277 × 0.0313 M = 0.00867 M Keq =AsF3[ ] F2[ ]AsF5[ ] =

0.00867[ ] 0.00867[ ]0.0226[ ] = 0.00333

(d) Remember mole fraction is part out of total. You can calculate mole fraction with moles or molarity values.




(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 .


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).





[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






mol AsF5 = 0.0226 M × 10.5 L = 0.237 mol


mol fraction F2 = 2

2 3 5


= 0.09100.0910 + 0.0910 + 0.237

= 0.217

OR

mol fraction F2 = 0.008640.00864 + 0.00864 + 0.0226

= 0.217


fraction of F2(g).





[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






mol AsF5 = 0.0226 M × 10.5 L = 0.237 mol


mol fraction F2 = 2

2 3 5


= 0.09100.0910 + 0.0910 + 0.237

= 0.217

OR

mol fraction F2 = 0.008640.00864 + 0.00864 + 0.0226

= 0.217


fraction of F2(g).

OR





[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






mol AsF5 = 0.0226 M × 10.5 L = 0.237 mol


mol fraction F2 = 2

2 3 5


= 0.09100.0910 + 0.0910 + 0.237

= 0.217

OR

mol fraction F2 = 0.008640.00864 + 0.00864 + 0.0226

= 0.217


fraction of F2(g).


C(s) + CO2(g) ⇄ 2 CO(g) 3. 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 to the right.

(a) Write the expression for the equilibrium constant, Kp, for the reaction. (aka: Write the equation that defines the equilibrium condition.)

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

(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

(ii) the value of the equilibrium constant, Kp.

(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.)

In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s), plus CO2(g) and CO(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.

2008 AP® CHEMISTRY FREE-RESPONSE QUESTIONS

© 2008 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

GO ON TO THE NEXT PAGE. -6-

CHEMISTRYSection II

(Total time—95 minutes)

Part A Time—55 minutes

YOU MAY USE YOUR CALCULATOR FOR PART A.

CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures.

Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the pink cover. Do NOT write your answers on the green insert.

Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent.

C(s) + CO2(g) →← 2 CO(g)

1. 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.

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


ANSWER # 6

(a) Kp =PCO( )2PCO2( )

(b) PV = nRT n = 5.00atm( ) 2.00L( )0.08206Latmmol−1K −1( ) 1,160K( )10.5L

= 0.105mol

(c i) Remember that the total pressure is the sum of the partial pressure of all gases present (Dalton’s Law of Partial Pressures).

AP® CHEMISTRY 2008 SCORING GUIDELINES


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

! ! ! One point is earned for the correct setup.

One point is earned for the correct answer.

(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.

(c ii)

AP® CHEMISTRY 2008 SCORING GUIDELINES


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 .

One point is earned for the correct answer with justification.

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.

(d) The total pressure of gases at equilibrium with a catalyst would be the same as the total pressure of the gases without a catalyst. Although a catalyst would cause the system to reach the equilibrium position more quickly, it would not have any effect on the extent of the reaction.

(e) Since we are given the initial partial pressures of both gases, we can calculate Q which will tell us if we are at equilibrium, and if not, in which direction will the reaction shift in order to achieve equilibrium.

Q =PCO( )2PCO2( ) =

2.00atm( )22.00atm( ) = 2.00 < Kp (27.9) Thus the reaction will shift to the right

and the pressure of the CO2 will decrease as the system approaches equilibrium


2 HI(g) ⇄ H2(g) + I2(g)

4. After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700. K, the reaction represented above occurs. The concentration of HI(g) as a function of time is shown to the right.

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

(b) What is [HI] at equilibrium?

(c) Determine the equilibrium concentrations of H2(g) and I2(g).

(d) On the graph above, make a sketch that shows how the concentration of H2(g) changes as a function of time.

(e) Calculate the value of the following equilibrium constants at 700. K.

(i) Kc

(ii) Kp

(f) At 1,000 K, the value of Kc for the reaction is 2.6 × 10−2. In an experiment, 0.75 mole of HI(g), 0.10 mole of H2(g), and 0.50 mole of I2(g) are placed in a 1.0 L container and allowed to reach equilibrium at 1,000 K. Determine whether the equilibrium concentration of HI(g) will be greater than, equal to, or less than the initial concentration of HI(g). Justify your answer.

AP* General Equilibrium Free Response Questions page 3

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ANSWER # 7

(a) Kc =H 2[ ] I2[ ]HI[ ]2

(b) From the graph, [HI] is 0.80 M

(c) Use a RICE box to determine the equilibrium concentrations of [I2] = [H2] = 0.10 M

(d) The graph should have three important characteristics.

[H2] starts at 0 M,

increases to 0.1 M,

reaches equilibrium at the same time that [HI] reaches equilibrium

(e i) Kc =H 2[ ] I2[ ]HI[ ]2

=0.10[ ] 0.10[ ]0.80[ ]2

= 0.016

(e ii) At equilibrium [H2] = [I2] = 0.10 mol, PH2 = PI2 =0.10mol( ) 0.08206( ) 700K( )

1L( )

and [HI] = 0.80 mol, PHI =0.80mol( ) 0.08206( ) 700K( )

1L( )

Kp =

0.10mol( ) 0.08206( ) 700K( )1L( )

⎛⎝⎜

⎞⎠⎟

0.10mol( ) 0.08206( ) 700K( )1L( )

⎛⎝⎜

⎞⎠⎟

0.80mol( ) 0.08206( ) 700K( )1L( )

⎛⎝⎜

⎞⎠⎟

2 =0.10( ) 0.10( )0.80( )2

= 0.016

Or perhaps you remember that Kp = Kc(RT )∆ n and Kc(RT )0 thus Kp = Kc = 0.016

(f) Q =H 2[ ] I2[ ]HI[ ]2

=0.10[ ] 0.50[ ]0.75[ ]2

= 8.9 ×10−2 > Kc

Thus to establish equilibrium the reaction must shift to the left, therefore [HI] will increase.


R 2 HI ⇋ H2 + I2

I 1.0 M 0.3 0.16

C -0.20 M +0.10 M +0.10 M

E 0.80 M 0.10 M 0.10 M

AP* General Equilibrium Free Response Questions page 3

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C(s) + H2O(g) ⇄ CO(g) + H2(g) ∆H° = +131 kJ

5. A rigid container holds a mixture of graphite pellets, C(s), water, H2O(g), carbon monoxide, CO(g), and hydrogen, H2(g) at equilibrium as shown by the reaction above. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a) Additional H2(g) is added to the equilibrium mixture at constant volume.

(b) The temperature of the equilibrium mixture is increased at constant volume.

(c) The volume of the container is decreased at constant temperature.

(d) The graphite pellets are pulverized.


ANSWER # 8

(a) The number of moles of CO will decrease because by LeChatelier’s Principle, adding the product H2 will cause the reaction to shift to the left to adjust to a new equilibrium position.

(b) The number of moles of CO will increase because the reaction is endothermic and adding heat will drive the reaction to the right.

(c) The number of moles of CO will decrease because there are more moles of gas on the right than the left, thus decreasing the volume which increases the partial pressure of all the gases, and by LeChatelier’s Principle, the reaction will shift to the left to adjust to a new equilibrium position.

(d) The number of moles of CO will stay the same because solids are not in the equilibrium expression and have no effect on the equilibrium position.

Of course if solids are in the reaction, they must be present to be at equilibrium. Pulverizing the solid will increase the speed of the reaction and equilibrium would be established more quickly, but will not affect the equilibrium position.


6. For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased.

PCl5(g) ⇄ PCl3(g) + Cl2(g)

(a) Predict the sign of ∆S° for the reaction? Justify.

(b) What change, if any, will occur in ∆G° for the reaction as the temperature is increased. Explain your reasoning in terms of thermodynamic principles.

(c) Predict the sign of ∆Hº. Justify.

(d) If He gas is added to the original reaction mixture at constant volume and temperature, what will happen to the partial pressure of Cl2? Explain.

(e) If the volume of the original reaction is increased at constant temperature to twice the original volume, what will happen to the number of moles of Cl2 in the reaction vessel? Explain.


ANSWER # 9

(a) ∆Sº is positive because there are more moles of products.

(b) ∆Gº will become more negative (or less positive). Since ∆Sº is positive, as T increases, T∆Sº becomes a larger positive number. Thus at higher temperatures, you would be subtracting higher positive numbers from ∆Hº to get the ∆Gº value, thus ∆Gº becomes more negative (decreases) with increasing temperature.

(c) ∆Hº must be positive. Since adding heat produced more products, the reaction must be endothermic.

(d) Adding He gas will cause no change on the partial pressure of Cl2. The total pressure of the container would increase, but the partial pressures of the individual gases will not change, thus there would be no change in the equilibrium equation, thus there will be no shift in the equilibrium position.

(e) The moles of Cl2 will increase. Increasing the volume of the container will decreases the partial pressure of all the gases. Since there are more moles of gases on the product side, and by LeChatelier’s Principle, the reaction will shift to the right to adjust to a new equilibrium position.


CO2(g) + H2(g) ⇄ H2O(g) + CO(g)

7. When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.

(a) Write the equilibrium expression for the reaction above.

(b) What is the mole fraction of CO(g) in the equilibrium mixture?

(c) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.

(d) Determine Kp in terms of Kc for this system.

(e) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature.

(f) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature.


[H2] = 0.20 mol/L

[CO2] = 0.30 mol/L

[H2O] = [CO] = 0.55 mol/L

ANSWER # 10

(a) Kc =H 2O[ ] CO[ ]CO2[ ] H 2[ ] or Kp =

PH2O( ) PCO( )PCO2( ) PH2( )

(b) Mole fraction can be done in moles or molarity. Remember: mole fration = molestotal moles

There are no units on the number.

mole frationofCO = 0.550.55 + 0.55 + 0.2 + 0.3

= 0.551.60

= 0.34

(c) Kc =H 2O[ ] CO[ ]CO2[ ] H 2[ ] Kc =

0.55[ ] 0.55[ ]0.30[ ] 0.20[ ] = 5.0

(d) Remember, Kp = Kc (RT )prod−react

Kp = Kc (RT )2−2

Kp = Kc (RT )0

Thus for this reaction, Kp = Kc = 5.0

(e) Consider making a RICE box. Conversion of 30% of CO back to product means that 0.30 × 0.55 = −0.165 M of CO is changed into reactants. Water must also react (thus be lost) at the same time.

R CO2(g) + H2(g) ⇄ H2O(g) + CO(g)CO2(g) + H2(g) ⇄ H2O(g) + CO(g)CO2(g) + H2(g) ⇄ H2O(g) + CO(g)CO2(g) + H2(g) ⇄ H2O(g) + CO(g)

I 0.30 0.20 0.55 0.55

C +0.165 +0.165 -0.165 -0.165

E 0.465 0.365 0.385 0.385

(f) Consider a RICE box again. Remember that Kc at 2.000 K = 5. The container is 3.0 L, thus we should convert to molarity.

R CO2(g) + H2(g) ⇄ H2O(g) + CO(g)CO2(g) + H2(g) ⇄ H2O(g) + CO(g)CO2(g) + H2(g) ⇄ H2O(g) + CO(g)CO2(g) + H2(g) ⇄ H2O(g) + CO(g)

I 0.167 0.167 0 0

C -x -x +x +x

E 0.167 − x 0.167 − x x x

5 =x[ ] x[ ]

0.167 − x[ ] 0.167 − x[ ] this is a perfect square, so we can square root both sides: 5 =x[ ] x[ ]

0.167 − x[ ] 0.167 − x[ ]

2.24 =x[ ]

0.167 − x[ ] 2.24 0.167 − x[ ] = x[ ] 0.374 − 2.24x = x 0.374 = 3.24x x = 0.116

Thus the molarity of CO would be 0.116 M


Kc =0.385[ ] 0.385[ ]0.465[ ] 0.365[ ] = 0.87

Molarity = 0.50mol3.0L

= 0.167M

2 NaHCO3(s) ⇄ Na2CO3(s) + H2O(g) + CO2(g)

8. Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation above.

(a) A sample of 100. grams of solid NaHCO3 was placed in a previously evacuated rigid 5.00-liter container and heated to 160ºC. Some of the original solid remained and the total pressure in the container was 7.76 atmospheres when equilibrium was reached. Calculate the number of moles of H2O(g) present at equilibrium.

(b) How many grams of the original solid remain in the container under the conditions described in (a)?

(c) Write the equilibrium expression for the equilibrium constant, Kp, and calculate its value for the reaction under the conditions in (a).

(d) If 110. grams of solid NaHCO3 had been placed in the 5.00-liter container and heated to 160ºC, what would the total pressure have been at equilibrium? Explain.


ANSWER #11

(a) Since the total pressure, 7.76 atm is caused by the two gases that formed in equal stoichiometric quantities, they must each be causing the half of the total pressure, 3.88 for H2O and 3.88 atm for CO2.

n = PVRT

n =3.88atm( ) 5.00L( )

0.0821atmLmol−1K −1( ) 433K( ) = 0.545mol H 2O

(b) 0.545molH 2O × 2molNaHCO3

1molH 2O= 1.09molNaHCO3 reacted. 1.09molNaHCO3 ×

84.01gNaHCO3

1molNaHCO3

= 91.6gNaHCO3 reacted.

Since there were 100 g of NaHCO3 in the container to start; 100. g − 91.6 g reacted = 8 g remaining.

(c) Kp = PH2O( ) PCO2( ) Kp = (3.88) (3.88) = 15.1

(d) When equilibrium was reached, the pressure would still be 7.76 atm since some solid remained when reaching equilibrium described in part (a), adding more solid would not change the equilibrium proportions of the gases, since there was no temperature change. Solids are not part of the equilibrium expression and adding even more than was necessary to reach equilibrium will not change the outcome.


9. At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by the following equation:

SbCl5(g) ⇄ SbCl3(g) + Cl2(g)

(a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0 liter container at 182ºC.

(i) What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?

(ii) What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?

(b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182ºC, calculate the value for either equilibrium constant Kp or Kc, for this decomposition reaction. Indicate whether you are calculating Kp or Kc.

(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00 liter container maintained at a temperature different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?


ANSWER #12

(a i) 89.7gSbCl5 ×1molSbCl5299gSbCl5

= 0.300molSbCl5 0.300molSbCl5

15.0L= 0.0200M

(a ii) P = nRTV

P =0.300mol( ) 0.0821atmLmol−1K −1( ) 455K( )

15L( ) = 0.747atm

(b) When 29.2 % decomposes, then 70.8 % of each of the two products must form in equal quantities, since they are stoichiometrically equal. Solve the problem the same way in either molarity or pressure.

[SbCl3] = [Cl2] = 0.0200 M × 0.292 = 0.00584 M and [SbCl3] = 0.0200 M × 0.708 = 0.0142 M

Kc =SbCl3[ ] Cl2[ ]SbCl5[ ] =

0.00584[ ] 0.00584[ ]0.0142[ ] = 0.00240

(SbCl3) = (Cl2) = 0.747 atm × 0.292 = 0.218 atm and (SbCl3) = 0.747 atm × 0.708 = 0.529 atm

Kp =PSbCl3( ) PCl2( )PSbCl5( ) =

0.218( ) 0.218( )0.529( ) = 0.0898

(c) This is a bit of a tricky question.

First we are being told that at this new temperature Kc = 0.117. This equilibrium was achieved by placing 1.00 mol of SbCl3 and adding Cl2. The reaction will of course proceed to the left.

When reaching the desired 0.700 mole quantity of SbCl3, 0.300 mol of SbCl3 must have reacted, producing 0.30 mole SbCl5 (since there is a 1:1 stoichiometric ratio) in the 2.00 L container, resulting in 0.15 M SbCl5

SbCl3[ ] = 0.700mol2.00L= 0.350M [Cl2] = x

Kc = 0.117 =0.350[ ] x[ ]0.15[ ] x = [Cl2] = 0.050 M 0.050 M × 2.00 L = 0.10 mol Cl2

and finally, the last bit.

The 0.30 mol of SbCl3 that reacted with the incoming Cl2, must have reacted with 0.30 mol of Cl2 to produce the 0.30 mol of SbCl5, thus a total of 0.40 mol of Cl2 must have been placed in the container with the 1.00 mole of SbCl3 to reach this equilibrium position.

(You could have set up a RICE box to help with these calculations.)


10. Sulfuryl chloride, SO2Cl2, is a highly reactive gaseous compound. When heated, it decomposes as follows:

SO2Cl2(g) ⇄ SO2(g)+ Cl2(g) This decomposition is endothermic. A sample of 3.509 grams of SO2Cl2 is placed in an evacuated 1.00 liter bulb and the temperature is raised to 375 K.

(a) What would be the pressure in atmospheres in the bulb if no dissociation of the SO2Cl2(g) occurred?

(b) When the system has come to equilibrium at 375 K, the total pressure in the bulb is found to be 1.43 atmospheres. Calculate the partial pressures of SO2, Cl2, and SO2Cl2 at equilibrium at 375 K.

(c) Give the expression for the equilibrium constant (either Kp or Kc) for the decomposition of SO2Cl2(g) at 375 K. Calculate the value of the equilibrium constant.

(d) If the temperature were raised to 500 K, what effect would this have on the equilibrium constant? Explain briefly.


ANSWER #13This problem is very similar to # 12

(a) 3.509gSO2Cl2 ×1molSO2Cl2135gSO2Cl2

= 0.02600molSO2Cl2

P = nRTV

P =0.02600mol( ) 0.0821atmLmol−1K −1( ) 375K( )

1.00L( ) = 0.800atm

(b) Remember that the pressure of each gas will add to the total pressure, and remember that the amount of each gas that forms is stoichiometrically related to the amount of SO2Cl2 gas that decomposes.

Thus the pressure of SO2Cl2 = (0.800 − x) and the pressure of SO2 = Cl2 = x

Thus the total pressure = SO2Cl2 + SO2 + Cl2 = 1.43 = (0.800 − x) + x + x

x = 0.63 atm which is the pressure of SO2 and Cl2, the pressure of SO2Cl2 = 0.800 − 0.63 = 0.17 atm

(c) Kp =PSO2( ) PCl2( )PSO2Cl2( ) =

0.63( ) 0.63( )0.17( ) = 2.3

Alternatively you could change pressures of each gas to moles:

n =0.63( ) 1.00L( )

0.0821atmLmol−1K −1( ) 375K( ) = 0.0205molSO2 = Cl2 n =0.17( ) 1.00L( )

0.0821atmLmol−1K −1( ) 375K( ) = 0.00552molSO2Cl2

Since the container is 1.00 L, then moles = molarity. Kc =SO2[ ] Cl2[ ]SO2Cl2[ ] =

0.0205[ ] 0.0205[ ]0.00552[ ] = 0.076

(d) The reaction is endothermic which will absorb heat during the decompositions, thus when heat (a stress) is placed on the system, the reaction will shift so as to relieve that stress, thus the reaction will shift to the right and cause a larger K at 500 K.


unit f fr practice general equilibrium (pg 1 of 26) h o

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