AP Practice Questions
1)
The tables above contain information for determining thermodynamic properties of the reaction below.
C2H5Cl(g) + Cl2(g) <===> C2H4Cl2 (g) + HCl(g)
(a) Calculate ΔH° for the reaction above, using the table of average bond dissociation energies.
(b) Calculate ΔS° for the reaction at 298 K, using data from either table as needed.
(c) Calculate the value for Keq for the reaction at 298 K.
(d) What is the effect of an increase in temperature on the value of the equilibrium constant? Explain your
answer.
2)
BCl3(g) + NH3(g) <---> Cl3BNH3(s)
The reaction represented above is a reversible reaction.
(a) Predict the sign of the entropy change, ΔS, as the reaction proceeds to the right. Explain your prediction.
(b) If the reaction is thermodynamically favored to proceed to the right, predict the sign of the enthalpy
change, ΔH. Explain your prediction.
(c) The direction in which the reaction is thermodynamically favored to proceed changes as the temperature is
increased above a specific temperature. Explain.
(d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific
temperature at which the thermodynamically favored direction of the reaction changes? Explain.
3)
Cl2(g) + 3 F2(g) ---> 2 ClF3(g)
ClF3 can be prepared by the reaction represented by the equation above. For ClF3 the standard enthalpy of
formation, ΔH°f, is - 163.2 kilojoules/mole and the standard free energy of formation, ΔG°f, is - 123.0
kilojoules/mole.
(a) Calculate the value of the equilibrium constant for the reaction at 298 K.
(b) Calculate the standard entropy change, ΔS°, for the reaction at 298 K.
(c) If ClF3 were produced as a liquid rather than as a gas, how would the sign and magnitude of ΔS for the
reaction be affected? Explain.
(d) at 298 K the absolute entropies of Cl2(g) and ClF3(g) are 222.96 joules per mole-Kelvin and 281.50 joules
per mole-Kelvin, respectively.
(i) Account for the larger entropy of ClF3(g) relative to that of Cl2(g).
(ii) Calculate the value of the absolute entropy of F2(g) at 298 K.
4)
Br2 (l) ---> Br2 (g)
At 25 °C the equilibrium constant, Kp, for the reaction above is 0.281 atmosphere.
(a) What is ΔG°298 for this reaction?
(b) It takes 193 joules to vaporize 1.00 gram of Br2(l) at 25 °C and 1.00 atmosphere pressure. What are the
values of ΔH°298 and of ΔS°298 for this reaction?
(c) Calculate the normal boiling point of bromine. Assume that ΔH° and ΔS° remain constant as the
temperature is changed.
(d) What is the equilibrium vapor pressure of bromine at 25 °C ?
5)
C(s)
H2(g)
Enthalpy of Combustion
ΔH° Kilojoules/mole
- 393.5
- 285.8
Absolute Entropy, S°
Joules/mole-K
5.740
130.6
C2H5OH(l)
H2O(l)
- 1366.7
------------
160.7
69.91
Substance
(a) Write a separate, balanced chemical equation for the combustion of each of the following: C(s), H2(g), and
C2H5OH(l). Consider the only products to be CO2(g) and/or H2O(l).
(b) In principle, ethanol can be prepared by the following reaction.
2 C(s) + 2 H2(g) + H2O(l) --> C2H5OH(l)
Calculate the standard enthalpy change, ΔH°, for the preparation of ethanol, as shown in the reaction above.
(c) Calculate the standard entropy change, ΔS°, for the reaction given in part (b)
(d) Calculate the value of the equilibrium constant at 25 °C for the reaction represented by the equation in part
(b).
6)
CO(g) + 2 H2(g) <===> CH3OH(l)
For this reaction, ΔH° = -128.1 kilojoules
ΔHf° (kJ mol¯1) ΔGf° (kJ mol¯1) S° (J mol¯1 K¯1)
CO(g)
CH3OH(l)
-110.5
-137.3
+197.9
-238.6
-166.2
+126.8
The data in the table above were determined at 25 °C.
(a) Calculate ΔG° for the reaction above at 25 °C.
(b) Calculate Keq for the reaction above at 25 °C.
(c) Calculate ΔS° for the reaction above at 25 °C.
(d) In the table above, there are no data for H2. What are the values of ΔHf°, ΔGf°, and of the absolute entropy,
S°, for H2 at 25 °C?
7)
Substance
C(s)
CO2(g)
H2(g)
Standard Heat of
Formation, ΔHf°,
in kJ mol¯1
0.00
Absolute Entropy
S°, in J mol¯1 K¯1
5.69
-393.5
0.00
213.6
130.6
H2O(l)
O2(g)
-285.85
0.00
69.96
205.0
C3H7COOH(l)
?
226.3
The enthalpy change for the combustion of butyric acid at 25 °C, ΔH°comb, is -2,183.5 kilojoules per mole. The
combustion reaction is
C3H7COOH(l) + 5 O2(g) <===> 4 CO2(g) + 4 H2O(l)
(a) From the data above, calculate the standard heat of formation, ΔHf°, for butyric acid at 25 °C.
(b) Write a correctly balanced equation for the formation of butyric acid from its elements.
(c) Calculate the standard entropy change, ΔSf°, for the formation of butyric acid at 25 °C. The entropy
change, ΔS°, for the combustion reaction above is -117.1 J K¯1 at 25 °C.
(d) Calculate the standard free energy of formation, ΔGf°, for butyric acid at 25 °C.
8)
2 H2S(g) + SO2(g) <===> 3 S(s) + 2 H2O(g)
At 298 K, the standard enthalpy change, ΔH°, for the reaction represented above is -145 kilojoules.
(a) Predict the sign of the standard entropy change, ΔS°, for the reaction. Explain the basis for your prediction.
(b) At 298 K, the forward reaction (i.e., toward the right) is thermodynamically favored. What change, if any,
would occur in the value of ΔG° for this reaction as the temperature is increased? Explain your reasoning using
thermodynamic principles.
(c) What change, if any, would occur in the value of the equilibrium constant, Keq, for the situation described
in (b)? Explain your reasoning.
(d) The absolute temperature at which the forward reaction is no longer thermodynamically favored can be
predicted. Write the equation that is used to make the prediction. Why does this equation predict only an
approximate value for the temperature?
1) a) two points
[delta]H = bonds broken minus bonds formed.
[delta]H =
C2H5Cl
+
Cl-Cl
--->
C2H4Cl2
+
HCl
(2794
+
243)
minus
(2757
+
431)
[delta]H = 3037 - 3188 = - 151 kJ mol¯1
OR
[delta]H =
CH
+
Cl-Cl
--->
C-Cl
+
HCl
(414
+
243)
minus
(377
+
431)
[delta]H = - 151 kJ mol¯1
b) four points
[delta]G = [[delta]G°f C2H4Cl2 + [delta]G°f HCl] - [[delta]G°f C2H5Cl + [delta]G°f Cl2]
= (- 80.3 - 95.3) - (- 60.5 + 0) = - 115 kJ
[delta]G = [delta]H - T[delta]S
[delta]S = ((- 151 kJ - (- 115 kJ)) / 298 = - 0.120 kJ mol¯1 K¯1
c) two points
[delta]G = - RT ln K
- ln K = - 11510 / (8.314 x 298)
ln K = 46.46
K = 1.50 x 1020
d) one point
Keq will decrease with an increase in T because the reverse (endothermic) reaction will be favored with
addition of heat
OR
[delta]G will be less negative with an increase in temperature (from [delta]G = [delta]H - T[delta]S) which will
cause K to decrease.
2) a) two points
ΔS will be negative. The system becomes more ordered as two gases form a solid.
b) two points
ΔH must be negative. For the reaction to be thermodynamically favored, ΔG must be negative, so ΔH must be
more negative than -TΔS is positive.
c) two points
As T increases, -TΔS increases. Since ΔS is negative, the positive -TΔS term will eventually exceed ΔH (which
is negative), making ΔG positive. (In the absence of this, ΔG = ΔH - TΔS and general discussion of the effect
of T and ΔS gets 1 point.)
d) two points
The equilibrium constant is 1. The system is at equilibrium at this temperature with an equal tendancy to go in
either direction.
OR
ΔG = 0 at equilibrium so K = 1 in ΔG = -RT ln K
(In the absence of these, ΔG = -RT ln K gets 1 point).
The above concludes the AP scoring standards published in 1991. The following is simply alternate ways of
answering which the AP readers may or may not have given full credit to.
a) The amount of entropy goes down, ΔS is negative.
b) ΔG = ΔH - TΔS. If ΔS is negative, then ΔH must also be negative to get a negative ΔG.
c) Let us say ΔG is positive when ΔH is positive and ΔS is positive. As T goes up - TΔS becomes more
negative until it makes ΔG (which equals ΔH - TΔS) become negative.
d) At the temperature when the direction changes, the rate forward = the rate reverse. Since K = kf / kr, this
equals 1.
3) average = 3.5 (Only 30 scores of nine; kids did not see stoichiometry in (b), had problems on which gas
constant to use, and a hard time in (c) in relating a more negative value.)
a) two points [delta]G° = - RT ln K; rearranging gives ln K = [delta]G° ÷ - RT ln K = - 246,000 J ÷ - ((8.31
J/mol K) (298 K)) = 1.32 x 1042
b) two points
[delta]G° = [delta]H° - T[delta]S°
- 246,000 J = - 326,400 J - (298) (x)
x = - 270 J / K
c) two points
[delta]S is a larger negative number
ClF3 (liquid) is more ordered (less disordered) than ClF3 (gas)
This was my answer before I saw the standard given above: A liquid is more ordered than a gas. There would
be a greater entropy change in gas + gas ---> liquid than in gas + gas ---> gas. Therefore, sign is the same, but
absolute magnitude is greater.
d) three points
i) ClF3 is a more complex molecule (i.e. more atoms) with more vibrational and rotational degrees of freedom
than Cl2
ii) Cl2 + 3 F2 ---> 2 ClF3; use Hess's Law
[sigma]Srxn = [sigma]S products - [sigma]S reactants
- 270 = [2 (281.50)] - [222.96 + 3x]
x = 203 J mol¯1 K¯1
4) Average score = 3.2
a) two points
[delta]G° = - RT lnK
= - (8.31 J mol¯1 K¯1) (298 K) (ln 0.281)
= 3.14 x 103 J / mol
b) four points
[delta]H° = (193 J/g) (160. g/mol) = 3.08 x 104 J / mol
[delta]G° = [delta]H° - T [delta]S°
[delta]S° = ([delta]H° - [delta]G°) / T
= [(3.084 x 104) - (3.14 x 103)] / 298
= (2.770 x 104 J / mol) / 298 K = 92.9 J / mol K
c) two points
At boiling pt, [delta]G° = 0 and thus,
T = [delta]H° / [delta]S° = 3.08 x 104 / 92.9 = 332 K
d) one point
Vapor pressure = 0.281 atm
Note added for student's benefit: this comes directly from the fact that Kp = PBr2.
5) Average score = 3.25
a) two points
C + O2 --> CO2
2 H2 + O2 --> 2 H2O
C2H5OH --> 2CO2 + 3 H2O
(Half credit for recognition that combustion is reaction with O2)
b) three points
2 C + 2 O2 ---> 2 CO2
[delta]H° = 2 (- 393.5) = - 787.0 kJ
2 H2 + O2 ---> 2 H2O
2 CO2 + 3 O2 ---> C2H5OH + 3 O2
[delta]H° = 2(- 285.8) = - 571.6 kJ
[delta]H° = - (- 1366.7) = 1.366.7 kJ
Sum of three equations above
2 C + 2 H2 + H2O ---> C2H5OH
[delta]H° = + 8.1 kJ
(1 point for correct [delta]H° for each of the first 3 reations)
OR
[delta]H° combustion C(s) = [delta]Hf° CO2(g)
[delta]H° combustion H2(g) = [delta]Hf° H2O(l)
C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O [delta]H° = -1,366.7kJ
[delta]H° reaction = [sigma] [delta]Hf° (products) - [sigma] [delta]Hf° (reactants)
= [2( -393.5) + 3(- 258.8) - [delta]Hf°C2H5OH - 0] kJ = - 277.7 kJ
2 C + 2 H2 + H2O ---> C2H5OH [delta]H° reaction = [sigma] [delta]H° (products) - [sigma] [delta]H°
(reactants)
= [- 277.7 - 0 - 0 - (- 285.8)] kJ = + 8.1 kJ
c) one point
[delta]S° = [sigma] S° (products) - [sigma] S° (reactants)
= (160.7 - 11.5 - 261.2 - 69.9) J / mol K = - 181.9 J / mol K
d) three points
(1) [delta]G° = [delta]H° - T [delta]S°
= 8,100 J - (298) (- 181.9) J = 8,100 J + 54,200 J = 62,300 J
(2) [delta]G° = - RTlnK
lnK = - [delta]G / RT
= - 62.300J ÷ [(8.31J/mol-K) (298K)] = - 25.2
K = 1.1 x 10¯11
OR
[delta]G° = - 2.303 RT logK
logK = - [delta]G / 2.303 RT
= - 62.000J ÷ [(2.303) (8.31) (298)] = - 10.9
K = 1.3 x 10¯11
If the terms are not rounded, K = 1.2 x 10¯11
correct substitutions in (1) and (2) earns one point
6)
a) two points
[delta]G° = [sigma] [delta]Gf° (prod) - [sigma] [delta]Gf° (react)
[delta]G° = - 166.2 - (- 137.3 + 2 (0))
[delta]G° = -28.9 KJ/mol
b) two points
[delta]G° = - RT ln K (or - 2.3 RT log K)
3
- 28.9 = - ( 8.31 x 10¯ ) (298) ln K
ln K = 11.67
K = 1.17 x 10
c) two points
5
[delta]G° = [delta]H° - T [delta]S°
- 28,900 = -128,100 - 298 [delta]S°
[delta]S° = - 99,200 / 298 = - 333 J / mol-K
d) three points
[delta]Hf° (H2) = 0 and [delta]Gf° (H2) = 0
deltaS° = [sigma]S° (prod) - [sigma]S° (react)
- 333 J / mol-K = 126.8 J / mol-K - 197.9 J / mol-K - 2 S° (H2)
S° = 131 J / mol-K
7) Average score 5.52
a) three points
From Hess's law:
[delta]H°f = [4 (393.5) + 4 (205.85)] - 2183.5 kJ = - 533.8 kJ
b) one point
4 C(s) + 4 H2(g) + O2(g) ---> C3H7COOH
c) two points
[delta]S°f butyric acid = S° butyric acid - 4 S° carbon - 4 S° H2 - S° O2
[delta]S°f butyric acid = [226.3 - 4(5.69) - 4(130.6) - 205] joules/K = - 523.9 joules/K
Note: If part c was based on the equation in part b, and was done correctly, credit was given for part c even if
part b was wrong.
d) three points
[delta]G°f butyric acid = [delta]H°f - T [delta]S°f
[delta]G°f = [- 533.9 - (298) (- 0.524)] kJ
[delta]G°f= - 377.7 kJ
Note: For each of the problems, a maximum of one point was subtracted for gross misuse of significant figures
and. or for a mathematical error if correct principles were used.
8) Note: for parts (a), (b), and (c), just writing an equation is not sufficient for the 'explanation" point. To earn
credit, the student must connect the equation to the issue to be explained.
a) two points
Statement that [delta]S° is negative
3 moles of gas ---> 2 moles of gas plus solid (3 moles ---> 2 moles earns no points), OR
2 gases ---> one gas + solid OR
use of [delta]G° = [delta]H° - T[delta]S° with [delta]G° = 0
b) two points
[delta]G° is less negative, goes to 0, goes positive, gets larger
Explanation using [delta]G° = [delta]H° - T[delta]S°
c) two points
Keq decreases (exponent goes more negative) as T increases
OR
Keq goes from > 1, to 1, to < 1, as T increases
Correct explanation using the equation
[delta]G° = - RT ln K (or ln(k1 / k2) = ([delta]H° / R) (1/T2 - 1/T1)
OR
higher T favors the reverse reaction (Le Châtelier) because the forward reaction is exothermic.
Note: if answer for (a) is that [delta]S° is positive, then statement that Keq will decrease or increase depending
on the relative magnitudes of T and [delta]G° change earns two points.
d) two points
Since [delta]G° = 0 at this point, the equation is T= [delta]H° / [delta]S°
([delta]G° = [delta]H° - T [delta]S°S is NOT sufficient without [delta]G° = 0.)
Prediction is not exact since [delta]H° and/or [delta]S° vary with T.