Download AP Chem Chapter 16 Review Packet

Chapter 16 Review Packet
AP Chemistry Chapter 16 Practice Multiple Choice Portion
1.
For which process is ΔS negative?
Note:
ΔS = Sfinal  Sinitial therefore, if ΔS is positive, Sfinal > Sinitial
if ΔS is negative, Sfinal < Sinitial
a.
b.
c.
d.
e.
evaporation of 1 mole of CCl4 (ℓ)
Ssolid < Sliquid << Sgas so ΔS = +
mixing 5 mL of ethanol with 25 mL of water
ΔS = + because larger volume and more positional disorder
compressing 1 mole Ne at constant temperature
ΔS =  because a decrease in volume is less positions
raising the temperature of 100 g Cu from 275 K to 295 K
T increases so S increases (greater motion) so ΔS = +
grinding a large crystal of KCl to powder
more positions so ΔS = +
2.
In what reaction is ΔS° expected to be positive?
a.
I2 (g)  I2 (s)
Sfinal < Sinitial so ΔS = 
b.
H2O (ℓ)  H2O (s)
Sfinal < Sinitial so ΔS = 
c.
CH3OH (g) + 1.5 O2 (g)  CO2 (g) + 2 H2O (ℓ)
2.5 gaseous moles  1 gaseous mole; 1 2.5 = 1.5 so ΔS = 
d.
2 O2 (g) + 2 SO (g)  2 SO3 (g)
4 gaseous moles  2 gaseous mole; 2 4 = 2 so ΔS = 
e.
none of these
3.
Which statement is true?
a.
All real processes are irreversible.
page 809
b.
A thermodynamically reversible process takes place infinitely fast.
c.
In a reversible process, the state functions of the system are always much greater
than those of the surroundings.
d.
There is always more heat given off to the surroundings in a reversible process
than in an unharnessed one.
e.
All statements (a-d) are true.
4.
A 100-mL sample of water is placed in a coffee cup calorimeter. When 1.0 g of an ionic
solid is added, the temperature decreases from 21.5°C to 20.8°C as the solid dissolves.
For the dissolving of the solid
Note that the temperature of the solution decreased from 21.5°C to 20.8°C, so the process is
endothermic!!
a.
ΔH > 0
ΔH = + for endothermic reactions
b.
ΔSuniv < 0
The solid dissovles so ΔSuniv must be positive.
c.
ΔSsys < 0
The solid dissolves so it has access to a larger volume (more positions) so ΔSsys
will increase.
d.
ΔSsurr > 0
ΔSsurr is negative for endothermic reactions.
e.
none of these
5.
Which of the following result(s) in an increase in the entropy of the system?
I.
II.
III.
IV.
V.
a.
b.
c.
d.
e.
6.
See diagram to the right. (Closed on reactant side/ open on product
side) (more positions, so increases S)
Br2(g)  Br2(ℓ)
(gas to liquid, so decreases S)
NaBr (s)  Na+1 (aq) + Br1 (aq)
(solid dissolves, more positions, so increases S)
O2 (298 K)  O2 (373 K)
(moving faster, greater disorder, so increases S)
NH3 (1 atm, 298 K)  NH3 (3 atm, 298 K)
(increase P, so decrease V, less positions, so decreases S)
I
II, V
I, III, IV
I, II, III, IV
I, II, III, V
A mixture of hydrogen and chlorine remains unreacted until it is exposed to ultraviolet
light from a burning magnesium strip. Then the following reaction occurs very rapidly:
H2 (g) + Cl2 (g)  2 HCl (g)
ΔG = 45.54 kJ
ΔH = 44.12 kJ
ΔS = 4.76 J/K
(spontaneous because negative)
(exothermic because negative)
Both ΔH and ΔS are negative, so the reaction is
spontaneous at low temperatures.
Select the statement below that best explains this behavior.
a.
b.
c.
d.
e.
7.
Ten identical coins are shaken vigorously in a cup and then poured out onto a table top.
Which of the following distributions has the highest probability of occurrence?
(T = Tails, H = Heads)
a.
b.
c.
d.
e.
8.
The reactants are thermodynamically more stable than the products.
Thermodynamic Stability : The enthalpy or potential energy of a compound
relative to a reference state. For exothermic reactions we may say that the
products are thermodynamically more stable than the reactants. The opposite
would be true for endothermic reactions.
The reaction has a small equilibrium constant.
Thermodynamics does not tell us about kinetics.
The ultraviolet light raises the temperature of the system and makes the reaction
more favorable.
The reaction is exothermic so raising the temperature is not necessary.
The negative value for ΔS slows down the reaction.
Thermodynamics does not tell us about kinetics.
The reaction is spontaneous, but the reactants are kinetically stable.
If reactants are said to be kinetically stable, they react extremely slowly or not at
all.
T10H0
T8H2
T7H3
T5H5
T4H6
Which of the following statements is always true for a spontaneous process?
I.
II.
III.
IV.
a.
b.
c.
d.
e.
ΔSsys > 0 (it can be  and the process be spontaneous at low temps)
ΔSsurr > 0 (not even in answer choices)
ΔSuniv > 0 (2nd law of thermodynamics)
ΔGsys > 0 (ΔG must be negative for spontaneity.)
I
III
IV
I and III
III and IV
9.
For a spontaneous endothermic process, which of the following must be true?
I.
II.
III.
IV.
a.
b.
c.
d.
e.
10.
11.
wmax = ΔG (TRUE!!)
ΔSsurr > 0 (represents exothermic!)
ΔS cannot be negative
(If ΔH is +, then ΔS must also be + in order for ΔG to be negative.)
ΔG is positive (cannot be spontaneous)
all are true
none are true
I and III
I, II, and IV
III and IV
OMIT
For the reaction A + B  C + D, ΔH° = +40 kJ and ΔS° = +50 J/K. Therefore, the
reaction under standard conditions is
a.
b.
c.
d.
e.
spontaneous at temperatures less than 10 K
spontaneous at temperatures greater than 800 K
spontaneous only at temperatures between 10 K and 800 K
spontaneous at all temperatures
nonspontaneous at all temperatures
(If ΔH is + and ΔS is +, then the reaction will be spontaneous at high temperatures.)
Therefore, T > ΔH / ΔS
T > 40,000 J / 50 J/K
T > 800 K
12.
In which case must a reaction be spontaneous at all temperatures?
a.
b.
c.
d.
e.
13.
ΔH is positive, ΔS is positive
ΔH = 0, ΔS is negative
ΔS = 0, ΔH is positive
ΔH is negative, ΔS is positive
none of these
CHART!!!!
Consider the dissociation of hydrogen:
H2 (g)  2 H (g)
One would expect that this reaction:
a.
will be spontaneous at any temperature.
b.
will be spontaneous at high temperatures.
c.
will be spontaneous at low temperatures.
d.
will not be spontaneous at any temperature.
e.
will never happen.
(ΔS is +, because there are more possible positions in the products than in the reactants.)
(ΔH is + because it will require added energy to break the bonds between the hydrogens.)
CHART!!!!
14.
At 1 atm, liquid water is heated above 100°C. ΔSsurr for this process is
a.
b.
c.
d.
greater than zero.
less than zero.
equal to zero.
cannot be determined.
ΔH is + for endothermic reactions.
ΔSsurr =  ΔH/T
Therefore, ΔSsurr =  (+ΔH)/T so ΔSsurr = 
15.
At constant pressure, the following reaction 2 NO2 (g)  N2O4 (g) is exothermic. The
reaction (as written) is
a.
b.
c.
d.
always spontaneous
spontaneous at low temperatures, but not high temperatures.
spontaneous at high temperatures, but not low temperatures.
never spontaneous.
ΔH =  because it is exothermic.
ΔS =  because there are more reactant gaseous moles than product gaseous moles.
Chart: If ΔH is  and ΔS is , then the reaction will be spontaneous at low temperatures.
16.
Consider the following processes (at 298 K):
H2O (l)  H2O (g)
(Gas has more disorder so ΔS = +)
II.
Cl2 (g)  2 Cl (g)
(More gaseous product moles than reactant gaseous moles so ΔS = +)
III.
2 Fe (s) + 1.5 O2 (g)  Fe2O3 (s)
(More reactant gaseous moles than product gaseous moles so ΔS = )
I.
(H2O)
IV.
NaCl (s)   NaCl (aq)
(Solid dissolves, so more disorder so ΔS = +)
V.
CaC2 (s) + 2 H2O (l)  C2H2 (g) + Ca(OH)2 (s)
(Gas has more disorder so ΔS = +)
VI.
N2 (g, 1 atm)  N2 (g, 2 atm)
(Increase P, so decrease V. Therefore, less positions, less disorder; so ΔS = )
For which is ΔS positive?
a.
I, II, IV, V
b.
I, IV, V
c.
I, IV, V, VI
d.
III, VI
e.
all six processes
17.
The third law of thermodynamics states
a.
the entropy of the universe is increasing.
b.
the entropy of the universe is constant.
c.
the entropy is zero at 0 K for a perfect crystal. (Definition!!!)
d.
the absolute entropy of a substance decrease with increasing temperature.
e.
the entropy of the universe equals the sum of the entropy of system and
surroundings.
18.
Which of the following is not a state function.
a.
b.
c.
d.
e.
19.
q (heat)
G
H
E
P
The standard free energy of formation of AgCl (s) is 110 kJ/mol. ΔG° for the reaction
2 AgCl (s)  2 Ag (s) + Cl2 (g) is:
a.
b.
c.
d.
e.
110 kJ
220 kJ
110 kJ
220 kJ
none of these
The reaction has been reversed from the formation reaction, so the sign is opposite.
110 kJ/mol * 2 mol = 220 kJ
20.
For the reaction, CO2 (g) + 2 H2O (g)  CH4 (g) + 2 O2 (g), ΔH° = 803 kJ
which of the following will increase K?
a.
b.
c.
d.
e.
decrease number of moles of methane
increase volume of system
increase the temperature of system
all of the above
none of the above
ΔG = RT ln K which means K = e ΔG/RT
The only way to increase K would be to increase T or to decrease ΔG. (To decrease ΔG,
you would have to increase T or increase ΔS. To increase ΔS would be to increase
disorder or change ΔS so that Sfinal < Sinitial. Answer A would violate this. Volume would
have no effect.
For questions 21-23, consider the following system at equilibrium at 25°C:
PCl3 (g) + Cl2 (g)  PCl5 (g) for which ΔH° =  92.5 kJ at 25°C.
21.
If the temperature of the system is raised, the ratio of the partial pressure of PCl5 to the
partial pressure of PCl3 will a.
increase.
b.
decrease.
c.
stays the same.
d.
impossible to tell without more information.
The reaction is exothermic so heat can be thought of as a product. According to
LeChatelier's Principle, the reaction will shift back to the left away from the added
substance (heat). Therefore, the partial pressure of PCl5 to the partial pressure of PCl3
will decrease.
22.
When some Cl2 (g) is added at constant volume and temperature, the ratio of the partial
pressure of PCl5 to the partial pressure of PCl3 will
a.
increase.
b.
decrease.
c.
stays the same.
d.
impossible to tell without more information.
Chlorine is added. According to LeChatelier's Principle, the reaction will shift to the
right away from the added substance (chlorine). Therefore, the partial pressure
of PCl5 to the partial pressure of PCl3 will increase.
23.
When the volume is decreased at constant temperature, the ratio of the partial pressure of
PCl5 to the partial pressure of PCl3 will
a.
increase.
b.
decrease.
c.
stays the same.
d.
impossible to tell without more information.
According to LeChatelier's Principle, when the volume is decreased, the reaction will
shift towards the side with the lowest gaseous moles. In this case, the reaction will shift
right. Therefore, the partial pressure of PCl5 to the partial pressure of PCl3 will increase.
You will need a calculator to answer the following practice questions.
These will not be on the multiple choice section, but they are extra
practice for the free response section.
24.
Assume that the enthalpy of fusion of ice is 6020 J/mol and does not vary appreciably
over the temperature range 270-290 K. If one mole of ice at 0°C is melted by heat
supplied from surroundings at 280 K, what is the entropy change in the surroundings, in
J/K?
a.
b.
c.
d.
e.
22.0
21.5
0.0
21.5
22.0
6020 J/mol * 1 mol/280 K = 21.5 J/K
The sign is negative because the surroundings are losing heat.
25.
If the change in entropy of the surroundings for a process at 451 K and constant pressure
is 326 J/K, what is the heat flow absorbed by for the system?
a.
326 kJ
b.
24.2 kJ
c.
147 kJ
d.
12.1 kJ
e.
147 kJ
Remember from Chapter 6 that ΔH = q at constant pressure!!!
ΔSsurr =  ΔH/T
Therefore, ΔSsurr =  q / T
ΔSsurr =  ( J · 451 K= +147,000 J = 147 kJ
The sign is positive because heat is absorbed (endothermic).
26.
The heat of vaporization for 1.0 mole of water at 100°C and 1.0 atm is 40.6 kJ/mol.
Calculate ?S for the process H2O (l)  H2O (g) at 100°C.
a.
109 J/mol·K
b.
109 J/mol·K
c.
406 J/mol·K
d.
406 J/mol·K
e.
none of these
ΔSsurr =  ΔH/T
ΔSsurr =  40,600 J/mol / 373 K
ΔSsurr =  109 J/mol·K
Because ΔSsurr and ΔSsys oppose each other,
ΔSsys = +109 J/mol·K
27.
For the process CHCl3 (s)  CHCl3 (l), ΔH° = 9.2 kJ/mol and ΔS° = 43.9 J/mol·K.
What is the melting point of chloroform?
a. 63°C
b. 210°C
c. 5°C
d. 63°C
e. 5°C
At the M.P. ΔG = 0 so T = ΔH/ΔS
T = 9200 J / 43.9 J/mol·K
T = 210 K which is °C
28.
Given the following data, calculate the normal boiling point for formic acid (HCOOH).
HCOOH (ℓ)
HCOOH (g)
a.
b.
c.
d.
e.
ΔHf° (kJ/mol)
410.
363
S° (J/mol·K)
130.
251
2.57 K
1730°C
388°C
82°C
115°C
HCOOH (ℓ)   Cg
 ΔHrxn° = Ʃ (np · ΔHprod°)   Ʃ (nr · ΔHreact°)
ΔHrxn° = [(1 mol HCOOH (g)) ( kJ/mol   molC ℓ (kJ/mol
 ΔHrxn° = 47 kJ = 47,000 J
ΔSrxn° = Ʃ (np · ΔSprod°)   Ʃ (nr · ΔSreact°)
ΔSrxn° = [(1 mol HCOOH (g)) ( J/mol·K   molC ℓ ( J/mol·K
 ΔSrxn° = 121 J/K
At B.P. ΔG = 0
Therefore, T = ΔH/ΔS
T = 47,000 J / 121 J/K
T = 388 K = 115°C
29.
ΔG
ΔG
ΔG
ΔG
At 699 K, ΔG° = 23.25 kJ for the reaction H2 (g) + I2 (g)  2 HI (g). Calculate ΔG
for this reaction if the reagents are both supplied at 10.0 atm pressure and the product is
at 1.00 atm pressure.
a.
3.5 kJ
b.
36.6 kJ
c.
36.6 kJ
d.
50.0 kJ
e.
50.0 kJ
=
=
=
=
ΔG° + RT ln Q
ΔG° + RT ln (PHI2 / PH2 · PI2)
 J  Jmol·K)(699 K) [ ln (12 / 10·10)]
 J =  kJ
AP Chemistry Chapter 16 Practice Free Response Portion
Answer the following questions on a separate sheet of paper. SHOW ALL WORK!! Do not
assume anything - explain it! Staple your work to this paper.
1.
There has been great interest in the splitting of water into its elements for the purpose of
harvesting the hydrogen for fuel. The free energy of this reaction is so positive that there
is no hope of causing the reaction to occur without coupling it to another process. For
example, it has been proposed that the reaction can be promoted by first reacting silver
with water to produce hydrogen and silver oxide.
2 Ag (s) + H2O (g)  Ag2O (s) + H2 (g)
and then decomposing the silver oxide at high temperature to give oxygen and recover
silver.
Ag2O (s) 
2 Ag (s) +  O2 (g)
a. Calculate H, S, and G for both reactions at 25C.
b. Write a balanced equation for the overall reaction which occurs on summing these two
equations. What law is illustrated by this process? What is the enthalpy change, H, for
the two reactions added together?
c. What is the G for the two reactions combined? Is the overall process spontaneous
under standard conditions?
d. At what temperatures, if any, is the second reaction spontaneous?
2.
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. What is the sign of S for the reaction. Explain.
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. If He gas is added to the original mixture at constant volume and temperature, what
will happen to the partial pressure of Cl2? Explain.
d. If the volume of the reaction mixture is decreased at constant temperature to half the
original volume, what will happen to the number of moles of Cl2 in the reaction
vessel. Explain.
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