NAME _____________________________________ UNIT 5 (3): THERMODYNAMICS: BRINGING IT
ALL TOGETHER: ∆H, ∆S & GIBBS FREE ENERGY
I-V) up through Hess’s Law
VI) A Structured Overview Re: Spontaneity
There are essentially 2 ways of determining the spontaneity of a process
∆Suniv = ∆Ssys + ∆Ssurr
∆G = ∆H - T∆S
where spontaneity of a process is
determined by the entropy change it
produces in the universe
where spontaneity of process is
determined by the interplay between
enthalpy and entropy, yielding a -∆G
summarized as:
partially summarized as:
Interplay of ∆Ssys, & ∆Ssurr, in Determining
the Sign of ∆Suniv (Zumdahl p. 800)
∆Ssys ∆Ssurr
+
+
-
∆Suniv
+
-
+
-
?
-
+
?
spontaneous
no, the reverse
rxn occurs
spontaneous only
if ∆Ssys is larger
than ∆Ssurr
spontaneous only
if ∆Ssurr is larger
than ∆Ssys
So, when the is +∆Suniv a reaction (written as
the forward) is spontaneous
but ….
it is difficult to quantitatively acquire ∆Suniv
∆H
+
+
∆S
+
+
-
Gibbs Free Energy
∆G
(-) Spontaneous at all temperatures
(±) Spontaneous at HIGH temperatures
(±) Spontaneous at low temperatures
(+) Not spontaneous, at any temperature
a -∆G indicates a
spontaneous reaction
quite valuable to chemists since these values
are collected essentially at constant T and P
& most biochemical and industrial processes
occur (or are tested at) constant T and P
+∆S &-∆G
both mean a spontaneous reaction
From Zumdahl 9th ed. 2014 (Cengage Publishing) p. 799
482
VII) The Second Law of Thermodynamics Governs the Thermodynamic Favorability of Processes
A) Recall … one of the goals of thermodynamics is to predict whether a particular reaction is
thermodynamically favored … (a process that occurs without ongoing outside intervention)
1) Characteristics of a thermodynamically favored process include:

any process with a +∆S or a -∆G …Each variable indicates that the process is
thermodynamically favored (a.k.a. spontaneous)

A process with a directionality meaning that products are favored at
equilibrium (K > 1) …hence favorability tells us how much work can be
done. Gibb’s free energy tells us the direction of a process, not its speed

a spontaneous process may be exothermic (-∆H) OR endothermic (+∆H)

The reverse of any spontaneous reaction is nonspontaneous. e.g. re-charging a
battery, electroplating a piece of jewelry

a spontaneous process is highly dependent upon temperature and changes in
entropy
2) Thermodynamically Favored does not mean a “fast” reaction (that’s kinetics)
a) The thermodynamic favorability of a reaction has *nothing to do with speed of a
reaction.
i) Processes that are thermodynamically favored, but do not proceed at a
measurable rate, are said to be * under “kinetic control.”
ii) *High activation energy
process to be under kinetic control.
is a common reason for a
b) The fact that a process does not proceed at a noticeable rate does not mean that the
chemical system is at equilibrium.
If a process is known to be thermodynamically favored (through qualitative
and/or quantitative analysis of ΔH° and ΔS°), and yet it is not occurring at a
measurable rate, then the conclusion is that the process is under kinetic control.
(The College Board: Course and Exam Description 2013)
i) Kinetics deals with the chemical potential of intermediate states and enables us
to determine why a reaction is slow or fast … where thermodynamics deals
with the relative chemical potentials of just the reactants and products.
ii) Hence at many levels, thermodynamics bridges kinetics and equilibrium …
providing a vocabulary, a set of tenets and a mathematical language to
communicate between the issues.
483
3) A nonspontaneous reaction is NOT an impossible reaction …. A nonspontaneous process
can be made to be spontaneous by coupling it to another (very spontaneous) reaction… or by
supplying energy from an external source.
i) For instance, there is always the ever-popular re-chargeable battery … a
nonspontaneous reaction, forced to occur, by the application of a superiorly large
electrical force.
ii) For instance, turning iron ore (rock having the oxidized form of iron) from rust back
to iron is NOT thermodynamically favored – but it happens every day in the furnaces
of steel mills … Of course, huge amounts of energy and appropriate chemicals are
required
CHECK THIS OUT: Article on: Reversing a boiled egg!
http://news.yahoo.com/no-yolk-scientists-unboil-egg-without-defying-physics-165413670.html
4) Most spontaneous reactions are exothermic (BUT NOT ALL) … Hence not
every reaction proceeds along a path of lower enthalpy … My “ladder of
potential energy” (from honors) is only an approximation, …
It is a broad tendency
Consider melting ice or evaporating water, or dissolving salt…
each is endothermic …. Each is spontaneous … each has an increase
in entropy ….AND(!) It is this last piece (ΔS) that seems so important.
Enthalpy alone will not explain why these forward reactions are
thermodynamically favored… there is a marriage with entropy as well.
*The increase in the entropy of the universe is the criterion absolutely necessary for
spontaneity.
5) For future work, you may wish to grasp some other vocabulary …although for our work
they play a small but necessary role.
a) Isothermal process: a process which occurs at a constant temperature … e.g. the
expansion of a gas, in contact with an energy source, which heats
the gas, as it expands, keeping the temperature constant.
b) Adiabatic process: Often associated with adiabatic cooling. As explained by
Richard Fitzpatrick at http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html
…imagine a gas is thermally isolated from its surroundings. If the gas were allowed
to expand quasi-statically under these so called adiabatic conditions then the gas does
work on its environment, and, hence, its internal energy is reduced, and its
temperature changes. …thus the expansion was adiabatic, and not isothermal….
484
c) Reversible process: A process that can go back and forth between states along exactly
the same path; a system at equilibrium is reversible if equilibrium
can be shifted by an infinitesimal modification of a variable such
as temperature. … In a reversible process, the change occurs in
such a way that the system and surroundings can be restored to its
original condition with no net change to either the surroundings or
system.
d) Irreversible process: a process that cannot simply be reversed to restore the system
and its surroundings to their original states…different values for
q and w are required. It is a process which demands a change to
the surroundings to restore the system.

an irreversible process cannot be called spontaneous process

e.g. any real process: baking … or recombining CO2 + H2O back into fossil
fuels or discharging a battery … because recharging (for example) requires
more work than the battery produced… Hence the surroundings (universe)
is (are) different after the recharge cycle.

The first law of thermodynamics implies that the flow of heat from a
hotter object to a cooler object irreversible …
AND NOW….….A drumroll ….puh-leeeze!
485
IX) Free Energy! (as in … Gibbs Free Energy!!!! …. J Willard Gibbs that is….)
Definition / Conditions Meaning/
Connection to Spontaneity & Chemical
Reactions (-∆G)
Free Energy and
Equilibrium
Free Energy and Electrochemistry
(Cell Potential) …done in
electrochemistry
Graduate student Daniel M. Dryden explains the differential thermodynamic equation tattooed on his shoulder
thusly: "For anything that has happened on the surface of Earth, from the formation of this ball of rock to
the first replication of RNA to me hitting 'save' at the end of this post, the thermodynamic driving force
behind it all was a reduction in Gibbs free energy." Credit: Daniel Dryden http://cen.acs.org/articles/91/i25/Chemistry-Tattoos.htmlhttp://
A) Beyond the relationships expressed in ∆Suniv = ∆Ssys + ∆Ssurr Gibbs Free Energy is particularly
useful in dealing with the temperature dependence of spontaneity.
B) Free Energy: symbol G
and Change in Free Energy: symbol: ∆G
1) Free Energy … *as in energy that is “free to do work”
2) Free Energy (G) or Gibbs Free Energy is the thermodynamic function whose value
describes whether or not a process is thermodynamically favored in the

forward direction. (Hence: directionality in the forward rxn, where K>1)
3) The beauty of the equation defining the free energy of a system is its ability to determine the
relative importance of the enthalpy and entropy terms as driving forces behind a particular
reaction.
a) The change in the free energy of the system that occurs during a reaction measures
the balance between the two driving forces that determine whether a reaction is
thermodynamically favored. Hence, the different sign conventions of enthalpy
and entropy really do matter. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php
b) Given the above: Gibbs Free Energy is dependent upon:



the change in enthalpy (±ΔH)
the change in entropy (±ΔS)
the Kelvin temperature (T)
486
4) These dependencies lead to an equation which is valuable for processes which occur at
constant temperature. Assuming T is in Kelvin, and the process is isothermal (that is,
when there is no change in temperature), then:
a) Equation: ∆G° = ∆H° - T∆S°
i) The quantities of the equation refer to the system.
b) Now, a process is thermodynamically favored when +∆Suniv, well … it is also true
to say that  * a process is thermodynamically favored with a -∆G , so
+∆Suniv and/or -∆G result in a spontaneous process
c) Stay with me here for a minute … for I hope to show you particular relationship.
That relationship is that -∆G = +∆Suniv & it ties these two ideas together.
Solve for ∆S: ∆G° = ∆H° - T∆S°
∆G° = ∆H +ΔS
T
T
Remember that at constant temperature and pressure: ∆Ssurr = ∆H
T
Substitute:
∆G° = ∆H +∆S = ∆Ssurr + ∆S = ∆Suniv
T
T
Thus: ∆Suniv= ∆G at constant temperature and pressure
T
This means that a process at constant T and P is spontaneous only if ∆G is
negative or rather: A process is spontaneous at constant T and P in the direction in
which the free energy decreases ….-∆G = +∆Suniv
d) The standard free energy, (∆G°), is the change in free energy at standard conditions of
1 atm and 25°C.
5) How ∆H and ∆S Help Predict the Thermodynamic Favorability of a Reaction
a) Keep in mind the equation: ∆G° = ∆H° - T∆S°
enthalpy total entropy term
term
i) This can be re-written as: ∆G° = ∆H° + (-T∆S°)
When you do this, you can see that the entropy part of the concept
depends directly on the absolute temperature. (This approach may be
valuable, interpretively speaking for you)
487
b) Why “free energy”? Your Brown & LeMay text does just about the best job I’ve read on this issue p. 836
i) The sign and magnitude of ∆G are both important
ii) When the magnitude is large and negative (e.g. burning a fossil fuel),
we are being informed that the reaction is thermodynamically favored and
capable of doing a great deal of work on the surroundings
Remember that work done by a system on the surroundings is negative.
 The change in free energy for a process equals the maximum useful work
that can be done by the system on its surroundings in a spontaneous process
occurring at constant temperature and pressure: ∆G = −wmax
iii) The above is an elegant entry then into understanding why it is called “free
energy” …. Recall that there are two parts to most energy exchanges
∆E = q + w (and we can envision q as ∆H, at constant pressure)
Thus free energy is the portion of the energy change of a spontaneous
reaction that is “free” to do useful work. … AND…
iv) For non-spontaneous reactions where ∆G >0 or a positive value, free energy
is the measure of the minimum amount of work that must be done to cause
the process to occur.
c) When the equation results in a -∆G the reaction is spontaneous
Summary for the Possible Conditions for Gibbs Free Energy
ΔH
―
+
―
+
ΔS
+
+
―
―
ΔG
―
±
±
+
Spontaneity
always thermodynamically favored
yes – but only at high temperature
yes- but only if the temperature is low enough
never spontaneous
6) Inferences: At constant T and P the relationship between the sign of ∆G and the
spontaneity of a process is as follows:
When: ∆G° < 0, [a negative], the reaction is spontaneous in the forward reaction
MEMORIZE
∆G° = 0, the reaction is at equilibrium … 0 is the minimum numerical value
for a reaction … just reaching spontaneity.
∆G° > 0, [a positive], the reaction in the forward direction is nonspontaneous
(work must be done to make it occur) BUT the reverse reaction is
spontaneous.
i) Note that second inference … Any idea as to WHY a ∆G° of 0 is a system at
equilibrium?
* Only at equilibrium are the factors of enthalpy change (per endo and exo),
entropy change and the temperature factor occurring at the same rates.
488
7) Calculating ∆G°
Calculate the standard free energy change for the formation of NO(g) from N2(g) and O2(g)
at 298 K, given that ∆H° = 180.7 kJ and ∆S° = 24.7 J/K
[please note the unit for the change in
entropy is J not kJ and we must take that
into account….]
N2(g) + O2(g) → 2 NO(g)
∆G° = ∆H° - T∆S°
∆G° = 180.7 kJ - (298 K) (0.0247 kJ/K) = 173.3 kJ
Is the reaction a spontaneous reaction based upon the calculation? * NO
Based upon the signs for ∆H° and for ∆S° at 298 K should it be spontaneous?
*NO… a + ∆H with a +∆S cannot generate a -∆G unless the temperature is
a high temperature (and 298 K is not a high temp) … refer to the chart
Assignment Trivedi 18.14
8) As with enthalpy of formation (ΔHf), there are standard free energies of formation (∆Gf)
∆Gf for a substance is the *free energy change for its formation from its elements under
standard conditions of 1 atm and 298 K.
a) for solutions standard conditions normally refer to concentrations of 1 M
b) the free energy of formation for an element = 0 J/mol
c) And of course….∆G° = ∑ npG°products - ∑ mrG°reactants
9) Now here’s an interesting application of the Gibbs Free Energy equation.
Given: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4H2O(ℓ)
∆H° = -2220 kJ
Using the information given predict whether ∆G° for this reaction is more negative
or less negative than ∆H°
(Brown & Lemay p. 835)
Using: ∆G° = ∆H° - T∆S°
The answer must depend upon the value of -T∆S
Notice the decrease in entropy …hence ∆S is negative. This makes
T∆S a negative value and given the negative value for ∆H we
have a (-) – (-) thus ∆G must be less negative than ∆H! Wow!!
489
10) Using the Gibbs Free Energy equation to find the temperature at which a process becomes
thermodynamically favorable:
ΔG° = ΔH° -TΔS°
∴ 0 = ΔH° -TΔS° at equilibrium
- ΔH° = -TΔS° (negative signs cancel…so…)
ΔH° = TΔS°
thus: T = ΔH°
ΔS°
TRY THIS! Recall: There is a challenge with the units for enthalpy and entropy…. THINK!
Find the thermodynamic boiling point of H2O(ℓ) →H2O(g)
Given the following information:
Hvap = +44 kJ/mol and Svap = 118.8 J/K
*T = ΔH°
ΔS°
= 44000 = 370K
118.8
ans: 370 K
TRY THIS! Recall: There is a challenge with the units for enthalpy and entropy…. THINK!
Calculate the thermodynamic boiling point, in °C, of ethanol C2H5OH(ℓ) → C2H5OH(g)
ΔHf
S
Hint 1: * Convert the entropy
(kJ/mol)
(J/K mol)
values to kJ (or the enthalpy
160.6
C2H5OH(ℓ) -277.7
values to joules …) For the
282.6
C2H5OH(g) -235.1
sake of the math …I opt to
change the entropy to kJ ….
*Entropy values = 0.1606 kJ and 0.2826 kJ respectively
*ΔH = -235.2 – (-277.7) or 42.6
* ΔS = 0.286 – 0.1606
Hint 2: * Find ΔH and ΔS
using “products – reactants”
Hint 3:* Apply ΔT = ΔH/ΔS
or 0.122
Ans: 76°C
*T = 42.6 = 349 K or approx.. 76°C
0.122
490
TRY THIS! The Haber Process for the manufacture of ammonia is represented by the balanced
chemical reaction:
2 N2(g) + 3 H2(g)  2 NH3(g)
ΔHrxn = -93 kJ/mol
ΔS = -198 J/K mol
At what temperature in Kelvin, will the reaction become spontaneous?
*Since we may assume a system at equilibrium, ΔG = 0 and thus T = ΔH/ΔS
* T = -93000/-198 = 470 K ….
The reaction becomes spontaneous at 470 K
D) Gibbs Free Energy, Non-Standard Conditions and the Equilibrium Constant
1) Okay … in a perfect world, all reactions would occur at standard conditions …Yeah, well
you’re relatively young …You’ll get over it. This ain’t no perfect world …. How boring an
all vanilla world would be anyway!
a) Equation: ∆G = ∆G° + RTlnQ
where R = 8.314 J/mol∙K,
T = Kelvin
Q = Reaction Quotient
Under standard conditions at equilibrium, the concentrations = 1 M
Q = 1 thus the ln Q = 0 …. and we have at equilibrium ∆G = ∆G°
At nonstandard conditions we must calculate Q in order to calculate ∆G
Recall that for Q the concentrations of gases are always expressed as partial pressures
in atmospheres and solutes are expressed in their molar concentrations.
2) In contrast to ΔG° which is a constant for a given reaction, (and whose value(s) you can get
off of tables), ΔG varies continuously as the composition of a reaction changes, finally
reaching zero at equilibrium. ΔG is the “distance” (in free energy) from the equilibrium state
of a given reaction.
Thus:
3) The key to understanding the relationship between
Go and K is recognizing that the

magnitude of

The smaller the value of
smaller K.

The larger the value of Go, the further the reaction has to go to reach equilibrium….
the larger K
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php
Go tells us how far the standard-state is from equilibrium.
Go, the closer the standard-state is to equilibrium…the
491

Equilibrium is essentially that point at which the system has fallen to its lowest energy
and that is symbolized by ΔG = 0
Take a look at some data: …. Recall that a negative value indicates a level of
thermodynamic favorability ….Note the size of K in these cases.
Values of ΔG° and K for Common Reactions at 25°C
ΔG° (kJ)
Reaction
2 SO3(g) ↔ SO2(g) + O2(g)
H2O(ℓ) ↔ H+(aq) + OH-(aq)
AgCl(s) + H2O(ℓ) ↔ Ag+(aq) + Cl-(aq)
141.7
1.4 x 10-25
79.9
1.0 x 10-14
55.6
1.8 x 10-10
CH3COOH(aq) + H2O(ℓ) ↔ H+(aq) + CH3COO-(aq) 27.1
N2(g) + 3 H2(g) ↔ 2 NH3(g)
K
1.8 x 10-5
-32.9
5.8 x 105
HCl(aq) + H2O(ℓ) ↔ H+(aq) + Cl-(aq)
-34.2
1.0 x 106
Cu2+(aq) + 4 NH3(aq) ↔ Cu(NH3)42+(aq)
-76.0
2.1 x 1013
-211.8
1.4 x 1037
Zn(s) + Cu2+(aq) 
Zn2+(aq) + Cu(s)
Notice how thermodynamically favored
this reaction is (a large value, with a
negative sign) …and the size of K … the
two go together.
4) Let me try to explain …. this REALLY COOL IDEA …
a) At equilibrium, because Q = K
∆G = ∆G° + RTlnQ
becomes
∆G = ∆G° + RTlnK
or as it is on your tables: ∆G° = - RTlnK
This is an important construct and beyond the normal AP work …Let me make a
pitch here …. ‘cause this is cool!
By relating K to ∆G° we can then relate K to entropy and enthalpy changes
for a reaction (B & L p. 840)
Ultimately we can rework the equation to: K = e-∆G°/RT
This can lead to the conclusion that:
*The more negative ∆G the larger the value of K
492
This makes soooooooo much sense! When ∆G is a negative, it means that the
forward reaction is spontaneous. When it is a large negative value it means that
it is a very spontaneous forward reaction and really “works it” ….
Hey, when the forward reaction is highly favored … it goes far to the right,
before achieving an equilibrium …. meaning…. the forward reaction has a
really large K!!!!
But recall that a reaction may NOT produce tons of product due to high
activation energy or the lack of proper orientation re: collision theory….
b) However, as a general concept: for a system at equilibrium:
MEMORIZE
∆G° < 0
∆G° = 0
∆G° > 0
K > 1 Products are favored over reactants at equilibrium
K=1
K < 1 Reactants are favored over products at equilibrium
C’mon that’s gotta capture the “coming full circle” person in you!!!!
Assignment Trivedi 18.16, 18.17, 18.19, 18.20 & 18.24
493
Just For Example: Calculate the equilibrium constant at 25°C for the reaction
H+(aq) + OH–(aq) → H2O(l) from the following data:
H+(aq)
OH-(aq)
H2O(ℓ)
ΔHf° (kJ mol-1)
0
-230.0
-285.8
S°
0
-10.9
70.0
(J K-1 mol-1)
from: http://www.chem1.com/acad/webtext/thermeq/TE5.html
Hint: You want to calculate K from thermodynamic data …hence: ∆G = ∆G° + RTlnK
re-worked to K = e-∆G°/RT
which is
Thus: We need ∆G°, and the Kelvin temperature.
To Get ΔG: We must find ΔH° and ΔS°
ΔH° = (∑ ΔHf°products) – (∑ ΔHf°reactants) = (–285.8) – (–230) = –55.8 kJ mol–1
ΔS° = (∑ ΔS°products) – (∑ ΔS°reactants) = (70.0) – (–10.9) = +80.8 J K–1 mol–1
ΔG° = ΔH° – TΔS° = (–55800) – (298) (80.8) = –79900 J mol–1
notice the value for ΔH has been converted to joules to accommodate entropy in joules
Using: K = e-ΔG°/RT …. K = e-79900/ (8.134) (298) = 1.01 x 10-14
E) One of the more esoteric learning objectives states: The student is able to explain how the application
of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause
processes that are not thermodynamically favorable to become favorable
1) This learning objective will be covered when we learn about electrolytic cells in our unit
on electrochemistry …. That is … by providing energy to a non-thermodynamically favored
system, we can drive that non-thermodynamically favored reaction….
a) The classic example is a battery re-charger. It is not thermodynamically favored to
the products of a battery to revert to their original reactive forms …. However, by
overwhelming the system with electrical energy from your household outlet, we can
force the chemicals to revert back to these more reactive forms, and thus re-constitute
the battery’s reactants ….and re-use the battery.
b) And while we will take a hard look at this idea later, the learning objective does
raise another issue …. coupled reactions.
2) Coupling reactions is a valuable means of making a non-thermodynamically favored reaction,
to occur.
494
3) Coupling reactions links a reactant-favored reaction (one which is not thermodynamically
favored) linked to a product-favored reaction (a thermodynamically favored reaction) so that
both reactions yield products.
That is, when reactions are “linked”, the reaction with a –ΔG drives the other reaction.
Very often, the product of one reaction is consumed as a reactant in a second reaction
A → B
B → C

If the standard free energy of the first reaction is positive but that of the second reaction
is sufficiently negative, then for the overall process will be negative and we say that the
first reaction is “driven” by the second one.

Similarly, the equilibrium constant of the overall reaction is the product of the
equilibrium constants of the two steps. [recall that: Koverall = (K1) (K2)

Study the following table to understand how two reactions may be linked:
1. Cu2S(s) → 2 Cu(s) + S(s)
ΔG° = + 86.2 kJ
ΔH° = +76.3 kJ
2. S(s) + O2(g) → SO2(g)
ΔG° = - 300.1 kJ ΔH° = +296.8 kJ
3. Cu2S(s) → 2 Cu(s) + SO2(g) ΔG° = -213.9 kJ ΔH° = -217.3 kJ
In the table we have two reactions (as in our example of A→B and B →C)
Reaction 1 is the first step in obtaining metallic copper from one of its principal ores
{copper(I) sulfide}. This reaction is endothermic and it has a positive free energy
change, so it will not proceed spontaneously at any temperature.
If Cu2S is heated in the air, however, the sulfur is removed as rapidly as it is formed by
oxidation (combination with oxygen) in the highly spontaneous reaction 2, which
supplies the free energy required to drive 1.
You will notice that a product of reaction 1, the sulfur, is consumed as a reactant in the
more thermodynamically favored, second reaction.
The combined process, known as roasting, (3) is of considerable industrial importance
and is one of a large class of processes employed for winning metals from their ores.
from: http://www.chem1.com/acad/webtext/thermeq/TE5.html
495
4) Given the decomposition of calcium carbonate:
CaCO3(s) → CaO(s) + CO2(g)
ΔGº = 130.40 kJ
You can see by the positive ΔG, the reaction is reactant-favored.
Does this mean that this reaction cannot be made to yield products?
No. In fact, if the temperature is raised above 837 ºC, this decomposition reaction
becomes product-favored.
How can the temperature be raised enough to cause this? One possibility is to heat
the calcium carbonate over a coal fire:
C(s) + O2 →CO2(g)
ΔGº = -394.359 kJ
The heat released by burning the coal raises the temperature enough that the
decomposition of calcium carbonate becomes product-favored.
If you add the two equations and apply Hess's Law:
CaCO3(s) + C(s) + O2 → CaO(s) + 2CO2(g)
ΔGº = -263.96 kJ
496
TRY THIS! From the 2006 AP Chemistry Exam … old syllabus …but good question!
497
_______________________________________________________________________________
_______________________________________________________________________________
_______________________________________________________________________________
________________________________________________________________________________
Answers to the 2006 exam question re: Thermodynamics
a)
b)
498
c)
d)
e)
TRY THIS! (NMSI)
Consider the reaction: 2 SO2(g) + O2(g) → 2SO3(g) carried out under standard conditions of
25°C and 1 atm. Calculate ΔH°, ΔS° and ΔG° using the following date:
Substance ΔH°f (kJ/mol) ΔS (J/mol)
SO2(g)
-297
248
SO3(g)
-396
257
O2(g)
0
205
Ans:
499
Questions: The Basics….
1)
For questions 2 – 8 select the most appropriate terms from the list
___2
___3
___4
___5
___6
___7
___8
9) Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction,
or in equilibrium:
(a) When a piece of metal heated to 150 °C is added to water at 40 °C, the water gets hotter.
(b) Water at room temperature decomposes into H2 (g) and O2 (g),
(c) Recall from Honors that: the normal boiling point is the temperature at which a vapor at 1 atm is in
equilibrium with its liquid. Now consider benzene vapor, C6H6 (g), at a pressure of 1 atm condenses to liquid
benzene at the normal boiling point of 80.1°C. Is this spontaneous as described …in the reverse direction or
in equilibrium?
500
10) Using the equation (slightly re-written): ∆G = ∆H + (-T∆S)
a) assuming a reaction with a -∆H and a+ ∆S what effect will raising temperature have on ∆G?
b) assuming a reaction with a -∆H and a+ ∆S what effect will raising temperature have on ∆G?
c) assuming a reaction with a -∆H and a - ∆S, under what conditions of temperature might this
reaction be spontaneous?
11)
12)
13)
14)
15)
501
16) If H and S are both negative values, or positive values, then G has a ___________ sign.
a)
b)
c)
d)
positive
negative
variable (+ or -)
large
17) Which of the following represents an increase in entropy?
a)
b)
c)
d)
e)
boiling of water
freezing of water
crystallization of salt from a supersaturated solution
the reaction 2 NO(g)  N2O2(g)
the reaction 2 H2(g) + O2(g)  2 H2O(g)
18) Of the following, a 1 mole sample of which of the gases has the largest entropy at standard conditions?
a)
b)
c)
d)
CH4
H2
He
C3H8
19) All of the following have Gf = 0 EXCEPT
a)
b)
c)
d)
e)
O2(g)
Hg(l)
H2(g)
Ca(s)
Br2(g)
20) Given the following information, calculate G for the reaction below at 25C:
2 H2O2(ℓ)  2 H2O(ℓ) + O2(g)
Compound H(kJ/mol) S(J/K·mol)
H2O2(ℓ)
H2O(ℓ)
O2(g)
a)
b)
c)
d)
e)
-187.8
-285.8

109.6
69.9
205.1
-37700 kJ
+342.6 kJ
-233.5 kJ
+233.5 kJ
-157.9 kJ
502
21) For the process at 25C I2(g)  I2(s) what are the signs of G, H, and S? (tricky…think of I2 & standard state)
a)
b)
c)
d)
e)
G
+



+
H


+

+
S


+
+
+
22) If a process is exothermic and not spontaneous, then what must be true?
a)
b)
c)
d)
e)
S > 0
H > 0
G = 0
S < 0
H = 0
23) For any reaction at equilibrium, which of the following is true?
a)
b)
c)
d)
e)
H < 0
S = 0
S < 0
H = 0
G = 0
24) When ΔH is a negative value and ΔS has a positive value, which conclusion must be made?
a)
b)
c)
d)
The reaction is not thermodynamically favorable.
The reaction is thermodynamically favorable.
The reaction is at equilibrium.
The reaction is spontaneous in the reverse.
25) The free energy change for a given reaction is -36.2 kJ. What is the equilibrium constant at 298 K?
a)
b)
c)
d)
e)
0.985
2.22 x 106
1.01
8.32 x 10-7
3.25 x 106
503
26) Which of the following is a reasonable conclusion give a reaction with a ΔG° = -386 J K-1 mol-1?
a)
b)
c)
d)
K<1
K=Q
K=0
K>1
27) At what temperature would a given reaction become spontaneous if H = +119 kJ and S = +263 J/K?
a)
b)
c)
d)
e)
452 K
2210 K
382 K
2.21 K
363 K
28) The enthalpy of vaporization of methanol (CH3OH) is 35.3 kJ/mol at the boiling point of 64.2 C.
Calculate the entropy change for methanol going from a liquid to vapor.
a)
b)
c)
d)
e)
600. J/K·mol
551 J/K·mol
105 J/K·mol
-105 J/K·mol
-551 J/K·mol
Answers to: The Basics
1) b
9)
2) a 3) d
4) b 5) e
6) f
7) g 8) c
(a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred
from the hotter object to the colder one. Thus, heat is transferred from the hot metal to the cooler water. The final
temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between
the initial temperatures of the metal and the water.
(b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gases
spontaneously bubbling up out of water! Rather, the reverse process—the reaction of H2 and O2 to form H2O—is
spontaneous.
(c) At 80.1 C it is in equilibrium… Again, by definition, the normal boiling point is the temperature at which a
vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. If the temperature were
below 80.1 °C, condensation would be spontaneous.
504
10)
a) Raising the temperature will increase the value of (-T∆S) … which effectively increases the entropy factor of the reaction.
The sign of ∆G under the given conditions is already (-), but with the increase in temperature, and the increase in the
entropy factor (more +), the magnitude of ∆G should become greater, yielding an even more negative value…
b) Lowering the temperature will decrease the value of (-T∆S). This will lower the entropy factor of the reaction, making
∆G smaller.
c) Low temperature could allow for the calculation of a -∆G and thus a spontaneous reaction. The temperature would need to
be low enough to make the variable (-T∆S) smaller than the change in enthalpy, so that when subtracted from the change
in enthalpy, the value of ∆G < 0
11) a It is a spontaneous reaction CO2(s) → CO2(g) …You have seen this and experience tells you this … but also recall that when ∆S
is + and ∆H is + at high temperature the reaction is spontaneous…because (-T∆S) will be large enough to give a -∆G
12) b This is equilibrium
13) b a mixture of gases shows the greatest random motion of particles (greatest # of possible
microstates or arrangements) 14) b
15) b This is a clever question … at these conditions ∆G = 0!!! Nice…
16) c
17) a
18) d
19) e
20) c
21) b
22) d
under these conditions of both negative or both positive … the favorability is very temperature-dependent.
greater complexity ….more electrons
the gaseous form of bromine (dibromine) is not the standard state…thus the ΔG°f cannot be assigned as 0
I2(s) is the standard state for the element …hence it is a thermodynamically favored reaction (-ΔG)
Tough to envision perhaps …but if it were exothermic and spontaneous .. then entropy could be + or – (at low temp) …
but this reaction is NOT spontaneous …and thus the change in entropy must be really quite negative.
23) e
24) b true at all temperatures.
25) b use: ΔG = -RTlnK
26) d A rather significantly negative change in Gibbs free energy indicates a strong probability for the forward reaction to be
thermodynamically favored, which correlates with a reaction moving far to the right … which means a K >1
27) a use: T = ΔH/ΔS
28) c use: T = ΔH/ΔS
505