Download Chapter 18 - Louisiana Tech University

CHEM 102 Instructional Objectives
Prof. Upali Siriwardane,
Chemistry Program, Louisiana Tech University, Ruston, LA 71272
CHAPTER 18
Chapter 18. Thermodynamics: Directionality of Chemical Reactions
18.1 Reactant-Favored and Product-Favored Processes
18.2 Probability and Chemical Reactions
18.3 Measuring Dispersal or Disorder: Entropy
18.4 Calculating Entropy Changes
18.5 Entropy and the Second Law of Thermodynamics
18.6 Gibbs Free Energy
18.7 Gibbs Free Energy Changes and Equilibrium Constants
18.8 Gibbs Free Energy, Maximum Work, and Energy Resources
18.9 Gibbs Free Energy and Biological Systems
18.10 Conservation of Gibbs Free Energy
18.11 Thermodynamic and Kinetic Stability
Objectives are as follows:
Basic Skills
Students should be able to:
1. Understand and be able to use the terms product-favored and reactant favored
(section 18.1)
2. Explain why there is a higher probability that both matter and energy will be
dispersed than that they will be concentrated in a small number of nano-scale
particles (Section 18.2).
3. Calculate the entropy change for a process occurring at constant temperature (Section
18-3).
4. Use qualitative rules to predict the sign of the entropy change for a process (Section
18-3).
5. Calculate the entropy change for a chemical reaction, given a table of standard molar
entropy values for elements and compounds (Section 18.4).
6. Use entropy and enthalpy changes to predict whether a reaction is productfavored
(Section 18.5).
7. Describe the connection between enthalpy and entropy changes for a reaction and the
Gibbs free energy change; use this relation to estimate quantitatively how
temperature affects whether a reaction is product-favored (Section 18.6).
8. Calculate the Gibbs free energy change for a reaction from values given in a table of
standard molar free energies of formation (Section 18.6).
9. Relate Gibbs free energy change and standard equilibrium constant for the same
reaction and be able to calculate one from the other (Section 18.7).
10. Describe how a reactant-favored system can be coupled to a product-favored system
so that a desired reaction can be carried out (Section 18.8).
11. Explain how biological systems make use of coupled reactions to maintain the high
degree of order found in all living organisms; give examples of coupled reactions that
are important in biochemistry (Section 18.9).
12. Explain the relationship between Gibbs free energy and energy conservation
(Sections 18.8 and 18. 10).
13. Distinguish between thermodynamic stability and kinetic stability and describe the
effect of each on whether a reaction is useful in producing products (Section 18. 11).
CHEM 102 CLASS PROBLEMS
Prof. Upali Siriwardane,
Chemistry Program, Louisiana Tech University, Ruston, LA 71272
CHAPTER 18
Chapter 18. Thermodynamics: Directionality of Chemical Reactions
Review of Thermochemistry (Chapter 6)
First Law of thermodynamics is expressed in several forms:
* Energy of the Universe is constant.
* Energy is neither created nor destroyed in the Universe.
* Internal energy of the Universe is constant.
Internal energy: Internal energy is energy associated with changes in atoms, molecules
and subatomic particles. This could include all known and unknown forms of energy in
the Universe including heat, mechanical (explosions or expansions), electrical, chemical,
and radiation (nuclear energy). For systems such as a piston in an engine where heat and
mechanical is the only forms of energy being inter-converted the change in internal energy
∆E is given by the following equation.
∆E = q + w q = heat, w = work
for a piston w is volume expansion work, therefore, w = -P∆V
∆E = q - P∆V
q = ∆E + P∆V
piston operating at constant pressure (P) and temperature (T)
q = ∆H
the heat, q, is called ∆H or enthalpy is equal to
∆H =∆E + P∆V
∆H (enthalpy) is also used for chemical reactions since heat changes involved are carried
out at constant pressure (P) and temperature (T). If the reaction is carried out at 1 atm
pressure and 25 oC temperature, The ∆H is called ∆H o, ( o) representing the standard
thermodynamic conditions: 1 atm pressure and 25 oC temperature.
∆H o for a reaction could be calculated using Hess's law in two ways:
1) ∆H o = ∆H 1o + ∆H 2o + ∆H 3o + ∆H 4o
(∆H 1o and ∆H 2o .... comes from several reactions given)
2) ∆H o = Σ∆H fo (products) - + Σ∆H fo (reactants)
Reactant-Favored and Product-Favored Processes
Based on Enthalpy, ∆H of some reactions they could be classified as product-favored or
reactant –favored. Hydrogen and oxygen combine to form water
Once started, this reaction continues until one reactant is consumed. This is an example of
a spontaneous reaction. This reaction proceeds to make primarily product with liberation
of heat (exothermic). The reaction is classified as product-favored.
We can also decompose water to give H2 and O2
This reaction requires constant input of energy. The reaction proceeds only as long as
electrical energy is supplied. The reaction is classified as reactant-favored.
Just because it’s reactant favored doesn’t mean we can’t “force”it to go.
One initiated, product-favored reactions proceed until one reactant is consumed. Usually
liberate energy as heat. Reactant-favored reactions require constant input of energy to
produce product.
The Gibbs Free Energy Equation is one of the most important equations in science. The
change in Gibbs Free Energy is a state function that only relates to the system. Know this
equation!!
∆G0system= ∆H0system - T∆S0system
1. A negative ∆G refers to a product favored process, ( spontaneous)
2. A positive ∆G refers to a reactant favored process, (not spontaneous)
You should be able to complete two of the boxes in the Table below. The Gibbs Free
Energy equation will help us complete two other boxes. Let's go to the worksheet.
∆H0system
∆S0system
Reaction exothermic
Reaction exothermic
+
Reaction endothermic
+
Reaction endothermic
+
Less order
More order
+
Less order
More order
Product Favored?
(aka spontaneous?)
√
√ at lower T
√ at higher temperature
Always reactant Favored
Probability and Chemical Reactions
Entropy or disorder is a function which describes the number of arrangements (positions
or energy levels) available to a system. Nature spontaneously proceeds towards states that
have the highest probability of existing.
How many ways can we place four individual molecules, a, b c, and d in two boxes or
compartments. Each of these arrangements:
a b
c d
a b
c d
b c
d a
c
b d
b c
d
a
a
b c
d
a
b
c d
a
d b
a
c
c d
a
b
c
d a
b
a d
b
c
d a
b
c
d
a b
c
c
a
b
a b
c
d
Number of microstates per arrangement:
1
4
d
6
4
a b
c d
1
Each of these ways is called a microstate and the total number of microstates is 16
Consider a gas expanding into a vacuum.
For similar reasons, entropy increases in going from a solid to a liquid to a gas.
As a solid, molecules have little freedom to move around (small distribution of positional
microstates) therefore solids have low entropy.
Positional probability is determined by the number of molecules.
• Fewer molecules means fewer possible positional configurations.
• More molecules means more possible positional configurations.
Compare 1 mole of H2O compared to a 1 mole sample of H2 at same temperature and
pressure.
One mole of H2O have higher disorder or entropy compared to H2 at the same
temperature and pressure because of increased positional probabilities available for H2O. The
additional positional probabilities comes from the extra rotations and vibrations H2O
molecule can adopt because of more atoms in the molecule compared to simple H2 molecule.
Measuring Dispersal or Disorder: Entropy
We have defined enthalpy, ∆H as the driving force for some changes. Entropy, ∆S is
also a driving force for physical and chemical changes (reactions). Entropy, symbol S, is
related to energy, but it a different aspect of energy. This concept was developed over a long
period of time. Human experienced chemical and physical changes that cannot be explained
by energy (mostly heat or enthalpy, ∆H in chemistry) alone. A different concept is required
to explain spontaneous changes such as the expansion of a gas into an available empty space
(vacuum) and heat transfer from a hot body into a cold body. These changes cause an
increase in entropy for the system under consideration, but energy is not transferred into or
out of the system.
The entropy concept is associated with the second and third laws of
thermodynamics. Entropy is related to the energy distribution of energy states of a
collection of molecules, and this aspect is usually discussed in previously as positional
probabilities, configurations or microstates.
How is entropy related to energy?
When a system receives an amount of energy q at a constant temperature, T, the entropy
increase ∆S is defined by the following equation.
∆S = q / T.
Entropy is the amount of energy transferred divided by the temperature at which the
process takes place. Thus, entropy has the units of energy unit per Kelvin, J K-1.
Entropy (∆S) normally increase for the following changes:
i) Solid ---> liquid (melting)
ii) Liquid ---> gas
iii) Solid ----> gas
iv) Increase in temperature
v) Increasing in pressure (at constant volume)
vi) Increase in volume ( at constant pressure)
vii) Increase in energy ( heat absorbed)
Three changes in temperature, pressure and volume are directly/indirectly related to
changes in either energy or positional probabilities available to the system.
What are phase transitions? How are entropies of phase transitions evaluated?
At 54 K, the normal melting point, oxygen goes from a solid to a liquid and positional
entropy is added. The increase in entropy is:
∆S°sys = - ∆S°surr ;
∆S°sys =
∆H ofusion
T
;
∆S°surr = -
∆H ofusion
T
From 54 K to 90 K, thermal entropy is added to liquid oxygen as the kinetic energy of the
molecules increase. At 90 K, the normal boiling point, oxygen converts from a liquid to a
gas, an increase in positional entropy. Lastly from 90 K to 298 K we have an increase in
the thermal entropy of the gas.
Qualitative Prediction of ∆S for Chemical reactions
Taking following chemical reactions as examples we could qualitatively predict the sign
of ∆S.
a) 2H2O (g)
------> 2 H2 (g) + O2 (g)
b) N2 (g) + 3 H2 (g) --------> 2 NH3 (g)
The prediction of ∆S of a chemical reaction is done by looking at the physical states, (g),
(l), or (g) or the ∆n value. ∆n is the change in stoichiometric coefficients (n) when reactants
go to products.
∆n = Σ n (gaseous products) - Σ n( gaseous reactants)
a) 2H2O (g)
------>
2 H2 (g)
+ 1 O2 (g)
All reactants and products are in gas phase. Use ∆n to predict ∆S.
If ∆n is positive ∆S is positive or increase.
∆n =
Σ n (gaseous products) - Σ n( gaseous reactants)
∆n = 3 - 2 = +1
The reaction have a positive or higher ∆S because ∆n is positive or more gaseous
particles are produced with higher disorder.
b) 1 N2 (g)
+ 3 H2 (g)
-------->
2 NH3 (g)
All reactants and products are in gas phase. Use ∆n to predict ∆S.
If ∆n is positive ∆S is positive or increase.
∆n =
Σ n (gaseous products) - Σ n( gaseous reactants)
∆n = 2 - 4 = -2
The reaction have a negative or lower ∆S because ∆n is negative or fewer more gaseous
particles are produced creating less disorder.
What is the entropy a perfectly crystalline substance at zero degree Kelvin? Zero.
How is standard molar entropy defined and measured? By heating a substance from 0 K
to particular temperature and diving it by that temperature.
How is entropy related to the structure and molar mass of molecules? More atoms in
the molecule more positive entropy. More molar mass more positive entropy.
Calculating Entropy Changes
Calculate the ∆S for the following reactions using:
∆So = Σ ∆So (products) - Σ ∆S o(reactants)
a) 2SO2 (g) + O2 (g) ------> 2SO 3(g)
∆So [SO2(g)] = 248 J/K mole ; ∆So [O2(g)] = 205 J/K mole;
∆So [SO3(g)] = 257 J/K mole
b) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l)
∆So[ NH3(g)] = 193 J/K mole ; ∆So [N2(g)] = 192 J/K mole;
o
∆S [N2O(g)] = 220 J/K mole; ∆S[ H2O(l)] = 70 J/K mole
a) 2SO2 (g) + O2 (g) ------> 2 SO 3(g)
∆So [SO2(g)] = 248 J/K mole ; ∆So [O2(g)] = 205 J/K mole;
∆So [SO3(g)] = 257 J/K mole
2SO2 (g)
2 x 248
∆So
+
O2 (g)
1 x 205
496
∆So =
------>
205
2 SO 3(g)
2 x 257
514
Σ ∆So (products) - Σ ∆S o(reactants)
∆So = [514] - [496 + 205]
∆So =
514 - 701
∆So =
-187 J/K mole
b) 2NH 3 (g)
+ 3N2O (g)
-------->
4N2 (g) + 3H2O (l)
∆So[ NH3(g)] = 193 J/K mole ; ∆So [N2(g)] = 192 J/K mole;
∆So [N2O(g)] = 220 J/K mole; ∆S[ H2O(l)] = 70 J/K mole
2NH 3 (g) +
3 N2O (g) -------->
2 x 193
3 x 220
386
660
∆So = Σ ∆So (products) - Σ ∆S o(reactants)
4 N2 (g) +
4 x 192
768
3 H2O (l)
3 x 70
210
∆So = [768 + 210] - [386 + 660]
∆So =
978 - 1046
∆So =
-68 J/K mole
Entropy and the Second Law of Thermodynamics
2nd law of thermodynamics:
• Entropy (∆S) of the Universe increase during spontaneous process.
•
For a spontaneous process the entropy of the Universe should increase.
Definitions:
Spontaneous Process: A process that occurs without outside intervention.
Entropy ∆S : A measure of randomness or disorder of a system.
The Universe: The sum of all parts under consideration.
System: Part of the Universe we are interested in and a change is taking place.
Surrounding: Part of the Universe we are not interested in or can not observe.
∆Suniv = entropy of the Universe
∆Ssys = entropy of the system
∆Ssurr = entropy of the surrounding
∆Suniv =
∆Ssys +
∆Ssurr
Describe the following terms derived from the 2nd law and their interrelations:
∆Ssys, ∆S surr, ∆Suni, and -∆H/T .
∆Suniv = entropy of the Universe
∆Ssys = entropy of the system
∆Ssurrr = entropy ogf the surrounding
∆Ssurr
∆Suniv = ∆Ssys +
∆Suniv
∆Ssys
+
+
+
+ (∆Ssys>∆Ssurr)
+
-
∆Ssurr
+
+ (Ssurr>∆Ssys)
The connection between ∆Ssys and ∆Ssurr is usually through exchange of heat. If a
system exchanges heat with its surrounding, the exact relationship of change in entropy of
surrounding, the heat change of the system (∆H) and the Kelvin temperature (T) of the heat
exchange is given by the following equation:
∆Ssurr = - ∆Hsys
T
This relationship connecting ∆Hsys to entropy change of surrounding ∆Ssur is a very
important equation thermodynamics later used to derive ∆G, Free energy (problem 5).
∆Suniv
=
∆Ssys +
∆Ssurr (-∆Hsys/T)
∆Suniv = ∆Ssys -∆Hsys/T
The Third Law of Thermodynamics
The Third Law of Thermodynamics:
• The entropy of a perfect crystalline substance is zero at absolute zero.
• The positional entropy of perfect crystalline substance is zero and at absolute zero
so is the thermal entropy.
Now we can determine the absolute entropies of substances, if we measure the increase
in entropy of a substance when heated from 0 K to 298 K, the change in entropy is given
by ∆S = Sfinal − Sinitial = Sfinal since Sinitial is zero.
Changes of State
At the normal boiling point of a pure liquid, the partial pressure of the gas is 1 atm (by
definition) and the system is at equilibrium.
If we use water as an example:
H2O(l) Æ H2O(g)
At 100°C (the normal boiling point) this reaction is at equilibrium and so:
∆Suniverse = ∆Ssystem + ∆Ssurroundings = 0
As the partial pressure of water is 1 atm, the system is under standard conditions so:
∆S°system −
or ∆S°system =
o
∆H system
T
o
∆H system
T
= 0
and the enthalpy for this reaction is the enthalpy of vaporization, thus
∆S°system =
∆H ovaporization
T
The same approach is applicable to the normal melting point of a pure solid, and
∆S°system =
∆H ofusion
T
Absolute Entropy
Let us now look at the change in entropy of oxygen molecules (O2) with temperature.
Entropy of Oxygen (J/K)
200
gas
150
100
50
0
liquid
solid
0
50
100
150
Temperature (K)
200
250
At 0 K, the entropy of pure oxygen is 0 J.K-1. As the temperature increases up to 54 K the
solid warms up, the vibrational energy of the atoms on the lattice increases, and thermal
entropy is added.
Factors that Determine the Absolute Entropy of a Substance
Look at the standard molar entropies in the appendix in the text and consider the
following:
a) Entropy is dependant on concentration – positional entropy. The enthalpy is relatively
independent of concentration.
b) Entropy of gases greater then liquids which are greater than solids
c) Note the entropies of the noble gases increase with atomic number (as do the halogens).
The higher the atomic number, the greater the number of electrons and the greater number
of ways in which the energy levels can be populated – thermal disorder (distribution of
energies).
d) the more complex the molecule (larger number of bonds), the greater the number of
ways it can distribute its energy in modes of vibration and rotation, the greater its entropy.
Gibbs Free Energy
The Gibbs Free Energy Equation is one of the most important equations in science. The
change in Gibbs Free Energy is a state function that only relates to the system. Know this
important equation!!
∆G0system= ∆H0system - T∆S0system
T o derive this equation we use an equation for ∆Suniv. We developed a relationship ∆Hsys to
entropy change of surrounding ∆Ssur ,∆Ssys and T.
∆Suniv = ∆Ssys -∆Hsys/T
This relationship is the starting point to derive ∆G Gibbs free energy, a single parameter
to predict the spontaneity of a chemical reaction. ∆G is given by the equation:
∆G = ∆H - T∆S.
∆G= Free Energy, ∆H= heat change of the system, T =Kelvin temperature of the system,
∆S = change in entropy of the system.
Let us drive this equation:
∆Suniv
=
∆Ssys +
∆Ssurr (-∆Hsys/T)
∆Suniv = ∆Ssys -∆Hsys/T
∆Suniv x T = T∆Ssys -∆Hsys : multiplying previous equation by T
-∆Suniv x T = -T∆Ssys +∆Hsys : multiplying previous equation by -1
-∆Suniv x T = ∆Hsys-T∆Ssys
-∆Suniv x T = ∆G;
∆G = ∆Hsys-T∆Ssys or ∆G = ∆H - T∆S.
∆G = -T∆Suniv; ∆G is directly proportional to ∆Suniv and T.
If ∆G is -, it makes ∆Suniv positive and the reaction will spontaneous.
If ∆G is 0, it makes ∆Suniv zero and the reaction will be an equilibrium reaction.
If ∆G is +, it makes ∆Suniv negative and the reaction will never take place.
Standard Molar Free Energies of Formation
The standard molar free energy of formation of a compound, ∆G°f is the free energy
change that occurs when 1 mole of compound is formed from its elements in their
standard states.
For ammonia we have:
½ N2(g) + 32 H2(g) → NH3(g); ∆G°rxn = ∆G°f (NH3) = −16.4 kJ.mol-1
As in the case of the standard molar enthalpy of formation, we define the standard molar
free energy of formation of any element in its standard state as zero.
We use the standard molar free energy of formation to determine ∆G° values for any
reaction:
∆G°rxn = Σn∆G°f(products) − Σn∆G°f(reactants)
Predict the spontaneity of the following processes from ∆H and ∆S at various
temperatures.
a) ∆H = 30 kJ ∆S = 6 kJ
T = 300 K
b) ∆H = 15 kJ ∆S = -45 kJ T = 200 K
∆H = 30 kJ
a)
∆S = 6 kJ
T = 300 K
∆G = ∆Hsys-T∆Ssys or ∆G = ∆H - T∆S.
∆H = 30 kJ
∆S = 6 kJ
T = 300 K
∆G = 30 kJ -(300 x 6 kJ)
= 30 -1800 kJ
∆G
= -1770 kJ
The processes with ∆H, T and ∆S should be spontaneous because ∆G value is negative.
b) ∆H = 15 kJ ∆S = -45 kJ T = 200 K
∆G = ∆Hsys-T∆Ssys or ∆G = ∆H - T∆S.
∆H = 15 kJ
∆S = -45 kJ
T = 200 K
∆G = 15 kJ -[200 (-45 kJ)]
= 15 kJ -(-9000) kJ
∆G = 15 + 9000 kJ = 9015 kJ
The processes with ∆H, T and ∆S should be non-spontaneous because ∆G value is
positive.
Calculate the ∆Go for the following chemical reactions using given ∆Ho values, ∆So
calculated above and the equation ∆G = ∆H - T∆S.
a) 2SO2 (g) + O2 (g) ------> 2 SO 3(g) ; ∆Go= ?
∆Hfo[SO2(g)] = -297 kJ/mole ; ∆Hfo [SO3(g)] = -396 kJ/mole
b) 2NH 3 (g) + 3 N2O (g) --------> 4 N2 (g) + 3 H2O (l) ; ∆Go= ?
∆Hfo[ NH3(g)] = -46 kJ/mole ; ∆Hfo [N2O(g)] = 82 kJ/mole;
∆Hfo[ H2O(l)] = -286 kJ/mole
There are two ways to calculate ∆G for chemical reactions.
i) ∆G = ∆H - T∆S.
ii) ∆Go = Σ ∆Gof (products) - Σ ∆G of (reactants)
iii) ∆Ho for a reaction is calculated by ∆Ho = Σ ∆Hof (products) - Σ ∆Hof (reactants)
iv) ∆So is calculated similar to previous problem using:
∆So = Σ ∆So(products) - Σ ∆So (reactants)
v) The ∆Hof, ∆Gof, and ∆So are usually provided to your or found in the appendix of the
book.
a) 2SO2 (g) + O2 (g)
------>
2 SO 3(g) ; ∆Go= ?
∆Hfo[SO2(g)] = -297 kJ/mole ; ∆Hfo [SO3(g)] = -396 kJ/mole
∆S for this reaction is calculated in previous problem.
∆S o = -187 J/K mole
To calculate ∆Ho, use ∆Ho = Σ∆Hof (products) - Σ ∆Hof (reactants)
o
2SO2 (g) + O2 (g)
------>
2 SO 3(g) ; ∆Ho= ?
∆H f
2 x (-297) 1 x 0
2 x (-396)
-594
0
-792
o
o
o
∆H = Σ ∆H f (products) - Σ ∆H f (reactants)
∆Hof = [-792] - [-594 + 0]
∆Hof = -792 + 594
∆Hof = -198 kJ/mole
To calculate ∆Go
Using ∆G = ∆H - T∆S. ∆H = -198 kJ/mole; ∆S = -187 J/K mole; T = 298 K
∆G = -198 - [298 (-187/1000)] kJ/mole
∆G = -198 + 55.7 = -142 kJ/mole
The reaction have a negative ∆G and the reaction is spontaneous or will take place
as written.
o
b) 2NH 3 (g) + 3 N2O (g) --------> 4 N2 (g) + 3H2O (l); ∆G = ?
∆Hfo[ NH3(g)] = -46 kJ/mole ; ∆Hfo [N2O(g)] = 82 kJ/mole;
∆Hfo[ H2O(l)] = -286 kJ/mole
2NH 3 (g) + 3 N2O (g) --------> 4 N2 (g) + 3 H2O (l);
∆So for this reaction is calculated in previous problem.
∆So = -68 J/K mole
To calculate ∆Ho, use ∆Ho = Σ∆Hof (products) - Σ ∆Hof (reactants)
2NH 3 (g) + 3N2O (g) --------> 4 N2 (g) +
3H2O (l);∆Ho= ?
∆Hf 2 x (-46) 3 x 82
4x0
3 x (-286)
-92
246
0
-853
o
o
o
∆H = Σ ∆H f (products) - Σ ∆H f (reactants)
∆Hfo = [0 +(-853] - [-92 + 246]
∆Hfo = -853 - 154
∆Hfo = -1012 kJ/mole
To calculate ∆Go
∆G = ∆H - T∆S. ∆H = -1012 kJ/mole; ∆S = -68 J/K mole= -68/1000 kJ/Kmole; T = 298 K
∆G = -1012 - [298 (-68/1000)] kJ/mole
∆G
= -1012 - 20.26
∆G = -1032.26 kJ/mole
The reaction have a negative ∆G and the reaction is spontaneous or will take place as
written.
o
Calculate the ∆G value for the following reactions using:
∆Go = Σ ∆Gof (products) - Σ ∆Gof (reactants)
N2O5 (g)
+ H2O(l) ------> 2 HNO3(l) ; ∆Go = ?
∆Gfo[ N2O5 (g) ] = 134 kJ/mole ;
∆Gfo [H2O(g)] = -237 kJ/mole;
o
∆Gf [ HNO3(l) ] = -81 kJ/mole
N2O5 (g)
∆Gfo 1 x 134
+
H2O(l) ------>
1 x (-237)
2 HNO3(l) ;
2 (-81)
∆Go = ?
134
-237
-162
o
o
∆G = Σ ∆G f (products) - Σ ∆G f (reactants)
o
∆Go = [-162] - [134 + (-237)]
∆Go = -162 + 103
∆Go = -59 kJ/mole
The reaction have a negative ∆G and the reaction is spontaneous or will take place as
written.
Equation Relating Gibbs Free Energy Changes, Equilibrium Constant (K) and
Reaction Quotient
Q
∆Grxn = +RTln   if Q = K , ∆Grxn = 0, an equilibrium reaction
K
Rearranging
∆Grxn = +RTlnQ − RTlnK
If the reaction is allowed to proceed under standard conditions, then all gaseous reagents
and products will have partial pressures of 1 atm and all solutions in the reactants and
products will have a concentration of 1 M.
Then ∆Grxn = ∆G°rxn and Q =
[products]
= 1 and lnQ = 0.
[reac tan ts]
Thus ∆Grxn = ∆G°rxn = +RTlnQ − RTlnK = 0 − RTlnK
Substitution gives
∆G°rxn = − RTlnK
∆Grxn = ∆G°rxn + RTlnQ
∆G = ∆Go + RT ln Q
The last expression allows us to calculate the free energy when the concentrations are not
1 M and the partial pressure are 1 atm, i.e. when the conditions are not standard state.
If we want to find ∆G at a temperature other than 25°C, we find ∆S° and ∆H° from the
thermodynamic tables (appendices) at 25°C. Then we assume these values do not change
significantly with temperature and calculate ∆G° at any temperature using the expression:
∆G°rxn = ∆H°rxn − T∆S°rxn
where T is the value of the temperature we require in Kelvin. Lastly we find ∆G from
∆Grxn = ∆G°rxn + RTlnQ
Calculate the ∆G for the following equilibrium reaction and predict the direction of
the change using the equation: ∆G = ∆Go + RT ln Q
Given ∆Gfo[NH3(g)] = -17 kJ/mole
N2 (g) + 3 H2 (g) → 2 NH3 (g); ∆G = ? at 300 K, PN2 = 300, PNH3 = 75 and
PH2 = 300
N2 (g) + 3 H2 (g) → 2 NH3 (g); ∆G = ?
∆G , free energy at non standard conditions (ie. at concentration or pressures not equal to 1 or the
temperature
not equall to 25.0 oC(298.15 K)) can be calculated using a equation:
∆G = ∆Go + RT ln Q; ∆G = Free energy at non standard conditions
∆Go= Free energy at standard conditions
R = Universal gas constant, 8.314 J/K mole
T = Kelvin temperature of the reaction
Q= Reaction quotient expressed in molarity or partial pressure
∆Go= can be calculated using ∆Go = Σ ∆Gof (products) - Σ ∆Gof (reactants)
R = 8.314 J/K mole
T = 273.15 + 300 = 573.15 K
Q= can be calculated from partial pressures of the chemicals
To calculate ∆Go
Using ∆Go = ∆Gof (products) - 3 ∆Gof (reactants)
∆Gfo[ N2(g) ] = 0 kJ/mole; ∆Gfo[ H2(g) ] = 0 kJ/mole; ∆Gfo[ NH3(g) ] = -17 kJ/mole
Notice elements have ∆Gfo = 0.00 similar to ∆Hfo
N2 (g) +
3 H2 (g) → 2 NH3 (g); ∆G = ?
o
∆Gf
0
0
2 x (-17)
0
0
-34
∆Go = ∆Gof (products) - 3 ∆Gof (reactants)
∆Go = [-34] - [0 +0]
∆Go = -34
∆Go = -34 kJ/mole
To calculate Q
Equilibrium expression for the reaction in terms of partial pressure:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
p2NH3
K=
_________
pN2 p 3H2
p2NH3
Q = _______ ; Q is when initial concentration is substituted into the equilibrium expression
pN2 p3H2
752
Q
=
_________ ; p2NH3= 752; pN2 =300; p3H2=3003
300 x 3003
Q = 6.94 x 10-7
To calculate ∆Go
∆G = ∆Go + RT ln Q
∆Go= -34 kJ/mole
R = 8.314 J/K mole or 8.314 x 10-3kJ/Kmole
T = 300 K
Q= 6.94 x 10-7
∆G =
∆G =
∆G =
∆G =
∆G =
∆G =
∆Go + RT ln Q
(-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln 6.94 x 10-7)
-34 + 2.49 ln 6.94 x 10-7
-34 + 2.49 x (-14.18)
-34 -35.37
- 69.37 kJ/mole
Gibbs Free Energy, Maximum Work, and Energy Resources
The magnitude of the free energy change, ∆G is the maximum useful work that can be
obtained from a chemical reaction. Another way of looking at ∆G, is that it is the potential
energy available in a chemical reaction that could be harnessed for work.
When ∆G is negative, the magnitude of ∆G is the maximum energy that can be used for
work.
When the Apollo astronauts traveled to the moon, electricity and pure water were
provided by a fuel cell that used hydrogen and oxygen:
2
H2(g) + O2(g) → 2 H2O; ∆G° = −474 kJ
The value of ∆G° is under standard conditions, so PH 2 = PO 2 = 1 atm and the temperature is
25°C, thus the maximum work that the cell could yield under these conditions is 474 kJ.
In fact the cell was 60 – 70 % efficient.
When ∆G is positive, the magnitude of ∆G is the minimum energy that must be supplied
to drive a reaction in this, the nonspontaneous direction.
The chloralkali process produces most of the world’s chlorine and sodium hydroxide.
Salt water is electrolyzed according to the reaction:
2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + Cl2(g) + H2(g); ∆G° = 422 kJ
The value of ∆G° is under standard conditions, so PH 2 = PCl 2 = 1 atm, [NaCl] = [NaOH] = 1
M and the temperature is 25°C, thus the minimum electrical energy that would yield
chlorine and sodium hydroxide under these conditions is 422 kJ.
Gibbs Free Energy and Biological Systems
Bioenergetics is the study of energy and energy transformations in biological
systems. A basic understanding of energy and its use in living organisms is invaluable.
Living organisms must perform work to stay alive, grow and reproduce. All living
organisms must possess the ability to obtain energy and to be able to transform that energy
into a form that can be used by its cells. Practically speaking, knowing the fundamentals
of bioenergetics aids in the understanding of cell function and allows us to understand
why and how the cell is able to harness energy.
Now, first and second laws thermodynamics are relatively easy to understand - but how
can we use them to answer questions dealing with biological reactions. Through our
experience, you have a feeling for how these laws apply to simple chemical reactions we
can see taking place. You do not, however, have an understanding about how these laws
apply to biochemical reactions that not only you can't see but you didn't even know were
occurring. For example, the following is a reaction that every cell performs, the adding of
a phosphate group (PO42-) to the sixth carbon of the monosaccharide, glucose.
PO42- + glucose <----------> glucose-6-phosphate
Is this reaction possible? Most likely, you have no idea as you have little practical
experience dealing with this reaction. You might assume that since every cell can
phosphorylate glucose, this reaction is thermodynamically permissible. But in actuality,
this reaction, as written, is (usually) impossible. This is because this reaction requires
energy to occur and there is no source for this energy (thus this reaction is analogous to
the pen moving to the ceiling). How were you to know this? You couldn't! At some time,
a scientist measured the change in the energy between the reactants (PO42- + glucose) and
the product (glucose-6-phosphate) and found that under certain conditions, the product
contained more energy. Note: the scientist's measurements took into account both actual
energy changes as well as entropy (disorder) changes. These measurements are reported
using a value known as Gibbs Free Energy, abbreviated ∆G. For a reaction in which the
products have more energy than the reactants (thus a gain in energy), the ∆G for that
reaction is positive. That means that a reaction with a positive ∆G cannot occur. For a
reaction in which the products have less energy than the reactants (thus a loss in energy),
the ∆G for that reaction is negative and the reaction is thermodynamically permissible.
You should remember this!
This should leave you wondering - the phosphorylation of glucose (as shown above) has a
positive ∆G and thus cannot occur, but every cell phosphorylates glucose! How can this
be? In order for the reaction to occur, an source of energy is needed. A common source of
energy that cells use is a compound call ATP (adenosine triphosphate). ATP can undergo
the hydrolysis (breaking off) of one of the phosphate groups to produce adenosine
diphosphate, ADP and a free phosphate. This is written as:
ATP ----------> ADP + PO42This reaction has a negative ∆G. Thus if the cell breaks up ATP at the same time that it is
adding the phosphate to glucose, the cell can use the energy released from ATP to fuel the
addition of the phosphate to glucose. This requires that the amount of energy released
from the hydrolysis of ATP is greater than the amount of energy needed to add the
phosphate. If this occurs, then these reactions are said to be coupled. If so, then the
reaction would be written as:
ATP + PO42- + glucose ----------> glucose-6-phosphate +ADP + PO42This reaction can be simplified (just like in Algebra) by noticing that there is PO42- on
each side of the reaction and can be cancelled.
ATP + glucose ----------> glucose-6-phosphate +ADP
Now, where do cells get the ATP? Obviously, if the breakdown of ATP has a negative
delta G (gives off energy), the formation of ATP must have a positive ∆G.
ADP + PO42- ----------> ATP
Where do cells get the energy to make ATP (remember, most of the cellular reactions that
require energy use the hydrolysis of ATP as the source - ATP is like money to the cell).
The ultimate source of this energy is from the nutrients eaten by organisms. The
breakdown of nutrients such as sugars and fats into carbon dioxide and water has a
negative delta G. The cells are capable of taking these nutrients, breaking them down and
using the energy in them to make ATP. You will know more about these processes later.
Conservation of Gibbs Free Energy
Conservation of Gibbs Free Energy applies to chemical reactions because:
∆G°rxn = Σn∆G°f(products) − Σn∆G°f(reactants)
Thermodynamic and Kinetic Stability
Dehydration of an alcohol such as 1-butanol (CH3CH2CH2CH2OH) produces a mixture of
butenes: 1-butene, cis- and trans- 2-butene.
1-butene
trans-2-butene
cis-2-butene
The activation barrier or the products are small. The relative amounts of each of these
products can be determined experimentally. By computing the energies of each isomer,
we can estimate their relative stabilities. If the thermodynamic stability of the isomers is in
agreement with the experimentally observed product distribution, the reaction is said to be
under thermodynamic control. If not, it is said to be under kinetic control. Usually kinetic
stability means large activation barriers that need to over come to participate in chemical
reaction even though thermodynamically stable products could be formed.
102 HOMEWORK 6
HOMEWORK FOR CHAPTER 18
1.Which of the following processes involve an increase in ∆Srxn?
1. Br2 (l)
-->
Br2 (g)
2. 2NO(g) + O2 (g) --> 2NO2 (g)
--> MgO(s) + SO3 (g) 4. 2HgO(s)
--> 2Hg(l) + O2 (g)
3. MgSO4 (s)
a. 1 and 2
b. 2 and 3
c. 3 and 4
d. 1, 3, and 4
e. 1, 2, 3, and 4
2. In which one of the following processes will ∆Srxn be negative?
a. 2NH3 (g)
-----> N2(g) + 3H2(g)
b. BaO(s) + CO2(g) -----> BaCO3(s)
c. H2O(g)
-----> H2(g) + 1/2O2(g)
d. I2(s)
-----> I2(g)
e. Hg(s)
-----> Hg(l)
3. The boiling point of mercury is 357°C. The heat of vaporization of mercury
at its boiling point is
59.1 kJ. The entropy of vaporization is
a. 0.0938 J/K b. 10.7 J/K c. 21.1 J/K d. 37.2 J/K e. 93.8 J/K
4. Which of the following processes would be expected to have a positive ∆S value?
a. 2H2(g) + O2(g) ---> 2H2O(g)
b. H2O(g) ---> H2O(l)
c. N2(g) + O2(g) ---> 2NO(g)
d. OF2(g) + H2O(g) ---> O2(g) + 2HF(g)
e. CO(g) + 1/2O2(g) ---> CO2(g)
5. Calculate the enthalpy change, ∆Ho, for the combustion of C3H6(g):
C3H6(g) + 9/2O2 (g) ------> 3CO2(g) + 3H2O(l); ∆Ho = ?
o
∆Hf values in kJ/mol are as follows: C3H6 (g) = 21; CO2(g) = -394; H2O(l)= -286 kJ/mol.
a. -2061 kJ
b. -2019 kJ c.
-701 kJ d.
2019 kJ e. 2061 kJ
6. Calculate ∆S° for the following reaction:
Fe2O3 (s) + 3H2 (g) -----> 2Fe (s) + 3H2O (g)
∆S°( Fe2O3(s)) =90 J/mol K ; ∆S°(H2(g)) =130.6 J/mol K; ∆S°(Fe(s)) = 27 J/mol K;
∆S°(H2O(g)) = 188.7 J/mol K.
a. 4.9 J/mol K b. 228.3 J/mol K c. 620.1 J/mol K d. 173.3 J/mol K e. 138.3 J/mol K
7. Consider the following chemical reaction which proceeds at 298 K (25°C). Standard free energies
∆G°f in kJ/mol are given in parentheses.
2H2S(g) + SO2(g) ----> 3S(s) + 2H2O(g) ∆G° = ?
∆G°f (-34.0) (300.0)
(0.0) (-229.0)
What is ∆ G° for this reaction?
a. 310 kJ/mol
b. -690 kJ/mol c. 100 kJ/mol
d. 300 kJ/mol
e. 405 kJ/mol
8. What is the ∆G°, if ∆H° = 72.75 kJ/mol and ∆S° = -145.3 J/K°mol, at 25°C for the following
reaction?
H2S(aq) --> 2H+(aq) + S2-(aq)
a. -72.7 kJ/mol b. -43.2 kJ/mol c. -29.5 kJ/mol d.116.1 kJ/mol
9. Which of the following compounds has the highest entropy in J/mol K at 298 K?
a. CH3OH(l) b. CO(g) c. SiO2(s) d. H2O(l) e. CaCO3(s)
10. Which of the following ionic compounds has the highest entropy in J/mol K at 298 K?
a. NaCl(s) b. KCl(s) c. NaF(s) d. CsCl(s) e. RbF(s)
11. The Ka experssion for the dissociation of acetic acid in water is based on the following
equilibrium:
HC2H3O2(l) <===> H+(aq) + C2H3O2-(aq)
What is ∆G° if Ka=1.8 x 10-5? [∆G°=-RTlnK]
a. -9.74 kJ/mol b. -27.1 kJ/mol c. 27.1 kJ/mol d. 9.74 kJ/mol
12. For a reactions for which ∆H<0, that is, H is negative, using ∆G = ∆H - T∆S, which of the
following is true?
a. are spontaneous under all temperatures
b. are spontaneous under all temperatures when ∆S>0.
c. are endothermic.
d. take place very slowly.
13. Identify the statement which best explains the term "coupled reaction."
a. A reactant-favored reaction consumes the products from a product-favored reaction.
b. The heat released from an exothermic reaction is used in a related endothermic reaction.
c. The Gibbs free energy released from a product-favored reaction is used to provide the energy
for a reactant-favored reaction to occur.
d. A product-favored reaction supplies the activation energy for another reaction.
e. The Gibbs free energy released from a product-favored reaction is released to the surroundings,
causing an increase in entropy.
14. Which set of conditions describes a reaction that is most likely to occur?
a. exothermic, increasing entropy, low activation energy
b. exothermic, increasing entropy, high activation energy
c. exothermic, decreasing entropy, high activation energy
d. endothermic, decreasing entropy, low activation energy
e. endothermic, decreasing entropy, high activation energy
15. Consider following reaction:
CO(g) + 2H2 (g) ---> CH3OH(l);
for which ∆G° = -29 kJ at 25°C, and reaction quotient, Q = 2.2 x 10-2
( Given ∆G =∆G° + RT lnQ, R = 8.314 x 10-3 kJ/mol K)
What is ∆G for this reaction?
a. -27 kJ
b. +27 kJ
c. -39 kJ
d. +23 kJ
e. + 270 J
102 Sample Test 6
SAMPLE TEST FOR CHAPTER 18
1. "The entropy of perfect crystalline substances is zero at 0 K " is
a. a statement of an experimentally determined value.
b. a statement of the first law of thermodynamics.
c. a statement of the second law of thermodynamics.
d. a statement of the third law of thermodynamics.
2. The second law of thermodynamics states that
a. mass is conserved.
b. energy is conserved.
c. the entropy of the universe increases in spontaneous changes.
d. the entropy of a perfect crystalline substance at 0 K is zero.
3. Which of the statements are true of ∆G,
a. when ∆G > 0 reactions are spontaneous.
b. at any value of ∆G, reactions are spontaneous.
c. when ∆G = 0 reactions go to equilibrium.
d. when ∆G < 0 reactions never take place.
4. Which of the following (assume the gases are at same T and P) would be expected to have the
largest entropy value per mole?
b. SO2Cl (l) c. SO2Cl (g) d. SO2 (g) e. Cl22(g)
a. SO2Cl (s)
5. What is ∆S°rxn for
2H2 (g) +
O2 (g) --> 2H2O(l) at 25°C?
S°, J/K mol: 130.68
205.03
69.91
a. -265.80 J/K b. -326.58 J/K c. 326.58 J/K d. 8.164 kJ/K
6. The normal boiling point of benzene is 80.1°C and heat of evaporation (∆H°vap)is 30.7 kJ/mol.
The ∆Ssurr (in J/K mol) for the evaporation of benzene is
a. 383 J/K mol. b. 111 J/K mol. c. -86.9 J/K mol. d. 49.4 J/K mol.
7. Calculate ∆S° for the production of ozone from oxygen.
3O2(g)
------>
2O3(g)
∆S°[O2(g)]= 205 J/(mol K)
∆S°[O3(g)]= 239 J/(mol K)
a. +137 J/(mol K) b. -137 J/(mol K) c. +34 J/(mol K) d. -34 J/(mol K) e. +444 J/(mol K)
8. The Ka experssion for the dissociation of acetic acid in water is based on the following
equilibrium at 25°C:
HC2H3O2(l)<===> H+(aq) + C2H3O2-(aq)
What is ∆G° if Ka=1.8 x 10-5? [∆G°=-RTlnK](R= 8.315 x 10-3 kJ/mol K)
a. -9.74 kJ/mol b. -27.1 kJ/mol c. 27.1 kJ/mol d. 9.74 kJ/mol
9. Given the reaction,CH4(g) + N2(g) ---> HCN(g) + NH3(g); ∆H°=+ 164 kJ; ∆G°= 159 kJ, calculate
∆S° for the reaction at 25°C.
a. 1.54 J/K b. 16.8 J/K c. 67.7 J/K d. 109 J/K e. 443 J/K
10. Which of the following processes involve an increase (positive)in ∆Srxn?
1. Br2 (l) --> Br2 (g)
2. 2NO(g) + O2 (g) --> 2NO2 (g)
3. MgSO4 (s) --> MgO(s) + SO3 (g)
4. 2HgO(s) --> 2Hg(l) + O2 (g)
a. 1 and 2 b. 2 and 3 c. 3 and 4 d. 1, 3, and 4 e. 1, 2, 3, and 4
11. What is the ∆G° (in kJ/mol) at 25°C for the following reaction?
Na2S(s) --> 2Na+(aq) + S2-(aq)
∆G°f, kJ/mol: -361.4
-261.87
86.31
a. -185.8 b. -76.0 c. 76.0 d. 185.8
12. What is equal to ∆H at equilibrium? (hint: use ∆G = ∆H-T∆S)
a. ∆S b. ∆H c. T d. ∆G e. T ∆S
13. The entropy of any element at room temperature, 298 K, is
a. zero. b. negative. c. positive. d. not defined.
14. Which process is product-favored?
a. mixing of oil and water
b. settling of solids from muddy water
c. separation of acetic acid from a vinegar solution
d. dissolving of diamonds in water
e. evaporation of water at -10 degrees Celsius
15. Which process is reactant-favored?
a. diffusion of the odor of cooking food
b. dissolving of sugar in coffee
c. decomposition of iron ore into pure iron
d. burning a candle
e. melting ice at 25 degrees Celsius
16. Which statement about reactant-favored and product-favored processes is not correct?
a. A product-favored process is the exact opposite of a corresponding reactant-favored process.
b. Once started, a product-favored process will continue on its own.
c. A reactant-favored process requires a continuous input of energy to continue.
d. The energy used by a reactant-favored process exceeds the energy difference between products and
reactants because of activation energy.
e. The term spontaneous is sometimes used to describe a reactant-favored process.
17. The "dispersal of energy" explanation of exothermic reactions says that nature favors exothermic
reactions because
a. the number of particles in the system remains constant while the energy is distributed to a select few of
the particles.
b. all of the energy of the reactants is transferred to the many particles of the surroundings.
c. the number of particles in the system increases while the energy of the system also increases.
d. the energy originally present in the reactants is distributed over a very large number of particles both in
the system and in the surroundings.
e. the energy and the number of particles remain the same, although the atoms may be rearranged.
18. Consider the options given for the distribution of three units of energy to three particles. Which
arrangement is least likely?
a. A*** b. A*B*C* c. A**B* d. A**C* e. B**C*
19. Consider the options given for the distribution of three units of energy to three particles. Which
arrangement is most likely?
a. A*** b. A*B*C* c. A**B* d. A**C* e. B**C*
20. If a coin is tossed five times, which of the arrangements shown is least likely?
a. THTHT b. HHTHT c. THHTT d. TTTHT e.HTTHT
21. Which process describes an increase in disorder?
a. adding sugar to water
b. distilling crude oil into gasoline, fuel oil, and jet fuel
c. isolating sea salt from ocean water
d. collecting nitrogen and oxygen from liquid air
e. filtering solid impurities out of a mixture
22. The boiling point of tin is 232°C. The heat of vaporization of tin at its boiling point is 247 kJ. The
entropy of vaporization is
a. 2.04 J/K b. 489 J/K c. 939 J/K d. 1065 J/K e. 2045 J/K
23. Which has the lowest entropy?
a. H2O(l) b. H2O(s) c. H2O(g) d. H2(g) e. O2(l)
24. Which has the lowest entropy?
a. S(s) b. O2(g) c. SO2(g) d. SO2(l) e. SO2(s)
25. Which has the highest entropy?
a. S(s) b. O2(g) c. SO2(g) d. SO2(l) e. SO2(s)
26. Which statement about entropy is not correct?
a. The entropy of a particular sample is greater in the gaseous state than in the solid state.
b. In considering a series of similar, but increasingly larger molecules all in the same state, the largest
molecule will have greatest entropy.
c. Because of the larger charges, the entropy of CaO is less than the entropy of KF.
d. Entropy decreases when ethanol, C2H5OH, dissolves in water.
e. A bottle of carbonated water has lower entropy than the same masses of carbon dioxide gas and
distilled water.
27. Which process illustrates an increase of entropy?
a. preparing an unknown containing five ions in one solution
b. writing a research paper from several sources
c. alphabetizing your chemistry books by author
d. compiling lab results from several sets of data
e. purifying water
28. Which process illustrates a decrease in entropy?
a. baking a cake from scratch
b. making tossed salad
c. producing aluminum foil from bauxite ore
d. shuffling a deck of cards
e. spreading grass seed on a lawn
29.Calculate the value of ∆S° for the reaction shown:
C3H8(g) + 5O2(g) → 4H2O(g) + 3CO2(g)
At 25°C the values of entropy in J/mol-K are propane, 270.2; oxygen, 205.15; water, 188.74; carbon
dioxide, 213.80.
a. -100.41 J/K b. -72.81 J/K c. 72.81 J/K d. 100.41 J/K e. 921.01 J/K
30. Calculate the value of ∆S° for the reaction shown:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
At 25°C the values of entropy in J/mol-K are sulfuric acid, 20.1; potassium hydroxide, 91.6; potassium
sulfate, 225.1; and water, 188.74.
a. 399.3 J/K b. 302.1 J/K c. 507.1 J/K d. 210.5 J/K e. 490.9 J/K
31. Calculate the value of ∆S° for the reaction shown:
N2(g) + 3H2(g) → 2NH3(g)
At 25°C the values of entropy in J/mol-K are nitrogen, 191.61: hydrogen, 130.68; and ammonia, 192.77.
a. -969.19 J/K b. -393.20 J/K c. -390.88 J/K d. -198.11 J/K e. -129.52 J/K
32. Calculate the value of ∆S° for the reaction shown:
2NH3(g) → N2(g) + 3H2(g)
At 25°C the values of entropy in J/mol-K are ammonia, 192.77; nitrogen, 191.612; and hydrogen, 130.68.
a. 129.52 J/K b. 198.11 J/K c. 390.88 J/K d. 393.20 J/K e. 969.19 J/K
33. Which combination of conditions describes a reaction that is product favored?
a. ∆H° negative; ∆S° negative
b. ∆H° negative; ∆S° positive
c. ∆H° positive; ∆S° negative
d. ∆H° positive; ∆S° positive
e. ∆H° zero; ∆S° zero
34. For a particular reaction, the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K. This reaction is
a. product-favored, because ∆S°universe is negative.
b. reactant-favored, because ∆S°universe is negative.
c. product-favored, because ∆S°universe is positive.
d. reactant-favored, because ∆S°universe is positive.
e. impossible to determine without further information.
35. According to the second law of thermodynamics,
a. any process in which entropy increases will be product-favored.
b. any process which is endothermic will be product-favored.
c. only processes in which the entropy of the universe is increasing will be product-favored.
d. heat will always be transferred from a warmer object to a cooler object.
e. the total amount of matter in the universe remains constant.
36.Which statement regarding the Gibbs free energy of a system is not correct?
a. It is defined as -T∆Suniverse.
b. In a product-favored reaction, its value is less than zero.
c. Its value is a function of ∆Ssurroundings and ∆Ssystem.
d. Its value is zero when ∆Hsystem = ∆Ssystem.
e. The previous four statements are all correct.
37. Use the data given to calculate the value of ∆Grxn for the reaction at 25°C.
AgCl(s) → Ag+(aq) + Cl-(aq)
a. 32.5 kJ b. 55.7 kJ c. 64.6 kJ d. 65.4 kJ e. 75.2 kJ
38. Use the data given to calculate the value of ∆Grxn for the reaction at 25°C.
Ag+(aq) + Cl-(aq) → AgCl(s)
a. -32.5 kJ b. -55.7 kJ c. -64.6 kJ d. -65.4 kJ e. -75.2 kJ
39. Use the data given to calculate the value of ∆Grxn for the reaction at 25°C.
2C(graphite) + H2(g) → C2H2(g)
a. -64.6 kJ b. -207.6 kJ c. -226.8 kJ d. -244.3 kJ e. -291.4 kJ
40.Use the data given to calculate the value of ∆Grxn for the reaction at 25°C.
2C(graphite) + O2(g) → 2CO(g)
a. -68.58 kJ b. +137.16 kJ c. -137.16 kJ d. +274.32 kJ e. -274.32 kJ
41. Use the data at 25°C given below to calculate the value of ∆Grxn for the reaction shown when it takes place
at 120°C.
C3H8(g) + 5O2(g) → 4H2O(g) + 3CO2(g)
a. -2083 kJ b. -2073 kJ c. -2055 kJ d. -2046 kJ e. -2004 kJ
42. Calculate the Celsius temperature at which a reaction with ∆Ηrxn = 177.8 kJ and ∆Srxn =160.5 J/K makes
the transition between reactant-favored and product-favored.
a. 111°C b. 630°C c. 835°C d. 903°C e. 1108°C
43. Which pair of values for ∆G°rxn and K indicates a product-favored system?
a. χ = 185.0 kJ; K = 10-8
b. ∆G°rxn = 0 kJ; K = 1
c. ∆G°rxn = -185.0 kJ; K = 10+8
d. ∆G°rxn = 0 kJ; K = 10-8
44.
45.
46.
47.
e. ∆G°rxn = -185.0 kJ; K = 10-8
The value of ∆G°rxn for a reaction at 25°C is -47.8 kJ. What is the value of the equilibrium constant for this
reaction?
a. 9.58 × 103 b. 2.39 × 108 c. 9.76 d. 1.26 e. 1.02
The value of ∆G°rxn at 45°C is -47.8 kJ. What is the value of the equilibrium constant for this reaction?
a. 1.05 × 1024 b. 1.40 × 109 c. 7.11 × 107 d. 1.14 e. 1.02
The value of the equilibrium constant for a reaction is 2.65 × 10-6 at 35°C. Calculate the value of ∆G°rxn .
a. 1.62 kJ b. 3.74 kJ c. 14.2 kJ d. 29.1 kJ e. 32.9 kJ
Use the data given to calculate the value of K for the reaction at 25°C.
AgCl(s) → Ag+(aq) + Cl-(aq)
a. 5.69 × 109 b. 1.76 × 10-10 c. 3.41 × 10-12 d. 4.75 × 10-12 e. 6.61 × 10-14
48. Use the data given to calculate the value of K for the reaction at 5°C.
AgCl(s) → Ag+(aq) + Cl-(aq)
a. 9.76 × 10-1 b. 7.68 × 10-7 c. 1.76 × 10-10 d. 2.61 × 10-11 e. 5.10 × 10-13
49. Which statement correctly describes the meaning of the value of ∆G°rxn for a product-favored reaction?
a. It is the maximum amount of useful work that can be obtained from the reaction.
b. It is the minimum amount of work which must be done on the system.
c. It is the amount of additional energy that must be supplied in order to initiate the reaction.
d. It is the minimum amount of heat that can be released from the system.
e. It is the amount of energy that is used to overcome the disorder of the system.
50.Which statement correctly describes the meaning of the value of ∆G°rxn for a reactant-favored reaction?
a. It is the maximum amount of useful work that can be obtained from the reaction.
b. It is the minimum amount of work which must be done on the system.
c. It is the amount of additional energy that must be supplied in order to initiate the reaction.
d. It is the minimum amount of heat that can be released from the system.
e. It is the amount of energy that is used to overcome the disorder of the system.
51. The molecule used to store Gibbs free energy in the human body in a readily available form is
a. C6H12O6 b. H2O c. CO2 d. ADP3- e. ATP452. The steps for converting the energy in nutrients into a biologically usable form are, in order,
a. digestion, production of acetyl groups, citric acid cycle
b. digestion, citric acid cycle, production of acetyl groups
c. production of acetyl groups, digestion, citric acid cycle
d. citric acid cycle, production of acetyl groups, digestion
e. citric acid cycle, digestion, production of acetyl groups
53. The conversion of glucose to glucose-6-phosphate is _____; the conversion of ATP to ADP is _____; and
the coupled reaction is _____.
a. endergonic; exergonic; endergonic
b. endergonic; exergonic; exergonic
c. endergonic; endergonic; endergonic
d. exergonic; endergonic; exergonic
e. exergonic; exergonic; exergonic
54.Since no chemical process is 100% efficient, some Gibbs free energy is always "wasted" and is converted to
_____ instead of doing useful work.
a. light
b. chemical energy
c. mechanical energy
d. heat
e. electricity
55. If a substance is thermodynamically unstable, but kinetically stable, it can be made to react by providing
a. more reactants
b. more products
c. activation energy
d. metals
e. lower temperature
56.Which set of conditions describes a reaction that is least likely to occur?
a. exothermic, increasing entropy, low activation energy
b. exothermic, increasing entropy, high activation energy
c. exothermic, decreasing entropy, high activation energy
d. endothermic, decreasing entropy, low activation energy
e. endothermic, decreasing entropy, high activation energy
Answers to Sample Test 6 (Chpater 18) Questions
1. d. a statement of the third law of thermodynamics.
2. c. the entropy of the universe increases in spontaneous changes.
3. c. when ∆G = 0 reactions go to equilibrium.
4. c. SO2Cl (g)
5. b. -326.58 J/K
6. c. -86.9 J/K mol.
7. b. -137 J/(mol K)
8. c. 27.1 kJ/mol
9. b. 16.8 J/K
10. d. 1, 3, and 4
11. b -76.0
12. e. T ∆S
13. c. positive.
14. b. settling of solids from muddy water
15. c. decomposition of iron ore into pure iron
16. e. The term spontaneous is sometimes used to describe a reactant-favored process.
17. d. the energy originally present in the reactants is distributed over a very large number of particles both
in the system and in the surroundings.
18. a. A***
19. b. A*B*C*
20. d. TTTHT
21. a. adding sugar to water
22. b. 489 J/K
23. b. H2O(s)
24. a. S(s)
25. c. SO2(g)
26. d. Entropy decreases when ethanol, C2H5OH, dissolves in water.
27. a. preparing an unknown containing five ions in one solution
28. c. producing aluminum foil from bauxite ore
29. d. 100.41 J/K
30. a. 399.3 J/K
31. d. -198.11 J/K
32. b. 198.11 J/K
33. b. ∆H° negative; ∆S° positive
34. b. reactant-favored, because ∆S°universe is negative.
35. c. only processes in which the entropy of the universe is increasing will be product-favored.
36. d. Its value is zero when ∆Hsystem = ∆Ssystem.
37. b. 55.7 kJ
38. b. -55.7 kJ
39. d. -244.3 kJ
40. e. -274.32 kJ
41. b. -2072 kJ
42. c. 835°C
43. c. ∆G°rxn = -185.0 kJ; K = 10+8
44. b. 2.39 × 108
45. c. 7.11 × 107
46. e. 32.9 kJ
47. b. 1.76 × 10-10
48. d. 2.61 × 10-11
49. a. It is the maximum amount of useful work that can be obtained from the reaction.
50. b. It is the minimum amount of work which must be done on the system.
51. e. ATP452. a. digestion, production of acetyl groups, citric acid cycle
53. b. endergonic; exergonic; exergonic
54. d. heat
55. c. activation energy
56. e. endothermic, decreasing entropy, high activation energy