Transcript - Salisbury University

Chapter 17
Free Energy
And
Thermodynamics
I. Directionality of Reactions
Question? Why do some reactions (processes)
occur in one direction while others in another?
A. Product-Favored Reactions (Processes)
1. Occurs spontaneously, as written from left to
right, without outside help or intervention.
2. Kc >1
3. Reaction occurs on its own
4. Tells us nothing about rate (speed of reaction)
B. Reactant-Favored Reactions (Processes)
1. Occurs nonspontaneously as written from left to
right.
2. Kc < 1
3. Reaction tends to proceed in opposite direction
as written.
4. Tells us nothing about the rate or speed of
reaction.
Examples of Each Type
Significance of being able to understand and
predict direction. (Is reaction spontaneous?)
II. Probability and Chemical Rxns
(Factors Affecting Direction of Rxn)
Dispersal of Energy
What is the likely flow of energy in a natural
process?
1. If a thousand bricks are dropped from an
airplane, would you expect them to land in the
form of a brick wall?
2. If you dropped a mixed deck of cards, would
you expect them to land in order?
3. Would you expect the elevated circle of water
around where a rock fell to remain motionless?
4. Would you expect one end of a metal rod to
spontaneously get hot while the other end gets
cold?
Dispersal of Energy occurs because the
probability is much higher that energy will
be spread over many particles than that it
will be concentrated in a few.
Entropy , “S”, is a measure of the
randomness or disorder of energy. It
relates macroscopically to the randomness
or disorder on the molecular scale.
Example – Melting of Ice At 250C
(Room Temp)
H2O (s) → H2O (l)
Is the system exothermic? (ΔH “+” or “-”?)
Is the process product-favored (spontaneous)?
Why?
Explanation
Both enthalpy (ΔH) and entropy (ΔS) for the
system are important in determining whether the
process is spontaneous (product-favored) or not.
Process (at 250C) is spontaneous because entropy
of system is more significant factor in this case.
III. Closer Look At Entropy
A. Qualitative Guidelines for Predicting
Entropy of Process or Reaction
1. Entropy of gases are generally much larger than
those of liquids, which in turn are larger than
those of solids.
Therefore, entropy increases when:
Solids melt to form liquids.
Solids or liquids vaporize to form gases.
2. Entropies of more complex molecules are
larger than those of simpler molecules,
especially in a series of closely related
compounds.
The more atoms in molecules (higher molar
mass), the greater is the entropy of the substance.
3. Entropies of ionic solids are larger the
weaker the attractions among the ions.
4. Entropy usually increases when a pure
liquid or solid dissolves in a solvent.
5. Entropy decreases when a gas dissolves in a
liquid.
6. Entropy increases when a chemical reaction
produces an increase in the number of gas
molecules.
Predict whether the entropy change will be
positive or negative for each reaction or
system. Explain your answer.
NaCl(s)  Na+ (aq) + Cl-(aq)
CO2(g) + MgO(s)  MgCO3(s)
H2O(g)  H2O(l)
N2(g) + 3 H2(g)  2 NH3(g)
B. Absolute Entropy Values
1. Entropy can be measured with calorimeter
for a process that takes place at constant
T and P.
ΔS = q/T (T in Kelvin scale)
2. Entropy is temperature dependent.
(predicted by Kinetic Molecular Theory)
3. What happens to entropy when T = 0 K?
4. Third Law of Thermodynamics- the entropy
of a perfect crystal = 0 at absolute zero.
5. Standard Molar Entropy Values can be
determined. See Appendix II or Table 17.2
S0 =
J / K mol (at 298 K)
6. Remember these are absolute values for
both elements and compounds.
( Not like ΔH values which were measured relative
to elements in their most stable state. Standard
enthalpies of formation (ΔH0f) of elements are
arbitrarily set to zero.)
C. Entropy Calculations
1. Determining the Entropy Change For a
Reaction (System)
“Since entropy is a state function and is
thus path independent, we can use a Hess=s
Law type calculation. Remember we are
working with absolute entropies.”
ΔS0rxn = S0products - S0reactants
ΔS0rxn = ∑{(moles of prod.)x So(prod.)} ∑{(moles of react.)x So(react.)}
Example Problem:
Calculate the standard entropy change (ΔS0) for
the following reaction. Standard molar entropies
are found in Appendix J or Table 18.1.
S0 (J K-1 mol-1)
2 CO(g) + O2(g)  2 CO2(g)
197.6
205.1
213.7
What is the standard entropy change for the
reaction: 2 CO2(g)  2 CO(g) + O2(g) ?
Example : Freezing of Water At -100C
(Below FP of Water)
H2O (l) → H2O (s)
Is the system exothermic? (ΔH “+” or “-”?)
Is ΔS for the system “+” or “-” ?
Is the process product-favored (spontaneous) at
-100C ?
Why?
Explanation
Both enthalpy (ΔH) and entropy (ΔS) for the
system are important in determining whether the
process is spontaneous (product-favored) or not.
Process (at -100C) is spontaneous.
Let’s explain why relative to total entropy of
universe.
D. Entropy and the Second Law of
Thermodynamics
“The total entropy of the universe (the system
plus its surroundings) always increases for a
spontaneous process.”
ΔSuniverse = ΔSsystem + Δssurroundings
Δsuniverse > O (or “+” )
We have calculated ΔSsystem = ΔSreaction by
using Hess’s Law. If we can determine
ΔSsurroundings , we can predict whether a
reaction is product favored (spontaneous).
Calculate ΔH for the reaction and assume that
this quantity of thermal energy is transferred
reversibly to or from the surroundings.
ΔSsurroundings = - ΔHrxn / T
Therefore:S
H
Hrxn

 TSTrxn
S universe  S rxn 
S rxn
Suniverse
universe
rxn
H rxn

T
We could use this information to predict if
reaction is product favored. But lets simplify.
S universe  S rxn
H rxn

T
Multiply both sides by –T(Kelvin scale) :
 TS universe  TS rxn
Simplify and rearrange:
H rxn
T
T
 TSuniverse  H rxn  TSrxn
Define Free Energy:
G  H rxn  TSrxn
IV. Gibbs Free Energy
A. Definition of Gibbs Free Energy
1. New thermodynamic function, G
2. ΔG = -TΔSuniverse
3. If entropy of universe increases, the Gibbs free
energy must decrease (have a “-” sign).
4. Plugging into previous equation (under standard
conditions gives:
ΔG0 = ΔH0 – TΔS0
ΔG0 = ΔH0 – TΔS0
If ΔG0 is negative (-), the reaction
is product-favored (spontaneous).
If ΔG0 is positive (+), the reaction
is reactant-favored (nonspontaneous)
To predict the direction of reaction:
Spontaneous or
ProductFavored?
ΔG0
ΔG0
=
ΔH0
ΔH0
-
TΔS0
ΔS0
Yes
-
-
+
Yes, low T
No, high T
?
-
-
No, low T
Yes, high T
?
+
+
No
+
+
-
B. Free Energy Calculations
Two Methods
1. From ΔH0 and ΔS0 Values
ΔG0 = ΔH0 - T ΔS0
2. From Hess’s Law Type Calculation
(from Table of Standard Free Energies of
Formation – see Appendix II or Table 17.3
ΔGo = ∑{(moles of prod.) Gof (prod.)}
- ∑{(moles of react.) Gof (react.)}
1. Calculate ΔG0 for the reaction below.
2 Fe2O3(s) + 3 C(gr)  4 Fe(s) + 3 CO2 (g)
ΔHof -824.2
o
0
-393.5
S0
87.4
5.74
27.78
ΔGof -742.1
0
0
(ΔHof and ΔGof in kJ/mol, S0 in J/K mol)
213.7
-394.4
a. Is the reaction endothermic or exothermic?
b. Does entropy increase or decrease for the
reaction? (Predict first, then verify)
c. Is the reaction product-favored or reactantfavored?
V. Temperature Effects and
Reaction Direction
A. Explanation
1. A reaction may be product-favored at one
temperature and reactant-favored at another.
2. How can we estimate that temperature?
Its where ΔGrxn equals zero. (ΔGrxn will switch from
“-” to “+’ or switch from “+” to “-” at that T.
ΔG0 = ΔH0 – TΔS0
0 = ΔH0 – TΔS0
T = ΔH0 / ΔS0
B. Problem
For the reaction previously discussed:
2 Fe2O3(s) + 3 C(gr)  4 Fe(s) + 3 CO2 (g)
the following thermodynamic data was determined.
ΔH0rxn = +467.9 kJ
ΔG0rxn = +301.0 kJ
ΔS0rxn = +560.2 J/K
1. What is ΔG0rxn at 250C ?
2. Determine (estimate) ΔGrxn at 500 Kelvin.
3. At what temperature will the reaction become
product-favored or exergonic?
VI. Free Energy & Equilibrium
A. ΔG and K Give Similar Information
** ΔG (Gibbs Free Energy) tells you whether a
reaction is spontaneous or not. It tells you
whether a reaction will proceed to products or
not.
** K (equilibrium constant) tells you how much
product or reactants will be present at
equilibrium. It tells you whether a reaction will
proceed to product or not.
B. Mathematical Relationship
ΔG = ΔG0 + RT lnQ
At equilibrium:
ΔG = 0
Q = Keq
Therefore:
ΔG0 = -RTlnKeq
ΔG0 = -RTlnKeq
1) T = temperature in Kelvin scale
2) R = gas constant (8.314 J / mol K)
3) Keq = Kc for reactions involving solutions
4) Keq = Kp for reactions involving gases
5) Keq = Ksp for equlibrium of slightly
soluble salts
ΔGo
Keq
Product -Favored
Positive
<1
no
Negative
>1
yes
zero
1
neither
C. Problem Solving
1. Determination of Free Energy
The following reaction conducted at a
temperature of 968K, was found to have an
equilibrium constant of Kp = 49.7.
H2(g) + I2(g)  2 HI(g)
Is the reaction product favored?
What is the value of the standard Gibbs Free
Energy for the reaction at this temperature?
2. Determination of Equilibrium Constant
Answer the following questions for the
reaction and information provided.
2 Cl2(g) + 2 H2O(g)  4 HCl(g) + O2 (g)
ΔHof
0
-241.82
-92.31
0
S0
223.07
188.83
186.91
205.14
ΔGof
0
-228.57
-95.30
0
(ΔHof and ΔGof in kJ/mol, S0 in J/K mol)
a. Determine the equilibrium constant for
the reaction at 298K.
b. Is the reaction reactant or product
favored at 298K?
VII. Significance of Sign and Size of ΔG0
Large, negative
G°; equilibrium
lies far to right.
Large, positive
G°; equilibrium
lies far to left.
Intermediate G°;
equilibrium lies in
intermediate position.
VIII. Gibbs Free Energy and
Maximum Work
ΔG represents the maximum useful work that
can be done by a product favored-reaction on the
surroundings.
Calculate the maximum useful work (energy) that
can be produced from the complete combustion of 8.0 g
of methane under standard conditions?
IX. Coupling Reactions and
Metabolism
Glucose is degraded in our body through a
series of chemical reactions which is called a
metabolic pathway (glycolysis).
glucose
glucose 6-P
glyceraldehyde 3-P
3-phosphoglycerate
fructose 6-P
fructose 1,6-BP
pyruvate
Consider the first reaction:
glucose + Pi  glucose 6-P + H2O ΔG0 = + 13.8 kJ
How can this reaction occur in the body?
By coupling of reactions
glucose + Pi  glucose 6-P + H2O ΔG0 = + 13.8 kJ
ATP + H2O  ADP + Pi
ΔG0 = - 30.5 kJ
glucose + ATP  glucose 6-P + ADP ΔG0 = - 16.7 kJ
Net coupled reaction is product-favored and
metabolic pathway may continue to occur!!
X. Thermodynamic vs. Kinetic
Stability
Thermodynamically Stable: if a substance does
not undergo product-favored reactions.
Kinetically Stable: if a substance may be involved
in a product-favored reaction, but it is so slow that
the substance is stable.
Example: Aluminum (in soda cans)
Thermodynamically unstable:
4 Al(s) + 3 O2(g)  2 Al2O3(s) ΔG0 = - 3164.6 kJ
Aluminum cans do not decompose readily, Why??