thermodynamics internal energy enthalpy entropy free energy chapter 17 (mcm) chapter 20 silberberg
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THERMODYNAMICS
Internal EnergyEnthalpyEntropyFree EnergyChapter 17 (McM)Chapter 20 Silberberg
Goals & Objectives
See the following Learning Objectives on page 914.
Understand these Concepts: 20.1-22.
Master the Skills: 20.1-10.
Thermodynamics the study of the changes in energy and
the transfers of energy that accompany chemical and physical processes.
Addresses three fundamental questions Will 2 or more substances react when they are
mixed under specified conditions? If they do react, what energy changes and
transfers are associated with their reaction? If a reaction occurs, to what extent does it
occur?
Thermodynamics
Used to determine if a reaction can occur under specified conditions. spontaneous reaction--can occur under
the specified conditions nonspontaneous reaction--do not occur
to a significant extent under the specified conditions
First Law of Thermodynamics
The internal energy of an isolated system is constant.
The total amount of energy in the universe is constant.
Some Thermodynamic Terms
System--the substances involved in the chemical and physical changes under investigation
Surroundings--the rest of the universe
Universe--the system and its surroundings
Types of Thermodynamic Systems
Open system--can exchange both matter and energy with its surroundings
Closed system--has a fixed amount of matter but can exchange energy with its surroundings
Isolated system--has no contact with its surroundings
Thermodynamic State of a System
defined by a set of conditions that completely specifies all the properties of the system
State functions--the properties of a system( pressure, temperature, energy, e.g.) whose values depend only on the state of the system
Changes in Internal Energy,E
Internal energy represents the total energy of a system.
E = q(heat flow) + w(work) Work is usually defined as PV If the work term is 0 (no work done)
then at constant volume E = q
Limitations of the First Law of Thermodynamics
E = q + w
Euniverse = Esystem + Esurroundings
Esystem = -Esurroundings
The total energy-mass of the universe is constant.
However, this does not tell us anything about the direction of change in the universe.
Esystem + Esurroundings = 0 = Euniverse
Enthalpy
The change in enthalpy () is measured at constant P.
At constant P: H = q
Figure 20.1 A spontaneous endothermic chemical reaction.
water
Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l).
H0rxn = +62.3 kJ
Enthalpy Change
Hfo (products) -
Hfo(reactants)
where Hfo is the standard molar
enthalpy of formation and H is the enthalpy change for the reaction.
Enthalpy Change
Calculate the enthalpy change for the following reaction at 298K.
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(l)
The Second Law of Thermodynamics
In spontaneous changes the universe tends toward a state of greater disorder.
In thermodynamics, entropy is a measure of the degree of disorder.
Entropy tends to increase.
The Second Law of Thermodynamics
Likely
The Second Law of Thermodynamics
Unlikely
The Concept of Entropy (S)
Entropy refers to the state of order.
A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.
solid liquid gasmore order less order
crystal + liquid ions in solution
more order less order
more order less order
crystal + crystal gases + ions in solution
Entropy
Entropy can be indirectly measured. Absolute standard molar entropy
values can be found in the textbook. An increase in entropy corresponds to
an increase in disorder. When S is _______, disorder
increases. When S is _______, disorder
decreases.
Entropy
The Third Law of Thermodynamics states that the entropy of a pure,perfect,crystalline substance at 0K is zero.
The following relationship applies to entropy changes.
S = So(products) - So(reactants)
Figure 20.4 Random motion in a crystal
The third law of thermodynamics.
A perfect crystal has zero entropy at a temperature of absolute zero.
Ssystem = 0 at 0 K
Changes in Entropy
Calculate the entropy change for the following reaction at 298K. Indicate whether disorder increases or decreases.
2NO2(g) -----> N2O4(g)
Free Energy Change, G
If heat is released in a chemical reaction, some of the heat may be converted to useful work. Some of it may be used to increase the order of the universe. If the system becomes more disordered, then more energy becomes available than indicated by enthalpy alone.
The Gibbs Free Energy Change
At constant T and P G = H - TS When G is > 0, the reaction is
nonspontaneous When G is = 0, the reaction is at
equilibrium When G is < 0, the reaction is
spontaneous
Gibbs Free Energy Change
The following relationship exists for standard molar Gibbs free energy Gibbs Free Energy Change changes:
Go = Gfo(products) -
Gfo(reactants)
Gibbs Free Energy Change
Calculate the Gibbs free energy change for the following reaction at 298K. Indicate whether the reaction is spontaneous or nonspontaneous under these conditions.
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(l)
Table 20.1 Reaction Spontaneity and the Signs of H0, S0, and G0
H0 S0 -TS0 G0 Description
- + - -
+ - + +
+ + - + or -
- - + + or -
Spontaneous at all T
Nonspontaneous at all T
Spontaneous at higher T;nonspontaneous at lower T
Spontaneous at lower T;nonspontaneous at higher T
The Gibbs Helmholtz Equation Calculate So for the following reaction at
298K. C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O(l) From previous examples we found Ho = -2219kJ and Go = -2107kJ Indicate whether disorder increases or
decreases
The Relationship Between Go and the Equilibrium Constant
The standard free energy change for a reaction is Go. This is the free energy change that would accompany the complete conversion of all reactants, initially present in their standard states, to all products in their standard states. G is the free energy change for other concentrations and pressures.
The Relationship Between Go and the Equilibrium Constant
The relationship between G and Go is
G = Go + RTlnQ where R = universal gas
constant(8.314J/moleK) T = temperature in K Q = reaction quotient
The Relationship Between Go and the Equilibrium Constant
When a system is at equilibrium, G = 0 and Q = K. Then: 0 = Go + RTlnK Rearranging gives Go = -RTlnK
FO
RW
AR
D R
EA
CT
ION
RE
VE
RS
E R
EA
CT
ION
Table 20.2 The Relationship Between G0 and K at 250C
G0(kJ) K Significance
200
100
50
10
1
0
-1
-10
-50
-100
-200
9x10-36
3x10-18
2x10-9
2x10-2
7x10-1
1
1.5
5x101
6x108
3x1017
1x1035
Essentially no forward reaction; reverse reaction goes to completion
Forward and reverse reactions proceed to same extent
Forward reaction goes to completion; essentially no reverse reaction
The Relationship Between Go and the Equilibrium Constant
Calculate the value for the equilibrium constant, Kp, for the following reaction at 298K.
N2O4(g) = 2NO2(g) At 25oC and 1.00 atmosphere
pressure, Kp=4.3x10-13, for the decomposition of NO2. Calculate Go at 25oC.