Transcript Chapter 17 - Richsingiser.com

Daniel L. Reger
Scott R. Goode
David W. Ball
http://academic.cengage.com/chemistry/reger
Chapter 17
Chemical Thermodynamics
Chemical Thermodynamics
• Chemical thermodynamics is the
study of the energetics of chemical
reactions.
Chapter 5 Review
• The system is the part of the
universe under examination (for us, it
would be a chemical reaction or other
process).
• The surroundings are the rest of the
universe.
• The change in enthalpy (DH) is the
heat absorbed or released by the
system at constant temperature and
pressure.
Work and Heat
• Under normal laboratory conditions, a
reaction (the system) usually exchanges
energy with the surroundings in two
forms, as heat and as work.
• Work is directed energy, and is associated
with moving a part of the system.
• Heat is random energy, and is associated
with the temperature of the system.
Work
• Work is an
application of a
force through a
distance:
work = force × distance.
• Work is also an
application of a
pressure causing a
volume change:
work = -PDV.
Pressure-Volume Work
• Express the work (in joules) when 20.0 L
of an ideal gas at a pressure of 12.0 atm
expands against a constant pressure of
1.50 atm. Assume constant temperature.
First Law of Thermodynamics
• First law of thermodynamics is the
law of conservation of energy; energy
can neither be created nor destroyed.
• Internal energy, E, represents the total
energy of the system and is a state
function (something that depends only
on the state of the system, not how the
system got to that state).
Internal Energy, E
• When a change occurs in a closed
system, the change in internal energy, DE,
is given by
DE = q + w
• The sign convention for q and w are that they
are positive if they transfer energy to the
system from the surroundings and negative if
they transfer energy from the system to the
surroundings.
Calculating ΔE, w, and q
A 7.56g sample of gas in a balloon that has a volume of 10.5 L.
Under an external pressure of 1.05 atm, the balloon expands to
a volume of 15.00 L. Then the gas is heated from 0.0 °C to 25.0
°C. If the specific heat of the gas is 0.909 J/g*°C. Calculate the
work, heat, and ΔE for the overall process.
Energy and Enthalpy
• Data about heats of reaction are
tabulated as standard enthalpies of
formation (DHf), as described in
Chapter 5.
• The equation that relates DH to DE is
DH = DE + PDV
• This equation is used to calculate the
enthalpy of reaction from heats measured
using constant-volume calorimetry.
Energy, Enthalpy, and PV Work
• The difference between changes in
enthalpy and internal energy is PV work
and is significant only for reactions that
involve gases.
• From the ideal gas law, PV = nRT, so
DH = DE + DnRT
Example: DH from DE
• Calculate DH at 25C and 1.00 atm
pressure for the following reaction.
2NO2(g) → N2O4(g) DE = -54.7 kJ
Randomness
• An increase in randomness (sometimes
referred to as disorder) is an important
driving force for many changes.
• Two different gases, initially separated by a
partition, will mix with the partition is
removed, increasing the randomness of the
system.
Entropy and Spontaneity
• Thermodynamics is able to relate
spontaneity to a state function called
entropy.
• Entropy (S) is the thermodynamic
state function that describes the
amount of randomness.
• A large value for entropy means a high
degree of randomness.
Entropy Change
• An increase in randomness results in an
increase in entropy. Some general
guides are:
• the entropy of a substance increase when solid
becomes liquid, and when liquid becomes gas.
• the entropy generally increases when a solute
dissolves.
• the entropy decreases when a gas dissolves in
a solvent.
• the entropy increases as temperature
increases.
Entropy
• The entropy of a system generally
increases when a molecular solid
dissolves in a liquid.
Second Law of Thermodynamics
• The second law of thermodynamics
states that in any spontaneous process,
the entropy of the universe increases.
• DSuniv = DSsys + DSsurr > 0
• If DSsys < 0 for a spontaneous process, then
a larger positive change in DSsurr must
occur.
Spontaneity and DSuniv
• When DSuniv > 0, the change occurs
spontaneously.
• When DSuniv < 0, the reverse change
occurs spontaneously.
• When DSuniv = 0, the change is not
spontaneous in either direction (the
process is at equilibrium).
Third Law of Thermodynamics
• The third law of thermodynamics states
that the entropy of a perfect crystal of a
substance at absolute zero is equal to 0.
• There is a minimum randomness in a perfect
crystal at 0 K.
• Unlike enthalpy and internal energy,
absolute values of entropy can be
determined.
Units for Entropy
• As heat is added to a perfect crystal at
0 K, the temperature rises and
randomness begins to increase.
• The change in entropy depends on the
amount of heat and on the
temperature, and is given by the
equation
DS = q/T
• Therefore, entropy has units of J/K.
Absolute Entropies
• In Appendix G, absolute entropies are given
for substances in their standard state at 298
K.
• The entropy change for a reaction is
DSrxn = SnS[prods] - SmS[reacts]
where n and m are the coefficients of the
products and reactants in the reaction.
• S for the free elements in their standard
states is not zero.
Example: Calculate DSrxn
• Calculate the standard entropy change
for the reaction
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
Substance
S, J/mol·K
C2H6(g)
229.49
CO2(g)
213.63
O2(g)
205.03
H2O(l)
69.91
Test Your Skill
• The combustion of ethane is
spontaneous, but DS is -620.21 J/K.
Explain why this does not violate the
second law of thermodynamics.
Free Energy and DSuniv
• J. W. Gibbs defined a state function
called the Gibbs free energy, G:
G = H – TS
• At constant temperature and pressure, this
becomes
• DG = DH – TDS
• DG is negative for a spontaneous change
that occurs at constant temperature and
pressure.
Free Energy and Spontaneity
• DG is a state function of the system.
• For any spontaneous change, DG < 0.
Spontaneous
Reaction
∆Suniv
∆Gsys
Forward reaction
positive (+)
negative (-)
At equilibrium
0
0
Reverse reaction
negative (-)
positive (+)
DG and Spontaneity
• From the equation DG = DH – TDS, a
negative DH and a positive DS favor
spontaneity.
• A minimum free energy occurs when the
system is at equilibrium, and DG = 0.
Standard Free Energy of Formation
• The standard free energy of formation is
the free energy change to form one mole
of a compound from its elements in their
standard states.
• D G f  D H f  T D S f , where DSf must be
calculated from absolute entropies.
• DGrxn = SnGf[prods] - SmGf[reacts]
where n and m are the coefficients of the
products and reactants in the reaction.
Temperature Dependence of DG
• The dependence of DG on temperature
arises mainly from the TDS term in the
definition of DG.
• The sign of DG is dominated by the sign of
DH at low temperatures and by the sign of
DS at high temperatures.
• Negative values of DH and positive values of
DS favor spontaneity.
Direction of Spontaneous Reaction
∆H
∆S
Temperature
∆G
-
+
-
-
+
+
+
-
All
Low
High
Low
High
All
+
+
+
Spontaneous
Direction
Forward
Forward
Reverse
Reverse
Forward
Reverse
Example: Temperature Dependence
• Assuming that DH and DS do not
change with temperature, calculate the
temperature for which DG is 0 for the
reaction
CS2(l) ⇌ CS2(g)
At 298 K, DH = 27.66 kJ and DS =
86.39 J/K.
Example: Calculate DG
• Given the following chemical reaction and
data at 298 K:
N2(g) + 3H2(g)  2NH3(g)
DHfo = 0
0
-46.1 kJ/mol
So = 191.5
130.6
192.3 J/mol.K
assuming that DHo and DSo do not change
with temperature, calculate DGo at 1000
K.
Concentration and Free Energy
• Concentrations of reactants and
products influence the free energy
change of a reaction according to the
equation
DG = DG + RTlnQ
where Q is the reaction quotient (see
Chapter 14).
Example: Concentration Dependence
• Substance
NO2(g)
N2O4(g)
DGfo kJ/mol
51.29
97.82
For the reaction
2NO2(g) ⇌ N2O4(g)
(a) calculate DGo at 298 K.
(b) calculate DG when PNO2 = 0.12 atm
and PN2O4 = 0.98 atm.
Free Energy and Keq
• For a system at equilibrium, DG = 0, and
Q = Keq, so
DGo = -RTln Keq
• DGo, calculated from the data in
Appendix G, can be used to calculate
the value of the equilibrium constant.
Example: Keq Calculation
• Evaluate the equilibrium constant at 298
K for
2NO(g) + Br2(g) ⇌ 2NOBr(g)
using the standard free energies of
formation at 298 K given below.
Substance
NO(g)
Br2(g)
NOBr(g)
∆Gf° (kJ/mol)
86.55
3.14
82.4
Temperature and Keq
• The temperature dependence of Keq is
derived from two equations given
earlier:
DH - TDS = DG = - RTln Keq
A graph of ln Keq vs. 1/T gives a line
with a slope = -DH/R and an intercept
of DS/R.
Useful Work
• The change in free energy is the maximum
work that can be performed by a
spontaneous chemical reaction at constant
temperature and pressure:
wmax = DG
• This is a limit imposed by nature, as we strive to
improve the efficiency of energy conversions
• When DG > 0 (spontaneous in the reverse
direction), it represents the minimum work
that must be provided to cause the change.