Download Thermodynamics

Thermodynamics
Spontaneous and Non-spontaneous
Processes
 Spontaneity in chemical processes = the direction in
which the reaction proceeds.
 Not speed of reaction (kinetics)
 A reaction can be thermodynamically favored but still be
slow.
Entropy
 Most spontaneous processes are exothermic, but
some are endothermic.
 Melting of ice above 0oC
 Dissolution of sodium chloride in water
 Why?
Entropy (S)
 Informally—the disorder or randomness of a system
 Formally—a measure of the molecular motional energy (plus
any phase change energy) that has been dispersed in a system at
a specific temperature.
 System with highest entropy has the greatest dispersal of energy
 For any spontaneous process, ΔSuniverse > 0
 Units = J/K
Phase changes and Entropy
 Ssolid < Sliquid < Sgas
 In gas, more ways to distribute energy than a solid.
 Energy in solid is due to vibrations between molecules.
 Energy in gas is due to translational, rotational, and
vibrational energy.
Predict the sign for ΔS for each process
and justify your answer
 Melting of ice to water
 Sublimation of carbon dioxide
 H2O(g)  H2O(l)
 2N2O(g)  2N2(g) + O2(g)
Heat Transfer
 Exothermic process increases entropy of
surroundings
 Endothermic process increases entropy of system
 ΔSuniv = ΔSsys + ΔSsurr
 Process is spontaneous as long as entropy of the
universe is positive.
Temperature dependence
 Temperature determines the magnitude of heat
flow to the surroundings.
 Impact of heat flow is greater at lower
temperatures
 Give money to a rich man vs. poor man
Entropy and Temperature
Heat Transfer
Practice
 Calculate the entropy change in the surroundings associated
with the combustion of propane gas occuring at 25oC (ΔHorxn
= -2044 kJ).
 Determine sign of the entropy change for the system.
 Determine the sign of the entropy change for the universe.
Will the reaction be spontaneous?
Practice
 A reaction has ΔHrxn = -107 kJ and a ΔSrxn = 285 J/K. At
what temperature is the change in entropy for the reaction
equal to the change in entropy of the surroundings?
Practice
 Which reaction is most likely to have a positive ΔS? Justify
your reasoning.
a. SiO2(s) + 3C(s)  SiC(s) + 2CO(g)
b. 6CO2(g) + 6H2O(g)  C6H12O6(s) + 6O2(g)
c. CO(g) + Cl2(g)  COCl2(g)
d. 3NO2(g) + H2O(l)  2HNO3(l) + NO(g)
Practice
 Elemental mercury is a liquid at room temperature. Its
normal freezing point is -38.9oC, and the molar enthalpy
of fusion is 2.29 kJ/mol. What is the entropy change of
the system when 50.0 g of Hg(l) freezes at the normal
freezing point?
Calculating Entropy
 Third Law of Thermodynamics
 In a perfect crystal (diamond), the entropy at 0K is 0 J/K.
 Can use third law to develop standard molar entropy values
with spectroscopy techniques.
 Factors that affect standard entropy:
 State of substance
 Molar mass of substance
 If substance is in a particular allotrope
 Molecular complexity
 Extent of dissolution
Practice--Whiteboard
Get into groups of three. On your whiteboard, answer the
following question.
 Arrange these gases in order of increasing standard
molar entropy: SO3, Kr, Cl2. Justify your reasoning.
Calculating ΔSorxn
ΔSorxn = ΣnpSo(products) – ΣnrSo(reactants) (same as ∆H)
 Calculate ΔSorxn for the reaction between ammonia and
oxygen producing nitrogen monoxide and water vapor.
Substance
So (J/mol K)
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8
Practice
 Determine the change in the standard entropy, ΔS⁰, for the
synthesis of carbon dioxide from graphite and oxygen:
C(s,graphite) + O2(g) → CO2(g)
Substance
So (J/mol K)
C(s, graphite)
192.8
O2(g)
205.2
CO2(g)
213.8
Practice
 For the previous example, the change in the standard
entropy, ΔS⁰, for the synthesis of carbon dioxide from
graphite and oxygen, use the previously calculated ΔS⁰sys and
standard enthalpy of formation values to determine S⁰surr and
ΔS⁰universe.
Substance
∆Hof (kJ/mol)
C(s, graphite)
0
O2(g)
0
CO2(g)
-393.5
Gibb’s Free Energy (G)
 A measure of whether or not a process will occur without the
input of outside energy.
 Criterion of spontaneity (thermodynamically favored or
unfavored)
 AKA chemical potential
Entropy and Free Energy
Summary
 ΔG proportional to –ΔSuniv
 ΔG < 0
= spontaneous
 ΔG > 0 = non-spontaneous
 ΔG = ΔH −TΔS
 Units for ΔG are generally kJ
Summary
 ΔG = ΔH −TΔS
 At low temperatures, enthalpy is dominant.
 At high temperatures, entropy is dominant.
∆H
∆S
+
+
−
+
−
−
+
−
∆G
Spontaneous at high temperature
Spontaneous at low temperature
Never spontaneous
Spontaneous at any temperature
Concept Check
 Which statement is true regarding the sublimation of dry ice
(solid CO2)? Justify your reasoning.
ΔH is positive, ΔS is positive, and ΔG is positive at low
temperature and negative at high temperature.
b. ΔH is negative, ΔS is negative, and ΔG is negative at low
temperature and positive at high temperature.
c. ΔH is negative, ΔS is positive, and ΔG is negative at all
temperatures
d. ΔH is positive, ΔS is negative, and ΔG is positive at all
temperatures.
a.
Calculating Standard Change in Free
Energy for a reaction
One of the possible initial steps in the formation of acid rain is the
oxidation of SO2 to SO3 via the following reaction:
SO2(g) + ½ O2(g)  SO3(g)
Calculate the ΔGorxn at 25oC and determine whether the reaction is
spontaneous.
Reactant or Product
ΔHof (kJ/mol)
So (J/mol K)
SO2(g)
-296.8
248.2
O2(g)
0
205.2
SO3(g)
-395.7
256.8
Hess’s law of Summation
Find ∆Horxn for the following reaction:
C + 2H2  CH4
C + O2  CO2
ΔH = -393 kJ
H2 + 1/2O2 H2O
CH4 + 2O2 CO2 + 2H2O
ΔH = -286 kJ
ΔH = -892 kJ
Hess’s Law for ∆G
Bond Energies
 Can utilize bond energies to calculate ∆H.
H = Σ Bond Energies broken – Σ Bond Energies formed
Calculate the change in energy that accompanies the following reaction
given the data below.
H2(g) + F2(g) → 2 HF(g)
Bond Type
H−H
F−F
H−F
Bond Energy
432 kJ/mol
154 kJ/mol
565 kJ/mol
Potential Energy Diagrams
Free Energy under non-standard
conditions
 ΔGrxn = ΔGorxn + R T lnQ
Q = reaction quotient (equilibrium concept)
T = Temperature in K
R = 8.314 J/mol K
For
aA + bB  cC + dD
Q = [C]c[D]d
[A]a[B]b