Download Thermodynamics (Part 2)

Thermodynamics
(Part 2)
Chapter 19
Questions to be answered...
Will a given reaction occur “by itself” at a particular
temperature and pressure without the exertion of any
outside force?
Why does a particular process occur in a given direction?
Spontaneous Processes
occurs without outside intervention
-sometimes may require a stimulus to get started, but continues
itself without further input of energy
ex: combustion, steel rusts when exposed to moisture & air,
water freezes below 0C, water boils above 100C, a gas fills a container
uniformly
-moves the system towards equilibrium
-can be fast or slow
-thermodynamics tells us if a reaction occurs; kinetics
explains speed
Nonspontaneous: do not occur at all, or require a constant supply of energy
ex: electrolysis of water
Spontaneous Processes
Predict whether each process is spontaneous as described, spontaneous in the
reverse reaction or in equilibrium
a. water at 40C gets hotter when a piece of metal heated to 150C is added
b. water at room temperature decomposes into H2(g) and O2(g)
c. benzene vapor, C6H6(g) at a pressure of 1atm condenses to liquid
benzene at the normal boiling point of benzene (80.1C)
Why are some reactions spontaneous?
At first, it was thought that reactions are spontaneous because they
are exothermic
-exothermic: energy is lost to surroundings and dH is
negative
-it is true that almost all exothermic reactions are
spontaneous at 298K and 1atm
What about reactions that are endothermic & nonspontaneous at one
temp, but are spont. when temp is raised? ex: ice melting at temps
above 0C
-there must be another factor besides the loss of energy to
predict the direction of spontaneity
Entropy - The Randomness Factor
Nature tends to move from a more ordered to a more
random state
-tendency to move towards states that have higher
probabilities of existing
once the stopcock is opened, the bottom picture has
a higher probability of occurring because there are
more positions available
Entropy and Microstates
Example: what are the possible outcomes of a toss of 4
coins?
What are the possible combinations to achieve the
macrostates?
Which macrostate is most probable?
Disorder is more probable than order because there are so
many more ways of achieving it.
Entropy and Motion
3 types of motion
1. translation: movement from one location to another
(all particles: monatomic and polyatomic)
2. rotation: polyatomic molecules only (although linear
molecules can only have 2 types of rotation; others
have 3)
3. vibrations: atoms periodically move toward and away
from eachother (polyatomic: increases as molecules
become more complicated/more atoms)
Putting it Together
The number of possible microstates increases with:
1. an increase in volume
2. an increase in temperature
3. an increase in the number of molecules
4. an increase in the complexity of the molecule
Increase in entropy is driven by the tendency of thermal
energy to spread into as many microstates of the system
and surroundings as possible
Molecular Motions
In which phase are water molecules least able to have rotational motion?
Entropy
Entropy is a state function
-depends only by the final and initial states, not the pathway
to those states (enthalpy is also a state function)
ΔS = Sfinal - Sinitial
+ΔS = more entropy = more random
-ΔS = less entropy = less random
Ssolids<Sliquids<Saqueous<Sgases
-generally, the higher the temperature, the greater the
entropy
-particles move around more freely
-more gas molecules = more entropy
Predicting the Sign of
ΔS
Predict whether ΔS is positive or negative, assuming each
occurs at constant temperature
a. H2O(l) → H2O(g)
b. Ag+(aq) + Cl-(aq) → AgCl(s)
c. 4Fe(s) + 3O2(g) → 2Fe2O3(s)
d. N2(g) + O2(g) → 2NO(g)
Predicting Entropy
Indicate whether each process produces an increase or
decrease in the entropy of the system.
a. CO2(s) → CO2(g)
b. CaO(s) + CO2(g) → CaCO3(s)
c. HCl(g) + NH3(g) → NH4Cl(s)
d. 2SO2(g) + O2(g) → 2SO3(g)
Predicting Entropy
In each pair, choose the system that has greater entropy and explain your
choice:
a. 1 mol of NaCl(s) or 1 mol of HCl(g)
b. 2 mol of HCl(g) or 1 mol of HCl(g) at 25C
c. 1 mol of HCl(g) or 1 mol of Ar(g) at 298K
Entropy & the 2nd Law of Thermodynamics
2nd Law of Thermodynamics: in any spontaneous process,
there is always an increase in the entropy of the universe
ΔSuniverse = (ΔSsystem + ΔSsurroundings)
ΔS > 0 = spontaneous
-entropy changes in surroundings are primarily
determined by the direction of heat flow
-exothermic: increase in the entropy of the surroundings
-endothermic: decrease in entropy of surroundings,
possible increase in entropy of the system
Calculating ΔS
-can assign absolute entropy values
3rd Law of Thermodynamics: the entropy of a
perfect crystal at 0K is zero (and has only one microstate)
-entropy values are listed in Appendix C
-elements have nonzero entropies
-standard molar entropies of pure substances
are always greater than zero
Calculating ΔS for Reactions
Standard entropy change for a reaction:
ΔS° = ∑S°products - ∑S°reactants
-entropy is an extensive property and depends on
the number of moles so coefficients must be used
Generally…
-a reaction that results in an increase in the number of
moles of a gas is accompanied by an increase in entropy
-the more complex a molecule, the higher the entropy
Calculating ΔS Practice 1
Calculate the change in the standard entropy of the
system, ΔS°, for the synthesis of ammonia from N2(g) and
H2(g) at 298K
Calculating ΔS Practice 2
Calculate the standard entropy change, ΔS° for the
following reaction at 298K:
Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)
Gibbs Free Energy
relates the enthalpy (H) and entropy (S) of a system
-for a reaction taking place at constant pressure and
temperature, ΔG represents the portion of the total energy change
that is free to do useful work
ΔG° = ΔH° - TΔS°
ΔG° < 0 reaction is spontaneous
ΔG° > 0 reaction is not spontaneous (reverse reaction is)
ΔG° = 0 reaction is at equilibrium (no tendency for the reaction to
occur in either direction)
Concept Comparison
ΔH, change in enthalpy
-a negative value tends to make a reaction spontaneous (maybe)
-exothermic
ΔS, change in entropy
-a positive value tends to make a reaction spontaneous (maybe)
-more disorder or randomness
ΔG, change in free energy
-a reaction at constant temperature and pressure will be
spontaneous if ΔG is negative (ALWAYS)
-decrease the free energy of a system
More on Free Energy
ΔG can be used to determine:
-the effect of temperature on reaction spontaneity
-the effect of pressure on reaction spontaneity
-the effect of concentration on reaction spontaneity
-ΔG can be used to calculate the equilibrium constant for a reaction
-when comparing reactions, the more negative ΔG, the further the
reaction will go to the right to reach equil.
-ΔG can be used to determine whether coupled reactions are
spontaneous (Hess’s Law)
Calculating ΔG° - Method 1
Determine ΔH and ΔS using reference tables, then use the Gibbs
equation to find ΔG
Example: Calculate ΔG° for a reaction for which ΔH° = 24.6kJ and ΔS°
= 132J/K at 298K. Is the reaction spontaneous under these
conditions?
Calculating ΔG° - Method 1
Calculate the standard free-energy change for the formation of NO(g)
from N2(g) and O2(g) at 298K
-given: S°O2 = 205J/molK, S°N2 = 191.5J/molK, S°NO =
210.62J/molK; ΔH°f for O2 and N2 = 0, ΔH°f for NO = 90.37kJ/mol
Calculating ΔG° - Method 2
Use standard free energies of formation from tables in appendix C
ΔG°rxn = ∑mΔG°products - ∑nΔG°reactants
Example: Use ΔGf values to calculate ΔG°rxn for this reaction:
4KClO3(s) → 3KClO4(s) + KCl(s)
Calculating ΔG° - Method 2
Use ΔG°f values to calculate ΔG°rxn at 298K:
a. 2NO(g) + Cl2(g) → 2NOCl(g)
b. 3H2(g) + Fe2O3(s) → 2Fe(s) + 3H2O(g)
Calculating ΔG° - Method 3
Use procedures to find ΔG like those used to find ΔH using
Hess’s Law.
-this is possible because free energy is a state function
remember: reverse equation→ reverse sign; equation
multiplied → multiply ΔG
Free Energy and Pressure
How does pressure affect the thermodynamic functions that
comprise free energy?
-for an ideal gas:
-enthalpy (H) is not pressure dependent
-entropy is because it depends on volume
-Slarge volume > Ssmall volume
SO: Slow pressure > Shigh pressure
-more room to move in large volumes
Free Energy and Pressure
To calculate free-energy at pressures other than 1atm:
ΔG°rxn = ΔG° + RTlnQ
R = gas constant: 8.3145J/Kmol
Q = reaction quotient
Free Energy and Equilibrium
Even though ΔG is negative, the reaction does not go to
completion
-the lowest possible free energy is at the equilibrium
point
At equilibrium ΔG = 0 and Q = K
SO…
ΔG° = -RTlnK
-we will use this equation for reactions at
equilibrium
-it’s in your reference tables
Free Energy and Equilibrium
Qualitative relationship between ΔG°f and the K for a
given reaction
ΔG°f = 0
and K = 0
then: ΔG°fp = ΔG°fr
ΔG°f < 0 and
K >1
system will shift right
ΔG°f > 0
and K < 1
system will shift left
then: ΔG°fp < ΔG°fr
then: ΔG°fp > ΔG°fr