Transcript Thermodynamics - Grand Junction High School

Thermodynamics
SPONTANEITY, ENTROPY, AND FREE ENERGY

ENTHALPY: ΔH is heat exchange—exothermic reactions are
generally favored.

ENTROPY: ΔS is the dispersal (disorder) of the matter and energy of
a system.

Thermodynamically favored processes are those that involve both a
decrease in the internal energy of the components (ΔHo < 0) and
an increase in entropy of the components (Δso >0). These processes
are thermodynamically favored if ΔGo <0 or negative.

Thermodynamically favored reactions occur without outside
intervention once the activation energy has been reached. Some
are slow like diamond turning to graphite and some are fast like the
combustion of methane.
First Law of Thermodynamics

Answers questions like:
 How
much energy is involved in the change?
 Does
energy flow into or out of the system?
 What
form does the energy finally assume?
 It
does not explain why a process occurs in a
given direction.

Thermodynamics lets us predict whether a process will occur but
gives no information about the amount of time required for the
process.

For example, according to the law of thermodynamics, a diamond
should change spontaneously into graphite. The fact that we don’t
see it happen doesn’t mean it won’t, its just very slow.

Consider the following:

A ball rolls down a hill but never spontaneously rolls back up the hill.

Wood burns spontaneously in an exothermic reaction to form CO2 and
H2O but wood is not formed from those two molecules when they are
heated together.

At temperatures below 0oC, water spontaneously freezes, and at
temperatures above 0oC ice spontaneously melts.

What thermodynamic principle will provide an explanation of why,
under a given set of conditions, each of these diverse processes occurs
in one direction but never the reverse?
Spontaneous Processes and
Entropy

A process is said to be spontaneous if it occurs without outside
intervention. Spontaneous processes may be fast or slow, they are
not instantaneous.

Thermodynamics can tell us the direction in which a process will
occur, but it can say nothing about the speed at which the process
occurs—that’s the job of kinetics.

Entropy can be viewed as a measure of molecular randomness or
disorder.

Entropy is a thermodynamic function that describes the number of
arrangements (positions and/or energy levels) that are available to
a system existing in a given state.
Entropy
and the Second Law of
Thermodynamics
(13:41)

The second law of thermodynamics is that in any spontaneous
process there is always an increase in the entropy of the universe.

OR: The entropy of the universe is increasing.

The change in the entropy of the universe can be expressed as

ΔS universe = ΔS system + ΔS surroundings

If the sign of ΔS universe is positive, the entropy of the universe
increases and the process is spontaneous in the direction written.

If ΔS universe is negative, the process is spontaneous in the opposite
direction.

If ΔS universe is 0, then the process has no tendency to occur and the
system is at equilibrium.

Consider the set of flasks above. In the first flask there are two gas
molecules confined to the right side. If the stopcock is opened,
there are then 4 possible places for the molecules to be, each
equally probable. The probability that one molecule being in the
right side is ½ while the probability that two molecules would be in
the right half would be ½ 2 = ¼ Three gas molecules would be ½ 3 =
1/8 etc.

If you were to consider a mole of gas molecules being in the righthand flask at the same time it would be1/2n where n = 6.022 x 1023 .

This is such a small number that essentially there is zero likelihood
that all the gas molecules will be found in the right-hand flask at the
same time. Not entirely impossible, but not probable. This is called
positional probability.
Positional entropy

For each of the following pairs, choose the substance with the
higher positional entropy (per mole) at a given temperature.

A. Solid CO2 and gaseous CO2

B. N2 gas at 1 atm and N2 gas at 1.0 x 10-2 atm

A. Since a mole of gaseous CO2 has the greater volume by far, the
molecules have many more available positions than in a mole of
solid CO2. Thus gaseous CO2 has a higher positional entropy.

B. A mole of N2 gas at 1 x 10-2atm has a volume 100 times that of a
mole of N2 gas at 1 atm. Thus N2 gas at 1x 10-2 atm has the higher
positional entropy. (Greater volume means more available
positions)
Predicting Entropy Changes

Predict the sign of the entropy change for each of the following processes:

A. Solid sugar is added to water to form a solution.

B. Iodine vapor condenses on a cold surface to form crystals.

A. The sugar molecules become randomly dispersed in the water when the
solution forms and thus have access to a larger volume and a larger number of
possible positions. The positional disorder is increased, and there will be an
increase in entropy. ΔS is positive, since the final state has a larger entropy than
the initial state. ΔS = Sfinal – Sinitial

B. Gaseous iodine is forming a solid. This process involves a change from a
relatively large volume to a much smaller volume, which results in lower positional
disorder. For this process, ΔS is negative (the entropy decreases).

In general, the greater the number of arrangements, the higher the entropy of
the system.
Temperature and Spontaneity

ΔS universe = ΔS system + ΔS surroundings

To determine the sign of ΔS universe consider this process:

H2O(l)  H2O(g)

A mole of liquid water has a volume of about 18 mL while a mole of water vapor at 1 atm and
100oC has a volume of about 31L. Therefore in this process the entropy of the system increases
and ΔS system increases.

The entropy of the surroundings are determined by the flow of energy in or out of the system. An
exothermic reaction releases energy into the surroundings, increasing the K.E. of the surrounding
atoms and increasing entropy. Thus ΔS surr is positive.

Endothermic reactions decrease entropy of the surroundings and ΔS surr is negative.

The sign of ΔS universe determines if the vaporization of water is spontaneous.

In the water to gas scenario, the ΔS surr is negative because the water absorbs heat to turn to
vapor but the ΔS system is positive.

What determines the ΔS universe is temperature. The impact of the transfer of a given quantity of
energy as heat to or from the surroundings will be greater at lower temperatures.

Driving force provided by energy flow (heat) = magnitude of the
entropy change of the surroundings = quantity of heat (J)
temperature (K)
Exothermic process: + quantity of heat (J) ÷ temperature (K)
Endothermic process: - quantity of heat (J) ÷ temperature (K)
ΔSsurr = -ΔH ÷ T

In the metallurgy of antimony, the pure metal is recovered via different reactions, depending
on the composition of the ore.

Iron is used to reduce antimony in sulfide ores:

Sb2S3(s) + 3Fe(s)  2Sb(s) + 3FeS(s)

Carbon is used as the reducing agent for oxide ores:

Sb4O6(s) + 6C(s)  4Sb(s) + 6CO(g)

Calculate ΔS for each of these reactions at 25oC and 1 atm.
ΔH = -125 kJ
ΔH = 778 kJ
ΔSsurr = -ΔH ÷ T

ΔSsurr = - (-125kJ ÷ 298K )= 0.419 kJ/K = 419 J/K (+ because its exothermic)

ΔSsurr = - (778kJ ÷ 298K ) = - 2.61 kJ/K = -2610 J/K ( - because its endothermic)

ΔSsurr is negative because heat flow occurs from the surroundings to the system.

If both ΔSsurr and ΔSsys are positive, then ΔS univ is positive and the process is
spontaneous.

If both ΔSsurr and ΔSsys are negative, then ΔS univ is negative and the process
does not occur in the direction indicated but is spontaneous in the opposite
direction.

If ΔS surr and ΔS sys have opposite signs, the spontaneity depends on the sizes
of the opposing terms.

Exothermicity is most important as a driving force in low temperatures.
ΔS sys ΔS surr ΔS univ Process spontaneous?
+
+
+
Yes
-
-
-
No (reaction occurs in opposite direction)
+
?
?
Yes, if ΔSsys has a larger magnitude than ΔS surr
-
?
?
Yes, if ΔSsurr has a larger magnitude than ΔS sys
Free Energy (G)

Free energy is another thermodynamic function related to spontaneity and
useful in dealing with temperature dependence of spontaneity.

G = H – TS

ΔG = ΔH – T ΔS
where H = enthalpy, T = K temp, S = entropy
At low temperatures, ΔH dominates
At high temperatures, ΔS dominates.
The calculation of ΔG ultimately decides if a reaction is thermodynamically
favored.
Case
Result
ΔS positive, ΔH negative
Spontaneous at all temperatures
ΔS positive, ΔH positive
Spontaneous at high temperatures (where exothermicity is relatively
unimportant)
ΔS negative, ΔH negative Spontaneous at low temperatures (where exothermicity is dominant)
ΔS negative, ΔH positive
Process not spontaneous at any temperatures (reverse process is
spontaneous at all temperatures)

Big Mamma equation:

ΔGrxn = ∑ ΔG 0(products) - ∑ ΔG 0(reactants)

Grandaddy equation:

ΔG 0 = ΔH 0 - TΔS 0 derives to T = ΔH0 ÷ ΔS0

Hess’s Law of Summation: same rules as you learned before for ΔH0 is used for ΔG 0.

Rat Link equation for calculating ΔG 0 at standard conditions using the given temperature and
equilibrium constant, K: ΔG 0 = -RT ln K where R = 8.1345 J/mol· K

Kp = P reactants ÷ P products

Solving for ΔG 0 using “minus nunfe” equation gien the standard cell potential, Faraday’s
constant and number of moles of electrons involved: ΔG 0 = - nFE0

Faraday’s constant: 96,485 Coulombs/mole e- and E0 is the standard cell potential for the
electrochemical process. Good to know that 1volt = 1 joule/coulomb
still raised to power of coefficients.
2H2O(l) + O2(g)  H2O2(l)

Calculate the free energy of formation for the oxidation of water to
produce hydrogen peroxide given the following information ΔG 0f
values:

H2O(l)
- 56.7 kcal/molrxn

O2(g)
0 kcal/molrxn

H2O2(l)

2(-27.2) – (2 x -56.7) = 59 kcal/mol
- 27.2 kcal/molrxn
2SO2(g) + O2(g)  2SO2(g)

Calculate the ΔH0, ΔS0, and ΔG0 using the following data at 250C and 1 atm.
ΔH0 = products – reactants
ΔS0 = products - reactants
ΔG0 = ΔH0 - T ΔS0
Both ΔG0 and ΔH0 = 0 for elements in their standard state.
Notice that S is not ΔS0 that’s because only a perfect diamond at absolute zero
will have an S value of 0. Now you know…

ΔH0 = -792 – (-594) = -198 kJ/mol
Substance
ΔH0 (kJ/mol)
S (J/K· mol)
SO2(g)
- 297
248
SO3(g)
- 396
257
O2(g)

ΔS0 = 2(257) – [ 2(248) + 205] = -187 J/K· mol

ΔG0 = -198kJ/mol – (298K)(-187J/K·mol x 1kg/1000 J) = -142 kJ/mol
0
205
Free energy and spontaneity
At what temperatures is the following process spontaneous at 1 atm?
 Br2(l)  Br2(g)
ΔSo = 93.0 J/K· mol , ΔHo = 31.0 kJ/mol

The vaporization process will be spontaneous where ΔGo is negative. ΔSo favors the
vaporization process because of the increase in positional entropy, ΔH favors the opposite
process, which is exothermic. These opposite tendencies will exactly balance at the boiling point
of Br2 since at this temperature liquid and gaseous bromine are in equilibrium and ΔGo = 0
Thus: 0 = ΔHo – T ΔSo
 ΔHo = T ΔSo Solve for T
 3.0 x 104J/mol ÷ 93.0 J/K· mol = 333K
Above 333K, the term ΔS controls. The increase in entropy when liquid Br2 is vaporized is dominant.

Below 333K, the term ΔH controls because the process is spontaneous in the direction in which it is
exothermic.
At 333K, and ΔG=0, the opposing driving forces are balanced (liquid and gaseous bromine coexist)
and this is the normal b.p.
Predicting Entropy Changes

Predict the sign of the entropy change (ΔS) ro each of the following
processes. Justify your answers.

Solid sugar is added to water to form a solution.

Positive

Iodine vapor condenses on a cold surface to form crystals.

Negative
Prediction

Predict the sign of ΔSo for each of the following reactions:

CaCO3(S)  CaO(S) + CO2(g)

2SO2(g) + O2(g)  2SO3(g)

Since in this reaction a gas is produced from a solid reactant, the
positional entropy increases, and ΔSo positive.

Here three molecules of gaseous reactants become two molecules
of gaseous products. Since the number of gas molecules
decreases, positional entropy decreases, and ΔSo is negative.
Third law of thermodynamics

The entropy of a perfect crystal at 0 K is zero. This is an unattainable ideal
that is taken as a standard but never actually observed.

With increased temperature the random vibrations increase and the entropy
also increases.

Because entropy is a state function of the system (not pathway dependent)
the entropy change for a given chemical reaction can be calculated with:

Δsoreaction = ∑npΔS0products - ∑nrΔS0reactants

Note that entropy is an extensive property and depends on the amount of
substance present. (number of moles)

ΔS is + when dispersal/disorder increases (favored)

ΔS is – when dispersal/disorder decreases
2NiS(s) + 3O2 (g)  2SO2(g) + 2NiO(s)

Calculate ΔS0 at 25o C for the above
reaction given the following entropy
values:
Substance
So (J/K · mol)
SO2(g)
248
NiO(s)
38
O2(g)
205
NiS(s)
53

Δsoreaction = ∑npΔS0products - ∑nrΔS0reactants

2S0 SO2(g) + 2S0 NiO(s) - 2S0 NiS(s) - 3S0 O2(g)

2 mol (248 J/K· mol) + 2 mol (38 J/K· mol)

-2mol (53 J/K· mol) - 3 mol (205 J/K· mol)

= 496 J/K + 76 J/K – 106 J/K – 615 J/K

= - 149 J/K

We would expect ΔSo to be negative because
the number (mols) of gaseous molecules
decreases in this reaction.
Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)


Calculate ΔS0 at 25o C for the
above reaction given the
following entropy values:
179 J/K
Substance
S0 (J/K · mol)
Al2O3(s)
51
H2(g)
131
Al(s)
28
H2O(g)
189
2SO(g) + O2(g) --> 2SO3(g)

Calculate the entropy change at
250C, in J/molrxn · K (not kJ!)

Given the following data:

SO2(g)
248.1 J/molrxn · K

O2(g)
205.3 J/molrxn · K

SO3(g)
256.6 J/molrxn · K

2(256.6) – [2(248.4) + 205.3] =

- 188.3 J/molrxn · K
Entropy changes as they relate to
reversible phase changes

ΔSsurroundings = heat transferred ÷ temperature at which change
occurs.
q/T = -ΔH/T expressed in J/molrxn· K
Where the heat supplied (endothermic) q > 0
Where heat is evolved (exothermic) q < 0
It is important to note if the reaction is endothermic or exothermic.
The temperature at which the process occurs is significant.
Example: water (liquid at 1000C) ↔ water (gas at 1000C)
The entropy will increase for the forward reaction (vaporizing) since the
reaction produces water in a less condensed state, thus the molecules
are more dispersed.
0
ΔH
Calculating
,
ΔG = ΔH – T ΔS


Calculate ΔG for the following reaction:
C(S) + O2(g)  CO2(g)
ΔH = - 393.5kJ /mol
ΔS= 3.05 J/K· mol
Temperature is held constant at 298K
ΔG0
=
ΔH0 - T ΔS0
ΔG0 = -393.5 kJ/mol – (298K)(3.05 x 10-3 kJ/K·
mol
- 394.4 kJ/mol of CO2
0
ΔG
,
0
ΔS and
Keq

Calculate ΔG, ΔH and ΔS for:

2SO2(g) + O2(g))  2SO3(g) at 1 atm
and constant temperature of 298K
Substance
ΔHf0 (kJ/mol) S0 (J/K· mol)
SO2(g)
--297
248.1
SO3(g)
--196
256.6
O2(g)
0
205.3
Find ΔH

ΔHoreaction = ∑npΔH0products - ∑nrΔH0reactants

ΔHf oreaction = 2ΔHf 0SO3 - 2ΔHf 0SO2 + ΔHf 0O2

= 2 mol (-396 kJ/mol) – 2 mol (-297 kJ/mol) - 0

= -792 kJ + 594 kJ

= -198 kJ
Find
0
ΔS and
0
ΔG

ΔSoreaction = ∑npΔS0products - ∑nrΔS0reactants

= 2 mol ( 256.6 J/K· mol) – 2 mol (248.1 J/K· mol) – 1 mol(205.3 J/K · mol)

= 513 J/K mol – (496.2 J/K mol+205.3 J/Kmol)

= -188 J/K/mol

You would expect ΔS0 to be negative because 3 molecules of gaseous reactants
give 2 molecules of gaseous products.

ΔG0 = ΔH0 - T ΔS0

= - 198kJ/mol – (298K) ( -187 J/K mol )(1kJ/1000J)

= - 198 kJ/mol + 55.7 kJ/mol = - 142 kJ/mol
Find Keq

ΔG0 = -RT ln K

ln K = ΔG0 ÷ - RT

lnK = -142 kJ/mol x 1000 J/kJ ÷ (-8.3145 J/mol· K)(298)

lnK = 57.3107303

7.76 x 1024 – huge number!
Calculate ΔG for the reaction
Cdiamond(s)  Cgraphite(s)

Cdiamond(s) + O2(g)  CO2(g) ΔG = -397 kJ

Cgraphite(s) + O2(g)  CO2(g) ΔG = -394 kJ

Reverse the last equation and change the sign.

CO2(g)  Cgraphite(s) + O2(g) ΔG = +394 kJ

Add them together (summative ΔG) = 394 kJ + (-397) = 3 kJ
4Fe(s) + 3O2(g)  2Fe2O3(s)
Use the data to the right to calculate
the equilibrium constant for the
above reaction at 25oC.
ΔG = ΔH - T ΔS
ΔH = 2(-826) = -1652 kJ/mol
Substance
ΔH0f (kJ/mol)
So (J/K · mol)
Fe2O3(s)
-826
90
Fe(s)
0
27
O2(g)
0
205
ΔS = 180 – [2(2.7) + 3(205)] = - 5.43 x 10-1 kJ/mol
ΔG = -1652kJ/mol – (298K)(-5.43 x 10-1 kJ/mol · K)= -1.49 x 103 kJ/mol
lnKeq = - (- 1.47 x 10-3 kJ/mol ÷ (8.3145 x 10-3 kJ/mol · K) = 601
K = e601 = 10261
H2O(l)  H2O(g)

Calculate the thermodynamic boiling point of water forming a gas
from a liquid.

ΔH = + 44 kJ/mol

ΔG = 0 at equilibrium so ΔH = T ΔS

ΔH = 44kJ/mol ÷ .1188 kJ/mol · K = 370 K
ΔS = 118.8 kJ/K· mol
2CO(g) + O2(g)  2CO2(g)

ΔG is – 2572 kJ/mol. Calculate the equilibrium constant at 25oC.

ΔG = - RT lnKeq

lnKeq = ΔG ÷ -RT

- 2.572 x 10-5 ÷ (-8.3145 J/mol · K)(298K)

1.27 x 1045
Calorimetry 11:47