CHEMISTRY 139 Chemistry: Physical Principles

Thermochemistry:
Chemical Energy
Chapter 8 and 16 (6-th Edition)

Thermodynamics deals with study of energy and its transformations

What is energy? It is the capacity to do work or supply heat
Types of energy:
KINETIC ENERGY:
energy of motion
Mechanical energy
Thermal energy
Electrical energy

POTENTIAL ENERGY
stored energy
Gravitational energy
Electrostatic energy
Chemical potential energy

Units: J = kgm2/s2 ; 1 calorie = 4.184 J

When energy is transferred from one object to another, it appears as work
and/or heat.

Each particle in a system has kinetic and potential energy sum of all these
energies is the internal energy, E, of the system.
2

Q. What is the system?

A. It is the part of the world in which we have a special

interest (all the rest are surroundings)
Examples:

a student

a reaction vessel
a plant

the sun
A simpler presentation

Thermodynamics

Closed System:: Only energy can be lost or gained.

Isolated System: No matter or energy is exchanged.
Open system? Both energy and matter can be flow in and out

Summary:
We define the system as the object of interest. The surroundings is
theoretically everything else.
Changes to the system and surroundings are EQUAL and OPPOSITE.
If system loses energy, surroundings must gain that energy.
Law of Conservation of Energy (the First law of thermodynamics):
energy is conserved (cannot be created or destroyed), the total energy
of the system and surroundings is constant.

Thermodynamic Changes
Change in internal energy

E = Efinal Einitial

E > 0, internal energy increases during a process.

E < 0, internal energy decreases during a process.
5
Magnitude of change is indicated by value of E, direction is indicated by sign.

Thermodynamic Changes:
Energy is transferred as heat or work.

E = q + w

q energy heat exchanged, negative if system heats surroundings.

w energy transferred by doing work, negative if work is done on surroundings

surroundings

Heat transfer in

q>0
Work transfer in

w>0

system
E = q + w

Heat transfer out

q<0
Work transfer out

w<0

The two examples where energy is transferred as heat only

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

The system releases heat

The system absorbs heat

Figure 6.4A

The two cases where energy is transferred

as work only.

Zn(s) + 2H+(aq) + 2Cl-(aq)

H2(g) + Zn2+(aq) + 2Cl-(aq)

The system does work on the surroundings.

8

State Function: property than depends only on the current state of the
system (the path doesnt matter).
The change is zero if the final state is the same as the original
Example: traveling between Toronto and Montreal:
(1) Toronto Montreal,
(2) Toronto Saskatoon Montreal

Change in position is a state function.

Distance travelled, jet fuel and ticket prices are not state functions.
9

E is a state function. E is also a state function.

q and w are not state functions.

State functions? P, V, T

10

Work (P-V work)

Pressure (P) is the force (F) per unit area, A

Force F
Pressure =

Area A
1 N/m2 = 1 Pa and 1 atm = 101,325 Pa
Work,
w, is a force that produces an objects movement,

times the distance moved (d):

Work = Force x Distance

1 N m = 1 J and 101 J = 1 atm L

Work
C3H8(g) + 5O2(g)
6 mol of gas

3CO2(g) + 4H2O(g)

w=Fxd

7 mol of gas
F=PxA
w =P x A x d

w =P V

w = -P V
negative because
system does work
on surroundings

2012 Pearson Education, Inc.

12

Example

How much work is done (in kilojoules), and in which

direction, as a result of the following reaction?

w= -PDV = - 2.5 atm * (3L- 2L) = - 0.25 kJ .

Work is done on surroundings
Conversion factor: 101 J = 1 atm L

HEAT
The amount of heat exchanged between the system and the
surroundings is given the symbol q.

DE = q + W; DE = q - PDV; q = DE + PDV
(General case: DE = q - PextDV ; q = DE + PextDV)
P Pext
At constant volume (DV = 0): qv = DE
At constant pressure P: qp = DE + PDV = DH
(enthalpy)

Exercise
Remember:

w = - PV ; q = DE + PDV

Gas compression has DE = 186 kJ.

Is the sign of DV
positive or negative?
(1)

(2) What is w?

(3) What is q ?

50 L

40 L

15

Exercise
A) The explosion of 2.00 mol of solid trinitrotoluene (TNT) with a volume of
0.274 L produces gases with a volume of 448 L at r. t. How much PV work is
done during the explosion? Assume P = 1 atm, T = 25C.
w = -PDV
= -1 atm (448 L- 0.274 L)
= -447.7 Latm (101 J/Latm)
= - 45.2 kJ
B) In the reaction between 2 moles of hydrogen and 1 mole of oxygen to yield
water vapor, 484 kJ is released into the surroundings.
(1) If the volume change is 5.6 L at atm. pressure, how much PV work is done?
w = -PDV = -(1 atm) (5.6L) (101 J/atmL) = -566 J
(2) What is the value of DE for this reaction?
DE = q + w
q = -484 kJ
DE = -484 kJ - 0.566 kJ = -484.6 kJ

16

Enthalpy

In a chemical reaction (or any other process) conducted at constant

pressure, heat is called enthalpy (H) DH = qp

Enthalpies are expressed as a difference: DH = Hproducts Hreactants

DH can be positive or negative, depending on the reaction

For some reactions, H E When?

1) no gases 2) ngases= 0; 3) qp >>PV
E = qp + wp
constant
pressure

E = qp -PextVsys

qp = E + PV

New quantity: Enthalpy H = E + PV; P=Pext=const

system open to atmosphere
H = E + PconstantV
H = (qp + wp) + PconstantV
H = (qp + -PextVsys) + PconstantV,

H = qp

17

Sign of Enthalpy

Exothermic: Heat flows out

of the system into the
surroundings and DH has a
negative sign.
Endothermic: Heat flows into
the system from the
surroundings and DH has a
positive sign.

Exo/endothermic stated another way

B

For an endothermic
reaction A B
DH > 0

For an exothermic
reaction A C
DH < 0

DH

C
reactants and products

18

Enthalpy Change in Reactions

Reversing a reaction changes the sign of DH for a
reaction.
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) DH = 2219 kJ
3 CO2(g) + 4 H2O(l) C3H8(g) + 5 O2(g) DH = +2219 kJ

Multiplying a reaction increases DH by the same

factor.
3 C3H8(g) + 15 O2(g) 9 CO2(g) + 12 H2O(l) DH = 6657 kJ
19

Exercise
How much heat (in kilojoules) is evolved or absorbed in the
following reaction?
Burning of 15.5 g of propane, given the reaction
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) , DH = 2219 kJ/mole

Solution

15.5 g propane = 15.5 g / 44 g/mole = 0.35 mole propane was burned

qp = DH
Hence DH0.35 mole = 0.35 mole * (-2219kJ/ mole) = -780 kJ
780 kJ of heat is evolved.

20

Extensive-Intensive Properties
Iceberg
Extensive Property: A property whose value
depends on the amount of material present
(e.g., size, volume, mass)
Intensive Property: A property whose value
does not depend on the amount of material
present (e.g., temperature, melting point,

Ice cube

density).
Density = M/V (relate two extensive properties)

Enthalpy is an extensive property

Molar enthalpy is an intensive property
21

Heat Capacity
the amount of heat (q) required to raise the temperature of an
q
object or substance by 1.00C
C=

DT

Specific Heat Capacity (Cmass): Heat required to raise

the temperature of 1.00 g of substance by 1.00C.

Molar Specific Heat Capacity (Cmoles): The amount of

heat required to raise the temperature of 1.00 mole of
substance by 1.00C.

Cmass

q
=
D Tmass

Cmoles

q
=
DT n

How much heat is required

to heat 5.00 g of water from
25.0 C to 35.0 C?
q = Cmass x m x T
= 4.184 J/(gC)(5.00 g)(10.0 C)
= 209 J
22

Equal masses of metals are placed in

boiling water. The hot metals were placed
on a block of wax. Which metal will melt
the most wax and travel the furthest
down the wax?
a) Al (Cmass = 0.902 J/(gC)
b) Cu (Cmass = 0.385 J/(gC)
c) Au (Cmass = 0.129 J/(gC)
d) Fe (Cmass = 0.450 J/(gC)
e) They will all melt the same amount of wax.

Al

Cu Au

Fe

wax

23

Heat Measurements: Calorimetry

Many reactions release heat:
Fe2O3 + 2Al Al2O3 + 2Fe
Many reactions consume heat:
Ba(OH)2.8 H2O(s) + 2NH4SCN(s) Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)
How can we measure the heat released or consumed in a process?
Calorimetry is the science of measuring the heat exchanged in chemical
reactions.
In calorimetry, the heat of reaction (qrxn) is measured indirectly by
means of a calorimeter.

calorimeter
(surroundings)

(desired)

Remember: energy is conserved, so

energy that is lost to the system
24
goes to the surroundings.

reaction
(system)

qrxn = -qcalorimeter
(measured)

Heat Measurements: Calorimetry

Two types of calorimetry:

Constant Pressure Calorimetry: A constant pressure calorimeter

measures the heat change at constant pressure qp

Constant Volume or Bomb Calorimetry: A bomb calorimeter

measures the heat change at constant volume qv

Coffee-cup calorimeter
constant pressure

Bomb calorimeter
constant volume

25

A manufacturer claims that its new dietetic

dessert has fewer than 10 Calories (10 Kcal)
per serving. To test the claim, a chemist at the
Department of Consumer Affairs places one
serving in a bomb calorimeter and burns it in O2.
The initial temperature is 21.862C and the
temperature rises to 26.799C. If the heat
capacity of the calorimeter is 8.151 kJ/K, is the
manufacturers claim correct?
Solution:
qsample = qcalorimeter
qcal = c DT

c = 8.151 kJ/K

DT = 26.799 oC- 21.862oC = 4.937 oC = 4.937 K

qcal = 8.151 kJ/K x 4.937 K = 40.24 kJ
1 Calorie = 4184 J

40240 J (1 calorie / 4.184 J) = 9617.6 calories = 9.6 Kcal < 10 cal

The claim "fewer than 10 Calories per serving is correct

26

50.0 mL of 0.500 M NaOH is placed in a

coffee-cup calorimeter at 25.00oC and 25.0
mL of 0.500 M HCl is carefully added, also
at 25.00oC. After stirring, the final
temperature is 27.21oC. Calculate qsoln (in
J) and the change in enthalpy, DH, (in
kJ/mol of H2O formed).
Assume that the total volume is the sum of
the individual volumes, that d = 1.00 g/mL
and c = 4.184 J/gK.

27

PLAN: Heat flows from the reaction (the system) to its surroundings (the solution
Since qrxn = qsoln, we can find the heat of the reaction by calculating the heat
absorbed by the solution.
SOLUTION:
(a) Find qsoln. Total mass (g) of the solution
(25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g

DTsoln = 27.21 C 25.00 C = 2.21 C = 2.21 K

qsoln = csoln x masssoln x DTsoln = (4.184 J/gK)(75.0 g)(2.21 K) = 693 J
To find DHrxn, need a balanced equation:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
The number of moles HCl involved in the neutralization
0.0250 L x 0.5 mol/L = 0.0125 mol
DH = -693 J / 0.0125 mol = -5.54 x104J = -55.4 kJ

28

Hesss Law
The heat of a reaction is constant whether the reaction is carried out in one
step or indirectly through a number of steps (H is state function).
The enthalpy change for a reaction is equal to the sum of the changes for
individual states of the reaction
, 3 H2 (g) + N2 (g) 2 NH3 (g) DH = 92.2 kJ
In practice, this reaction
follows a two-step path:

(1)

(2)

3 H2 (g) + N2 (g) H2 (g) + N2H4 (g)

H2 (g) + N2H4 (g) 2 NH3 (g)

If DH (step 1)
is 95.4 kJ
DH (step 2)
must be
-187.6 kJ

So if you can add up the equations to get the overall reaction, you
can also add up the Hs of the steps to get the overall H.

29

Hesss Law
Additional Rules:
Reversing a reaction changes the sign of DH for a reaction.
(rationale: state property)
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) DH = 2219 kJ
3 CO2(g) + 4 H2O(l) C3H8(g) + 5 O2(g) DH = +2219 kJ
Increasing the moles of reactants/products changes DH by the same
factor.
(rationale: extensive property)
3 C3H8(g) + 15 O2(g) 9 CO2(g) + 12 H2O(l) DH = 6657 kJ
What is H for the reaction, 2C(graphite) + 2H2(g) C2H4(g), given the
following data:
a) C(graphite) + O2(g) CO2(g)
b) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)
c) H2(g) + 1/2O2(g) H2O(l)

H = -393.5 kJ
H = -1410.9 kJ
H = -285.8 kJ

30

Thermodynamic Standard State: Most stable forms of a substance:

1 atm pressure, 1 mole of pure substances, and 1 M concentration for
substances in solution.
These are indicated by adding a superscript to the symbol of the quantity
reported.
Standard enthalpy change is indicated by the symbol DH.
Standard Heats of Formation (DHf): The enthalpy change for the formation
of 1 mole of substance in its standard state from its constituent elements in
their standard states.
The standard heat of formation for any element in its standard state is defined
as being ZERO. DHf = 0 for an element in its standard state
H2(g) + 1/2 O2(g) H2O(l)
3/

2 H2(g)

+ 1/2 N2(g) NH3(g)

2 C(s) + H2(g) C2H2(g)

DHf = 286 kJ/mol

DHf = 46 kJ/mol
DHf = +227 kJ/mol

31

Standard Enthalpies of Formation

It is NOT practical to list H of all known reactions, so Hf are used.
Hesss Law can be used to find the standard enthalpy change for a
reaction by adding up standard enthalpies of formation.
Hrxn = nHf(products) nHf(reactants)
For a balanced equation, each heat of formation must be multiplied by the
stoichiometric coefficient.

aA + bB cC + dD
DH = [cDHf (C) + dDHf (D)] [aDHf (A) + bDHf (B)]

What is the Hrxn for CaCO3(s) CaO(s) + CO2(g)?

Hf(CaCO3) = -1206.9 kJ/mol
Hf(CaO) = -635.1 kJ/mol
Hf(CO2) = -393.5 kJ/mol

32

Exercise
The industrial degreasing solvent methylene chloride
(CH2Cl2, dichloromethane) is prepared from methane by
reaction with chlorine:
CH4(g) + 2 Cl2(g)

CH2Cl2(g) + 2 HCl(g)

Use the following data to calculate DH (in kilojoules) for

the above reaction:
CH4(g) + Cl2(g)

CH3Cl(g) + Cl2(g)

CH3Cl(g) + HCl(g)

DH = 98.3 kJ

CH2Cl2(g) + HCl(g) DH = 104 kJ

CH4(g) + Cl2(g)
CH3Cl(g) + HCl(g)
DH = 98.3 kJ
+
CH3Cl(g) + Cl2(g)
CH2Cl2(g) + HCl(g) DH = 104 kJ
___________________________________________________
_
CH4(g) + 2Cl2(g)

CH2Cl2(g) + 2HCl(g) DH = 202 kJ33

Bond Dissociation Energies

Bond energies can be measured by looking at the energy difference
between a molecule and the fragments:

H2(g) H(g) + H(g) H = D = + 435 kJ/mol

Bond energy depends on the atoms in the bond and also the rest of the
molecule.
437 kJ/mol
492 kJ/mol
CH3CH2O-H
Mean bond enthalpies used.

H-O-H

Bond enthalpies can be used to estimate Hrxn in the same way as Hf, but it is
not as accurate.
DH = D (Bonds Broken) D (Bonds Formed)
Example:
Cl2 + H2 2 HCl

DHrxn = {D(ClCl) + D(H-H)} -2D(HCl) = [(1mol)(243 kJ/mol) + (1mol)(436 kJ/mol)]

34
- [(2mol)(432 kJ/mol)]
= -185 kJ (exothermic)

To Summarize
A2 + 2B C2

DH = ?

Three ways to calculate:

(1) If you know DH of intermediate reactions, Hesss

Law DH1 = 1 kJ
A2 + 2B D
DH2 = 2 kJ +
D C2
A2 + 2B C2 DH = DH1 + DH2 = 3 kJ
(2) Enthalpy of Formation
DHf(A2) = 2 kJ
DH = DHf(C2) [DHf(A2) + 2DHf(B)] = 3
DHf(B) = 1 kJ
DHf(C2) = 7 kJ kJ
(3) Bond Dissociation
D(A-A) = 6 kJ
D(C-C) = 3 kJ

DH = D(A-A) D(C-C) = 3 kJ

35

Lets see how youre doing

If a reaction vessel becomes cooler as consequence of a reaction,
which of the following must be true?
A) the reaction is exothermic
B) heat flows into the system
C) work is done on the surroundings
D) DH is negative
What is DH for the reaction NO2(g) 1/2N2(g) + O2(g) if
N2(g) + 2O2(g) 2NO2(g) has DH = +66 kJ?
A) -33 kJ
B) +66 kJ
C) -66 kJ
D) +132 kJ
If I am looking for the enthalpy of formation for O2(g) and notice that it is not on
my table, I should
A) panic B) cry C) panic, then cry D) yell at my Prof
E) Remember that DHf = 0 because O2(g) is the most stable form of the 36
element, oxygen.

Introduction to Entropy
Second Law of Thermodynamics: Reactions
proceed in the direction that increases the
entropy of the system plus surroundings.
A spontaneous process is one that proceeds on
its own without any continuous external influence.
A nonspontaneous process takes place only in
the presence of a continuous external influence.
37

Spontaneity deals with directionality of processes (e.g., reactions)

A spontaneous change occurs by itself under specified conditions
without an ongoing input of energy from outside the system.
Under a given set of reaction conditions, if a reaction is spontaneous
in one direction, the reverse reaction is not spontaneous.

38

Until the mid 19th century, the sign H alone was thought to
determine spontaneity.
For example, the formation of rust is exothermic and spontaneous:
2Fe(s) + 3/2 O2

Fe2O3(s)

H = - 827 kJ

Endothermic processes can also be spontaneous. Ice melting is

endothermic and spontaneous at room temperature.
H2O(s)

H2O(l)

H = 6.02 kJ

How DO we determine if a (process) reaction is

spontaneous?

39

Entropy
The measure of molecular disorder in a system is called the
systems entropy and denoted S.
(1) Ssolid < Sliquid < Sgas

(2) More
molecules
have
higher
entropy
than fewer
molecules
40
The decomposition of N2O4 (O2NNO2) is accompanied by an increase in
randomness.

40

Entropy has units of J/K (Joules per Kelvin).

DS = Sfinal Sinitial

Positive value of DS indicates increased disorder.

Negative value of DS indicates decreased disorder.
Second Law of Thermodynamics:
Any spontaneous process increases the entropy of the
universe (system + surrounding)
or

Any process will proceed in the direction that increases

the entropy of the system plus surroundings.
41

1) Predict whether DS is likely to be positive (a) or negative (b)

for each of the following reactions.
A. 2 CO(g) + O2(g) 2 CO2(g)
B. 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)
C. C2H4(g) + Br2(g) CH2BrCH2Br(l)
D. 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)
2) Consider the gas phase reaction of A2 molecules (red) with B
atoms (blue).
(a) Write a balanced equation for the reaction.
(b) Predict the sign of S for the reaction.

42

Entropy and Probability

Entropy can be defined in terms of Boltzmans equation:

S = k ln(W)

W is the number of ways a state can be

achieved
(# of ways particles can be arranged)
= (# states)(# molecules)
k = R/NA = 1.38 10-23 J/K

43

Entropy and Probability

Entropy can be defined in terms of Boltzmans equation

S = k ln(W)

W is the number of ways a state can be achieved

(# of ways particles can be arranged)
= (# states)(# molecules)
k = R/NA = 1.38 10-23 J/K

Consider the a crystal containing 1 mole of CO molecules

(6.02 1023 molecules).

(a) Perfectly oriented

(b) Randomly oriented

W = 1(6.02 10 ) = 1
S = k ln(1) = 0

W = 2(6.02 10 )
23
S = k ln(2(6.02 10 ))
~ 5.7 J/K

23

23

44

Entropy and Temperature

Entropy is associated with molecular motion and available
configurations (freedom of motion).
As temperature increases, entropy increases.
The third law of Thermodynamics: the entropy of a perfectly
ordered substance is zero at 0 K.

45

discontinuous jump at phase changes

Standard Molar Entropies

Standard Molar Entropy, S: The entropy of 1 mole of the pure substance
at 1 atm pressure and a specified temperature, usually 25C.
Standard molar entropies are absolute entropies measured against an
absolute reference point.

Standard entropy of reaction

S

= nS(products) nS(reactants)

Entropy, like enthalpy, is a state property, and an extensive property

Example: Calculate the standard entropy of reaction at 25C for the
decomposition of calcium carbonate
CaCO3(s) CaO(s) + CO2(g)
S(J/(molK)

91.7

38.1

213.6

S = nS(products) nS(reactants)
S =1mol S(CaO(s)) + 1molS(CO2)) 1molS(CaCO3)
= [(1 mol)(38.1 J/(Kmol) + (1mol)(213.6 J/(Kmol)] -[(1 mol) (91.7 J/(Kmol)]
46
=160 J/K

Entropy and Spontaneity

Second Law of Thermodynamics: Spontaneous reactions proceed in the
direction that increases the entropy of the system plus surroundings.
A spontaneous process is one that proceeds on its own without any
continuous external influence.
Spontaneous:

Stotal = Ssys + Ssurr > 0

((or Suniverse > 0)
A nonspontaneous process takes place only in the presence of a
continuous external influence.
Nonspontaneous:

Stotal = Ssys + Ssurr < 0

(or Suniverse < 0)

47

 BUT we are mostly interested in the system

It is useful to re-state the second law (derivation p. 656)

Gibbs Free Energy Change (DG): evaluates spontaneity as a
function of enthalpy and entropy of the system.

DGsys = DHsys TDSsys

DG < 0
DG = 0
DG > 0

Process is spontaneous
Process is at equilibrium
Process is non-spontaneous

48

When is a reaction spontaneous?

G = H TS

G is negative, reaction spontaneous

G = H TS

G is positive, reaction nonspontaneous.

G = H TS

G could be or + depends on temperature.

At high temperature, the reaction is spontaneous.

G = H TS

G could be or + depends on temperature.

At low temperature, the reaction is spontaneous.

Example:
Is the following reaction spontaneous (standard state conditions)?
2Na(s) + Cl2(g) 2NaCl(s)
H = -822.2 kJ
Reaction will be
S positive or negative?
spontaneous at lower
G = H TS
temperatures.
49
Bottom line: negative DH and positive DS favor spontaneity

Iron metal can be produced by reducing iron(III) oxide with hydrogen:

Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(g)
Hf(kJ/mol)
S(J/(Kmol)

-824.2
87.4

-241.8
130.6

27.3

188.7

Is this reaction spontaneous at 25C? (under standard state conditions)

DH = [2DHf(Fe) + 3DHf(H2O)] [DHf(Fe2O3) + 3DHf(H2)]
= [0 + 3(- 241.8 kJ)] [-824.2 kJ + 0] = 98.8 kJ
DS = [2S (Fe) + 3S (H2O)] [S(Fe2O3) + 3S(H2)]
= [2(27.3 J/K) + 3(188.7 J/K)] [87.4 J/K) + 3(130.6 J/K)] = 141.5 J/K
DG = DH TDS
= (98.8 kJ) (298 K)(141.5 J/K) = 56.6 kJ
(Not Spontaneous)
Is this reaction spontaneous at any temperature? (standard state conditions)
In spontaneous reactions, DG = DH TDS < 0
DH TDS < 0
T > DH/DS> (98.8 kJ)/(141.5 J/K) > 698.2 K

(Yes, spontaneous at T > 698.2 K)

50

Gibbs Free Energy

Standard Free Energy (Grxn) is the free energy for a reaction occurring
under standard state conditions. Reactants in their standard states are
converted to products in their standard states.
G is the standard free energy of formation of a compound from its
elements in their standard states.
As is the case for DH and DS, DG is a state property and an extensive
property, such that:
Grxn = n G (products) n G (reactants)

Example
Calculate the standard free-energy change for the following reaction
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l)
Gf(kJ/mol)

-16.5

87.6

-237.2

Grxn = n G (products) n G (reactants)

= [4DGf(NO)+ 6DGf (H2O)] [4DGf(NH3) +5DGf (O2)]

DGrxn = [(4mol)(87.6kJ/mol)]+(6mol)(-237.2 kJ/mol)]

- [(4mol)(-16.5 kJ/mol)+ (5mol)(0 kJ/mol)]
= -1006.8 kJ

52

Gibbs Free Energy

DG tells us that a reaction will proceed, in principle (thermodynamics)
Ea (activation energy barrier) tells us how fast a reaction will proceed (kinetics)
Consider iron iron oxide (rust), which is spontaneous

Ea Can be reduced by
a catalyst to make the
reaction faster

53

Gibbs Free Energy

The energy graph is one of the most important ideas in chemistry

For the reaction, A B

Actually:
A A*
A* A**
A** B**
B** B*
B* B

energy

Questions:
A

B*

A**
A*

B**

reaction progress

(1) What is DG for B A

(2) Is B A possible?

54

Gibbs Free Energy

What are the signs (+, , or 0) of H, S, and G for the following
spontaneous reaction of A atoms (red) and B atoms (blue)?

G
S
H

55

Free Energy and Chemical Equilibrium

aA + bB cC + dD,

Q = ([C]c[D]d/[[A]a[B]b]

Q applies for any condition

Q = K in the special condition called equilibrium

K > 1 means
K < 1 means

Q < K means
Q > K means

56

Free Energy and Chemical Equilibrium

Fundamental Relationship between thermodynamics and equilibrium

G = G + RT lnQ
At equilibrium, DG = 0 and Q = K such that:

G = -RT lnK

57

Free Energy and Chemical Equilibrium

Using free energy data, calculate the equilibrium constant (K) for the
following, under standard conditions:
CO(g) + 2 H2(g) CH3OH(g)
DGf(kJ/mol)

-137.2

-161.9

G = nG (products) n G (reactants)
= [DGf(CH3OH)] [DGf(CO)+ 2DGf(H2)]
= [(1mol)(-161.9 kJ/mol)] [(1mol)(-137.2 kJ/mol) + (2mol)(0 kJ/mol)]
= -24.7 kJ
At equilibrium, G = G + RT lnQ becomes
G = -RT lnK
K = exp(-G/RT)
= exp[(24.7 kJ/mol)(1000J/kJ)/((8.314 J/(Kmol))298K)]
= 2.1 104
Using the solubility product of silver chloride at 25C (1.6 x 1010), calculate
G for the process: AgCl(s) Ag+(aq) + Cl(aq)
G = -RT lnK
58
= 8.314 J/(Kmol) (298K) ln(1.6 x 1010)
= 56 x 103 J or 56 kJ

Free Energy and Chemical Equilibrium

Consider the decomposition reaction:
N2O4 NO2
N2O4 2 NO2
Under standard conditions
o (kJ/mol) 9.16
33.18
DH
f
(1) What is the equilibrium temperature?
o (J/K-mol)
S
304.2 240.0
(2) What is K at 298 K?
eq

(3) What happens to the spontaneity as T ?

Answer (1)
At equilibrium, DG = 0, and T = DHo/DSo
DHo = [SDHfo(prod)] - [SDHfo(react)] = 2(33.18) - 9.16 kJ = 57.2 kJ
DSo = S[So(prod)] - S[So(react)] = 2(240.0) - 304.2 = 175.8 J/K
T = 57200 J/(175.8 J/K) =325 K.
Answer (2)
DGo = DHo 298KDSo
= 57200 298K(175.8) J/K = 4810 J
Keq = exp(-DGo/RT) = exp(-4810J/mol/(8.314J/molK)(298K)) = 0.143

Answer (3)
DH and DS are both positive
Increasing T makes DS term dominant
Thus, increasing T increases the spontaneity

59

Free Energy and Chemical Equilibrium

DG tells whether a reaction will go from a particular set of prevailing

conditions
As products build up (and equilibrium is approached), DG will approach
zero
The reaction will continue only if one or more products are removed
60

Free Energy and Chemical Equilibrium

Where is Q smaller, equal to, and greater than K?

Q = [products]n/[reactants]m
Q<K

free
energy

G of
reactants

pure
reactants

Q=K

G of
products

Q>K

pure
products

61

Lets see how youre doing

A spontaneous reaction under standard state conditions:
A) has a large, negative K
B) has a positive DG
C) must be a fast reaction
D) has K > 1
For a reaction with DG < 0 and Q = K, there will be:
A) net formation of products
B) net formation of reactants
C) no net change in products/reactants
D) formation of products until equilibrium is reached
For a reaction with DG > 0 and 100% reactants
A) no products will be formed
B) products will be formed until DG = 0
C) products will be formed until K = 0
D) products will be formed until DG = 0
62

For a reaction with Q > K, which of the following must be true?

A) DG must be positive
B) DG must be negative
C) [products] > [reactants]
D) reactants will be formed

63