Download Slajd 1

22/03/2015

The thermodynamic standard state of a
substance is its most stable pure form under
standard pressure (one atmosphere) and at
some specific temperature (25°C or 298 K
unless otherwise specified).
Thermodynamics part 2
THERMODYNAMIC STANDARD
STATE
Examples of elements in their standard states at
25°C are:
 hydrogen, gaseous diatomic molecules, H2(g);
 mercury, a silver-coloured liquid metal, Hg(l);


sodium, a silvery white solid metal, Na(s);
carbon, a grayish black solid called graphite,
C(graphite).
THERMODYNAMIC STANDARD
STATE
Q.1.Could you listen other elements in their
standard state (conditions):
Q.2. The values associated with the term,
"standard reference conditions or shortly standard
condition“ in a thermochemical meaning refer to:
a. temperature: 0.00 K; pressure: 1.000 atm
b. temperature: 0.00 oC; pressure: 1.000 atm
c. temperature: 273.15 K; pressure: 1.000 Pa
d. temperature: 298.15 K; pressure: 1.000 atm
e. temperature: 298.15 K; pressure: 1.000 Pa
f. temperature: 25oC; pressure: 1 atm

Quiz
1
22/03/2015
•
•
•
1. For a pure substance in the liquid or solid
phase, the standard state is the pure liquid or
solid.
2. For a gas, the standard state is the gas at
a pressure of one atmosphere; in a mixture
of gases, its partial pressure must be one
atmosphere.
3. For a substance in solution, the standard
state refers to one-molar concentration.
TSS - RULES
For ease of comparison and tabulation, we often
refer to thermochemical or thermodynamic
changes “at standard states” or, more simply, to
a standard change.
 To indicate a change at standard pressure, we
add a superscript zero; e.g. ∆H0 .

STANDARD CHANGE

Choose the answer/s which present TSS
conditions:

Answer
Solute/
mixture
Condition
a)
NaOH(aq)
C = 0.1 M/L
b)
Mixture of O2
and N2
Po2 = 0.7atm; PN2 =
0.3atm
c)
He and Ar
Po2 = 0.8 atm; PN2 =
1.3atm
d)
HClaq
C = 1 M/L
Quiz

The standard enthalpy change, ∆H0 rxn, for
reaction:
reactants  products
refers to the H when the specified number of
moles of reactants, all at standard states,
are converted completely to the specified
number of moles of products, all at standard
states.
Standard enthalpy change, ∆H0
rxn, for reaction
2
22/03/2015

We allow a reaction to take place, with
changes in temperature or pressure if
necessary; when the reaction is complete,
we return the products to the same
conditions of temperature and pressure
that we started with, keeping track of
energy or enthalpy changes as we do so.
T, P = T,P
It is not possible to determine the total enthalpy
content of a substance on an absolute scale.
 We need to describe only changes in this state
function, however, so we can define an arbitrary
scale as follows.

STANDARD MOLAR ENTHALPIES OF
FORMATION, ∆H
When we describe a process as taking place “at
constant T and P,” we mean that the initial
and final conditions are the same.
 Because we are dealing with changes in state
functions, the net change is the same as the
change we would have obtained hypothetically
with T and P actually held constant.

At constant T and P
The standard molar enthalpy of formation,
∆Hf0 , of a substance is the enthalpy change
for the reaction in which one mole of the
substance in a specified state is formed from its
elements in their standard states.
 By convention, the ∆Hf0 value for any element
in its standard state is zero.

Standard molar enthalpy of
formation
3
22/03/2015

The standard molar enthalpy of formation of
ethanol, C2H5OH(l), is - 277.7 kJ/mol.
Write the thermochemical equation for the
reaction for which Hf0 rxn - 277.7 kJ/mol rxn.
•
•
•
Interpretation of Hf0
2 C(graphite) + 2H2(g) + ½ O2(g) → C2H5OH(l)
ΔHf0 = -277.7 kJ/mol rxn
Solution
The definition of Hf0 of a substance refers to
a reaction in which one mole of the
substance is formed.
We put one mole of C2H5OH(l) on the right
side of the chemical equation and put the
appropriate elements in their standard states
on the left.
We balance the equation without changing
the coefficient of the product, even if we
must use fractional coefficients on the left.
Plan
Q.3. The standard molar enthalpy of formation of
aluminium oxide, Al2O3(s), is -1675.7 kJ/mol. Write
the thermochemical equation for the reaction for
which Hf0 rxn -1675.7 kJ/mol rxn.
Example to solve
4
22/03/2015

Q. 3.

2Al(s) + 3/2 O2(g) → Al2O3(s)

ΔHf0 = -1675.7 kJ/ mol
Solution

Q.4. The standard molar enthalpy of
formation of hydrogen iodide, HI(g), is
26.48 kJ/mol. Write the thermochemical
equation for the reaction for which Hf0
rxn 26.48 kJ/mol rxn.
Quiz
In 1840, G. H. Hess (1802–1850)
published his law of heat summation,
which he derived on the basis of
numerous thermochemical observations.
 The enthalpy change for a reaction is the
same whether it occurs by one step or by
any series of steps.

Q.4.
½ H2(g) + ½ I2(g) → HI(g)
ΔHf0 = 26.48 kJ/mol
Solution
HESS’S LAW
5
22/03/2015
As an analogy, consider traveling from
Kansas City (elevation 884 ft above sea
level) to Denver (elevation 5280 ft).
 The change in elevation is (5280 - 884) ft
= 4396 ft, regardless of the route taken.

INTERPRETATION
Hess’s law lets us calculate enthalpy
changes for reactions for which the
changes could be measured only with
difficulty, if at all.
 In general terms, Hess’s Law of heat
summation may be represented as: (next
slide)

Hess’s Law
The standard enthalpy change of a reaction is equal to the
sum of the standard molar enthalpies of formation of the
products,each multiplied by its coefficient ,n, in the
balanced equlation, minus the corresponding sum of the
standard molar enthalpies of formation of the reactants.
Useful form of Hess’s Law.
A schematic representation of
Hess’s Law
6
22/03/2015

C
Consider the following reaction.
(graphite)
∆H0

+ 1/2 O2(g) CO(g)
rxn __ ?
The enthalpy change for this reaction cannot
be measured directly.
Even though CO(g) is the predominant product
of the reaction of graphite with a limited
amount of O2(g), some CO2(g) is always
produced as well.
C(graphite) + O2(g) → CO2(g)
ΔH0rxn = -393.5 kJ/mol rxn (1)
CO(g) + ½ O2(g) → CO2(g)
ΔH0rxn = -283.0 kJ/mol rxn (2)
EXAMPLE

Revise equation (2) to give (-2) thus we
obtain the CO(g) from carbon dioxide,
when the equation is revised, the sign of
ΔH0 is changed because the reversed of
exothermic reaction is endothermic
reaction, then add to eq.(1).
Plan
C(graphite) + O2(g) → CO2(g)
ΔH0rxn = -393.5 kJ/mol rxn (1)
CO2(g) → CO (g) + ½ O2(g)
ΔH0rxn = -(-283.0 kJ/mol rxn) (2)
C(graphite) + ½ O2(g) → CO(g)
ΔHrxn0 = - 110.5 kJ/mol rxn
EXAMPLE
7
22/03/2015

Calculate heat of reaction at 298K.
C2H4(g) + H2O(l) → C2H5OH(l)
use the following thermochemical equations
ΔH0
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
-1367kJ/mol (1)
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
- 1411kJ/mol (2)
Example
INTERPRETATION
ΔH0
2CO2(g) + 3H2O(l) → C2H5OH(l) + 3O2(g)
+1367kJ/mol (-1)
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
- 1411kJ/mol (2)


Revise equation (1) to give (-1) when the
equation is revised, the sign of ΔH0 is
changed because the reversed of
exothermic reaction is endothermic
reaction, then add to eq.(2).
Plan
C2H4(g) + H2O(l) → C2H5OH(l)
ΔH0 = -44 kJ/mol rxn
Solution
8
22/03/2015


What is the value for ΔH for the following
reaction?
CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g)
Given:
C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol
S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol
C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol
Calculate the standard enthalpy of
combustion, ΔHf0, for burning β-Dglucose, C6H12O6.
The required values of ΔHf0 are presented
above:
 β-D-glucose = -1268 kJ/mol
 CO2(g) = -393 kJ/mol
 H2O(l) = -285,5 kJ/mol
 O2(g) = 0 kJ/mol

Quiz
The internal energy, E, of a specific
amount of a substance represents all
the energy contained within the
substance.
It includes such forms as:
 kinetic energies of the molecules;
 energies of attraction
 repulsion among subatomic particles,
atoms, ions, or molecules;
 other forms of energy.

CHANGES IN INTERNAL ENERGY, E
The internal energy of a collection of molecules
is a state function.
 The difference between the internal energy of
the products and the internal energy of the
reactants of a chemical reaction or physical
change, E, is given by the equation

CHANGES IN INTERNAL ENERGY, E
9
22/03/2015
Δ E = Efinal - Einitial - Eproducts – Ereactants - q + w
CHANGES IN INTERNAL ENERGY, E
Sign conventions for q and w.

Inside a pile of oily rags or a stack of hay that has not
been thoroughly dried, decomposition causes heat to build
up. When heat cannot escape, the temperature can
become high enough to cause a fire.
Enthalpy, free energy, spontaneous
and non-spontaneous reaction

What was the character of the process?

Answer - spontaneous

You will learn about the conditions that will produce
a spontaneous chemical reaction.
Entropy and Free Energy
10
22/03/2015
• What are two characteristics of spontaneous
reactions?
• A spontaneous reaction occurs naturally and
favours the formation of products at the
specified conditions.
Any process which occurs without outside
intervention is spontaneous.
When two eggs are dropped they
spontaneously break.
The reverse reaction (two eggs leaping into
your hand with their shells back intact) is not
spontaneous.
Spontaneous reaction
We can conclude that a spontaneous process has
a direction.
A process that is spontaneous in one
direction is not spontaneous in the opposite
direction.
 The direction of a spontaneous process can
depend on temperature.
 Ice turning to water is spontaneous at T > 0°C.
 Water turning to ice is spontaneous at T < 0 °C
A reversible process is one which can go back and
forth between states along the same path.
When 1 mol of water is frozen at 1 atm at 0 °C to
form 1 mol of ice, q = ∆Hvap of heat is removed.
To reverse the process, q = ∆Hvap must be added to
the 1 mol of ice at 0°C and 1 atm to form 1 mol of
water at 0 °C
Spontaneous reactions
11
22/03/2015
 A nonspontaneous reaction is a reaction that does
not favour the formation of products at the specified
conditions.
 Photosynthesis is a nonspontaneous reaction that
requires an input of energy.
Nonspontaneous reaction

The major concern of thermodynamics is
predicting whether a particular process
can occur under specified conditions to
give predominantly products.
Q. 6. Non-spontaneous changes are:
 rusting of an iron
burning a piece of paper
metling of ice at room temperature
electrolysis
photosynthesis
2H2O→2H2 +O2
producing the aniline from nitrobenzene
Quiz

A change for which the collection of
products is thermodynamically more
stable than the collection of reactants
under the given conditions is said to be
product favoured, or spontaneous,
under those conditions.
Product favored
12
22/03/2015

A change for which the products are
thermodynamically less stable than the
reactants under the given conditions is
described as reactant-favoured, or
nonspontaneous, under those
conditions.

The concept of spontaneity has a very
specific interpretation in thermodynamics.
A spontaneous chemical reaction or
physical change is one that can happen
without any continuing outside influence.
Reactant- favored
Any spontaneous change has a natural
direction, like:
 the rusting of a piece of iron,
 the burning of a piece of paper,
 or the melting of ice at room temperature.
We can think of a spontaneous process as
one for which products are favoured over
reactants at the specified conditions.
Although a spontaneous reaction might
occur rapidly, thermodynamic spontaneity
is not related to speed. The fact that a
process is spontaneous does not mean
that it will occur at an observable rate.
 It may occur rapidly, at a moderate rate,
or very slowly.
The rate at which a spontaneous reaction
occurs is addressed by kinetics (see in
future lectures).

Spontaneus change
13
22/03/2015
Many product-favoured reactions are
exothermic.
 the combustion (burning)reactions of
hydrocarbons such as methane and octane
are all exothermic and highly productfavoured (spontaneous).

The enthalpy contents of the products
are lower than those of the reactants
Not all exothermic changes are spontaneous,
however, nor are all spontaneous changes
exothermic.
As an example, consider the freezing of
water, which is an exothermic process (heat
is released).
This process is spontaneous at temperatures
below 0°C, but it certainly is not
spontaneous at temperatures above 0°C.
Enthaply of exothermic reaction
Likewise, we can find conditions at which
the melting of ice, an endothermic
process, is spontaneous.

Spontaneity is favoured but not required
when heat is released during a chemical
reaction or a physical change.

Another factor, related to the disorder of
reactants and products, also plays a role
in determining spontaneity.
The dissolution of ammonium nitrate,
NH4NO3, in water is spontaneous,
although it is endothermic reaction.
14
22/03/2015

The system (consisting of the water, the

Nevertheless, the process is spontaneous
solid NH4NO3, and the resulting hydrated
because the system becomes more
NH4 and NO3 ions) absorbs heat from
disordered as the regularly arranged ions
the surroundings as the endothermic
of crystalline ammonium nitrate become
process occurs.
more randomly distributed hydrated ions
+
-
in solution.

An increase in disorder in the system
favours the spontaneity of a reaction.
In this particular case, the increase in
disorder overrides the effect of
endothermicity.
Two factors affect the spontaneity of any
physical or chemical change:
 1. Spontaneity is favoured when heat
is released during the change
(exothermic).
 2. Spontaneity is favoured when the
change causes an increase in
disorder.

Factors affect the spontaneity
15
22/03/2015
In spontaneous changes, the universe
tends toward a state of greater disorder.
 The Second Law of Thermodynamics is
based on our experiences.
 State of entropy of the entire universe, as
a closed isolated system, will always
increase over time. The second law also
states that the changes in the entropy in
the universe can never be negative.

Second law of thermodynamics
introduction
The thermodynamic state function
entropy, S, is a measure of the
disorder of the system.
 The greater the disorder of a system, the
higher is its entropy.


The reverse of any spontaneous change is
nonspontaneous, because if it did occur,
the universe would tend toward a state of
greater order. This is contrary to our
experience.
Spontaneous vs. nonspontaneous
 For a given substance, the entropy of the gas is
greater than the entropy of the liquid or the solid.
Similarly, the entropy of the liquid is greater than
that of the solid.
Now lets show some examples of systems
of increasing entropy.
Entropy
ENTROPY - SOLID-LIQUID-GAS
16
22/03/2015
 Entropy increases when a substance is divided into
parts.
INCREASING ENTROPY
 Entropy tends to increase when temperature
increases. As the temperature increases, the
molecules move faster and faster, which increases
the disorder.
INCREASING ENTROPY
 Entropy tends to increase in chemical reactions in
which the total number of product molecules is
greater than the total number of reactant molecules.
ENTROPY IN CHEMICAL REACTION
Q. Which ones of the processes below are not
accompanied by an increase in entropy?
a. Cooking the soup
b. decomposition of SO3
2SO3→ 2SO2 + 2O2
d. Cooling the tea in cup
e. restocking a canned goods shelf display
in a supermarket
f. Forming a snowball
Entropy quiz
17
22/03/2015
Q. Which system has the greater enthropy:

A) granulated sugar or icing sugar

B) wall of bricks or several bricks

C) desk of pedant person or desk of
 Spontaneous reactions produce substantial
amounts of products at equilibrium and
release free energy.
 Free energy is energy that is available to
do work.
messy person

D) the lecture room before or after the
lecture
Quiz
• What part does entropy play in chemical
reactions?
• Entropy is a measure of the disorder of a
system.
◦ Physical and chemical systems attain the
lowest possible energy.
◦ The law of disorder states that the natural
tendency is for systems to move in the
direction of maximum disorder or
randomness.
ENTROPY
Free Energy and Spontaneous
Reactions
• An increase in entropy favours
the spontaneous chemical
reaction.
• A decrease favours the
nonspontaneous reaction.
ENTROPY
18
22/03/2015
• What two factors determine the spontaneity of a
reaction?
• The size and direction of enthalpy changes and
entropy changes together determine whether
a reaction is spontaneous;
• that is, whether it favours products and releases
free energy.
Enthalpy, Entropy, and Free
Energy
Combustion of acetylene
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Enthalpies of formation
∆Horxn = -1238 kJ
Standard molar entropies
∆Sorxn = -97.4 J/K or -0.0974 kJ/K
∆Gorxn = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ
Reaction is product-favored in spite of negative
∆Sorxn.
Reaction is “enthalpy driven”
Gibbs energy in questions
• The Gibbs free-energy change (ΔG) is the
maximum amount of energy that can be coupled
to another process to do useful work.
Is the Gibbs free-energy change positive or
negative in a spontaneous process?
• The numerical value of ΔG is negative in
spontaneous processes because the system
loses free energy.
Gibbs free-energy
NH4NO3(s) + heat ---> NH4NO3(aq)
From tables of thermodynamic data we find
∆Horxn = +25.7 kJ
∆Sorxn = +108.7 J/K or +0.1087 kJ/K
∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is product-favored in spite of positive
∆Horxn.
Reaction is “entropy driven”
Calculating ∆Gorxn
19
22/03/2015
:
In the Haber process for the manufacture of ammonia
N2 + 3H2→ 2 NH3
At what temperature will the reaction above become
spontaneous?
If the enthalpy and entropy of this proces
ΔG = -93000 - (T x -198)
note that the enthalpy is given in kilojoules
if ΔG = 0
then the system is at the limit of reaction spontaneity
When ΔG = 0 then (T x -198) = -93000
ΔH = -93 kJ/mol; ΔS= -198 J/ mol K
The fact that both terms are negative means that the Gibbs
free energy equation is balanced and temperature
dependent:
ΔG = ΔH - TΔS
and T = 93000/198 Kelvin
therefore the reaction becomes spontaneous when
T = 469 K (196 ºC)
below this temperature the reaction is spontaneous.
Influence of enthalpy and entropy changes on reaction spontaneity
Enthalpy
change
Entropy
Spontaneity
reaction
Decreases
(exothermic)
Increases (more
disorder in products
than in reactants)
Yes
Increases
(endothermic)
Increases
Only if unfavourable
enthalpy change is
offset by favourable
entropy change
Decreases
(exothermic)
Decreases (more
disorder in reactants
than in products)
Only if unfavourable
entropy change is
offset by favourable
enthalpy change
Increases
(endothermic)
Decreases
No
20
22/03/2015
21
22/03/2015
True and False
The equation, Horeaction = Hof (products) - Hof (reactants)
is a statement of the second law of thermodynamics. ___
Quiz
1. Free energy from a reaction is the amount of
True/False:
ΔG > 0, the process is spontaneous
True/False:
A nonspontaneous process cannot occur with external
intervention.
energy that is
 absorbed by an entropy decrease.
 equal to the enthalpy change.
 wasted as heat.
 avaiable to do work
QUIZ
22
22/03/2015
2. Free energy is always available from
3. Choose the correct words for the spaces:
reactions that are
Spontaneous reactions produce ________
 endothermic.
and substantial amounts of _________ at
 nonspontaneous.
equilibrium.
 at equilibrium.
 free energy, products
 spontaneous.
 no free energy, reactants
 free energy, reactants
QUIZ
 no free energy, products
QUIZ
4. Which of the following involves a decrease in
entropy?
 Natural gas burns.
 A liquid freezes.
 Dry ice sublimes.
 Water evaporates.
6. A reaction is spontaneous if
 enthalpy
 enthalpy
 enthalpy
 enthalpy
decreases and entropy increases.
increases and entropy increases.
decreases and entropy decreases.
increases and entropy decreases
5. Both the enthalpy and the entropy are
combined to calculate the
a. reaction time
b. free energy
c. bound energy
QUIZ
QUIZ
23
22/03/2015
6. Choose the correct words for the spaces:
Gibbs free-energy change is the _________
amount of energy that can be ___________
another process to do useful work.
 maximum, coupled to
 maximum, duplicated by
 spontaneous, coupled to
 minimum, duplicated by
QUIZ
24