1. Topic # 9
Thermochemistry

2. Chemical Reactions involve:
reactants turned into products
an energy change in form of heat
Energy changes accompany all chemical reactions and are due to rearranging of chemical bonding.

3. The addition of energy is always a requirement for breaking of bonds but the breaking of bonds in and of itself does not release energy.
Energy release occurs when
NEW BONDS are formed.

4. If more energy is released when new bonds form than
was required to break existing bonds, then the
difference will result in an overall release of energy….
EXOTHERMIC
If more energy is required to break existing bonds then
is released when new bonds form,
then energy is being absorbed…..
ENDOTHERMIC

5. the study of energy transformations
THERMODYNAMICS:
the study of energy transformations
To study thermodynamics (thermochemistry)
situations need to be viewed a little differently……..
System : All substances under going a physical or
chemical change
Surroundings : everything else that is not a part
of the system

6. System: all of the reactants and products
Concentrated Sulfuric acid is added to
distilled water in a beaker.
H2SO4(l) + H2O(l)  HSO4-(aq) + H3O+ (aq)
System: all of the reactants and products
Surroundings: water molecules NOT being used,
beaker, air…… etc.
( this is an exothermic process, heat is given off….
what does that mean? )

7. Energy is a property that is determined by specifying
the condition or “state” ( temperature, pressure, etc)
of a system or substance.
(is NOT really solid, liquid or gas….)
That’s why its called a STATE FUNCTION
(where its value is determined only by its
present condition and not on its history)
Ohio

8. EXOTHERMIC REACTION Energy is – Given Off! Activation Energy Energy
Reaction Coordinate
Activation Energy
H < 0
Energy is –
Given Off!

9. Energy is + Absorbed! ENDOTHERMIC H > 0 Activation Energy Products
Reaction Coordinate
Activation Energy
H > 0
Reactants
Products
ENDOTHERMIC
Energy is +
Absorbed!

10. Internal Energy of a System
Internal energy can be increased by:
an increase in temperature
a phase change
the initiation of a chemical reaction
Internal energy can be Decreased by:
a decrease in temperature

11. ∆E = Efinal - Einitial Total Energy of a System cannot be determined,
but the CHANGE may be measured.
∆E = Efinal - Einitial

12. The Laws of Thermodynamics
First Energy cannot be created or destroyed but only transferred from one body to another or changed from one form to another. Second Every spontaneous change increases the entropy of the universe/system. Third The entropy of a perfect crystalline substance (no disorder) is zero at 0 K.

13. ∆Esystem = q + w In a chemical system, the energy exchanged between a
system and its surroundings can be accounted for by
heat (q) and work (w).
∆Esystem = q + w
∆E < 0 (negative) means system gave away energy to surroundings
∆E > 0 (positive) means system absorbed energy from surroundings
If q < 0 then system gave away heat to surroundings (-)
If w < 0 then system did work on surroundings (-)
ALL ENERGY MEASURED IN JOULES!

14. 2NH3(g)  N2(g) + 3H2(g)
More moles of gas are produced than reacted. This
system has done work on its surroundings. The system
is ‘pushing back’ to make room for the expanding gas.

15. VERY IMPORTANT AND USEFUL:
Only pressure/volume work (expansion or contraction
of a gas) is of importance in a chemical reaction
system, and only when there is an increase or decrease
in the amount of of gas present!

16. For q: + means the system gained heat - means the system lost heat w = ∆E -
Depends on values of q and w -
For q: + means the system gained heat
- means the system lost heat
For w : + means work is done ON system
- means system does work ON surroundings

17. Useful when volumes of gases are shown in a
More on Work; w ………
The expansion/contraction of a gas against constant
external pressure can be expressed:
w = -P∆V
Useful when volumes of gases are shown in a
chemical reaction.
conversion needed: J/liter atmosphere
**** If there is no change in the total volume of gas
before and after a reaction, then there is no work
done by OR on the system!****

18. ∆E = q Remember! To DO work, something MUST MOVE!
If a reaction occurs in a rigid container, only the pressure,
NOT the volume may change. If no volume change
occurs, nothing moved…… NO WORK is done on or
by the system!
That means that any ∆E change is due ONLY to heat
transfers. So……
∆E = q

19. B) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Example:
For the following reactions performed at
constant external pressure, is work done on the
system by the surroundings, by the system
on the surroundings or is the amount of work negligible?
A) Sn(s) + 2F2(g) SnF4 (s)
B) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
C) C(s) + O2(g)  CO2(g)
D) SiI4(g) + heat  Si(s) + 2I2(g)

21. ∆H = q (at constant pressure)
Most times chemical reactions occur where the external
pressure is constant….
most often at atmospheric pressure and in aqueous forms
Here, a pressure change or volume change does not
happen. So the heat exchange is the important
thing to consider…….
∆H = q (at constant pressure)
∆H is ENTHALPY, the exchange of heat between
a system and its surroundings at constant pressure
Commonly called: heat of reaction

22. The amount of ∆H in any given reaction
is directly proportional to the quantity of
reactants AND depends on the conditions
that the reaction takes place.
So… ∆H is associated with a balanced
chemical equation and the temperature and
pressure are specified and units of enthalpy
change is given in
Joules/mole

23. A thermochemical equation:
2HgO(s) + heat  2Hg(l) + O2(g) ∆H = kJ
(endo)
K2O(s) + CO2(g)  K2CO3(s) ∆H = kJ (exo)
Also:
HgO(s) + heat  Hg(l) + ½ O2(g) ∆H = kJ
2K2O(s) + 2CO2(g)  2K2CO3(s) ∆H = kJ
What’s going on here with these diff equations?

24. Standard temperature = 25oC
One special set of conditions used with
chemical equations are the Standard State
Conditions:
Standard temperature = 25oC
Standard pressure = 1 atm
Standard concentration = 1 molar
When these conditions are met, then the ∆H
measure is called
Standard Enthalpy of Reaction: ∆Horxn
(has an o by the letter)

25. Standard Enthalpy of Reaction: ∆Horxn
The amount of energy associated with
with reactions at standard states has been
calculated for many reactions and can be
found on lists of ∆Horxn
(some equation data can be made available
from the q equation at standard conditions)
If the sign of the kJ/mol is negative then
the reaction is exothermic.

26. First, let’s look at heat used/given off
If there is no list available to you, then these
numbers can be calculated in a variety of ways.
First, let’s look at heat used/given off
by the formation of compounds from
their elements.

27. Standard Enthalpy of Formation (∆H°f )
First Way:
Standard Enthalpy of Formation (∆H°f )
change in enthalpy (heat) that accompanies the formation of 1 mole of that substance from its elements, with all substances in their standard states

28. The standard enthalpy of formation
for ethanol (C2H5OH):
2C(s) + 3H2(g) + 1/2O2(g)  C2H5OH(l)
Hof = kJ/mol
only elements as reactants
only 1 mole of product made

29. Remember: the standard enthalpy of
the formation of element has got
to be zero.
Why is that?

30. H = Hproducts - Hreactants
Second Way:
Another way of calculating H is to use
combinations of Hof of multiple compounds
H = Hproducts - Hreactants
Fe2O CO  2Fe + 3CO2
Calculate the standard enthalpy of above
reaction using this Hof data:
CO = kJ/mol CO2 = kJ/mol
Fe2O3 = kJ/mol

31. Fe2O3 + 3 CO  2Fe + 3CO2 H = Hproducts - Hreactants
Hrxn = [ 3(-393.7)] – [3(-110.4) + (-822.2)]
No #’s needed for the 2Fe!

32. According to Hess’s law, we can do some
Third Way:
What if asked to find Hof for SrCO3?
But there isn’t a listing for enthalpy of
formation for SrCO3 from its elements?
According to Hess’s law, we can do some
creative equation manipulation!

33. Add these equations together and get:
Sr + 3/2 O2 + C  SrCO3

34. Hess's Law
The enthalpy change for any reaction depends on the products and
reactants and is independent of the pathway or the number of steps between the reactant and product.

35. Find other reactions that are easier to perform and when summed up, equal the original desired reaction.
According to Hess's Law,
the sum of the heat flows for these reactions is equal to the heat flow for the original reaction.

36. Find the standard enthalpy
of formation of N2H4
N2 + 2O2  2NO Horxn= +67.4
N2H4 + 3O2  2NO2 + 2H2O Horxn=
2H2 + O2  2H2O Horxn =
We need:
N2 + 2H2  N2H4
(coefficient of 1 before N2H4)

37. N2 + 2H2  N2H4 N2 + 2O2  2NO2 Horxn= +67.4
2H2 + O2  2H2O Horxn =
2NO2 + 2H2O N2H4 + 3O2 Horxn=
Needed to flip one equation, change the sign
N2 + 2H2  N2H4

38. Another kind of equation that can be used
Fourth Way:
Another kind of equation that can be used
is the heat of combustion: Hoc
Combustion is the adding of oxygen to
a compound or element. When using a
hydrocarbon, usually CO2 and H2O are formed.
Combustion of CO: CO + 1/2O2  CO2
Combustion of Al : 2Al + 3/2O2  Al2O3
You may have to create this equation if given
the kJ/mol of heat.

39. Must write the equations for combustion for each of these items
Calculate the standard enthalpy of formation
of ethane, C2H6, given the following
combustion data; C = -393, H2 = -286 and
C2H6 = -1560kJ/mol
Must write the equations for combustion
for each of these items
2. Arrange those elements to give you:
2C + 3H2  C2H6
3. Add up the correct kJ’s

40. 2(C + O2  CO2) ( -393 kJ)
3(1/2O H2  H2O) (-286 kJ)
(Flip!)C2H6 + 7/2O2  3H2O + 2CO kJ
2C + 3H2  C2H6

41. Hrxn = ΣHbonds broken - ΣHbonds formed
Fifth Way:
Still another way to describe energy changes in compound includes the energy changes from breaking and forming bonds
Hrxn = ΣHbonds broken - ΣHbonds formed
add up the bonds broken and formed in
a given equation and that can equal the
heat of reaction.
(pgs in text)

42. Bond Energy (kJ/mol) C=O 743 H--H 436 H-C 412 C--C 348 H-N 391 N--N
170
H-O
463
O--O
145
H-F
565
F--F
154
H-Cl
427
Cl-Cl
239
H-Br
366
Br-Br
193
H-I
299
I--I
151
C-C
C=C
612
C--N
305
CºC
837
C--O
360
C--S
272
C--F
485
O=O
498
C--Cl
339
C--Br
276
C--I
238
NºN
945
C=O

43. (Broken – Formed)
CH3CH=CH2 + H2  CH3CH2CH3
Broken Formed
C=C
H-H
2 C-H 2(412)
824 kJ
700 kJ
Hrxn = 700 – 824 = kJ/mol
NOT the same as
H = Hproducts - Hreactants

45. Spontaneous Processes and Entropy
Why does a chemical reaction go naturally in
one particular direction?
A spontaneous process is one that, once initiated,
continues to proceed in the forward direction
with no further input of energy.
Using certain characteristics of a reaction, we can
describe the spontaneity of the reaction.

47. What are the forces that could
lend a reaction to become
spontaneous or nonspontaneous?

48. Entropy
measure of disorder
Movement from order to disorder
[Many times overcoming a high positive
activation energy (high + H)]

49. Degree of disorder of a system = S
Called Entropy
Increase in disorder (entropy) will
result in + quantity.
Decrease will be – quantity.
Spontaneous reactions usually have + value
for Entropy

50. Entropy can take many forms:
In terms of Solid, Liquid and Gas Phases, the
phases that provide more random motion would
have greater entropy.
Within a gas sample, the sample that has the higher
temperature or volume would provide more
random motion and have greater entropy.
Substances that are made of larger particles compared
to substances of smaller particles, have greater
entropy.
Mixtures of gases, solids and solutions are usually
viewed has having greater entropy.

51. Entropy is a STATE function……. Means it depends
on the quantity and conditions
of the substance.
More substance reacting = more entropy

52. ∆Srxn = Sproducts – Sreactants
Calculating Entropy
There are starting points of Entropy for substances. Measured at the standard state conditions. These are listed in appendix of the text. (So )
H2O(l)  H2O(g)
Is liquid water more or less ordered than gaseous water?
∆Srxn = Sproducts – Sreactants
∆Srxn = ( ) J/mol = 119 J/mol
+ Entropy means greater disorder!

53. Consider this spontaneous reaction:
N H2  2NH3
Why is this Spontaneous,
Is the product more or less ordered?

54. N2 + 3H2  2NH3 Using : Srxn = Sproducts – Sreactants
= J/mol
Negative sign says the product is more ordered.
Entropy is not the reason this reaction occurs.
so what DID cause this reaction to be spontaneous?

55. Suniv = Sreaction + Ssurr
To determine the spontaneity of a process,
more than a few things need consideration.
Second (Law of Thermodynamics)
Every spontaneous change increases the total
entropy of a system and its surroundings.
For a chemical process to be spontaneous,
it must increase the entropy of the universe!
Suniv = Sreaction + Ssurr
+ Suniv = Spontaneous Suniv = NOT Spontaneous
(What is meant by the term….. Universe, here?)

56. -H Ssurr = T The entropy of the surroundings can be calculated:
We know how to calculate the Entropy of the reaction….
The entropy of the surroundings
can be calculated:
-H
Ssurr =
(Sometimes q) T

57. Suniv = Srxn + Ssurr Putting these two ideas together…..
Suniv = [Sproducts -Sreactants] + -H T
(let’s see this in action…….)

58. Consider this spontaneous situation:
2H2(g) + O2(g)  2H2O(l) kJ/mole
∆So for the following are:
H2(g) = 131 J/mol O2(g)= 205J/mol H2O(l) = 70 J/mol
S = Sproducts – Sreactants
∆Sorxn = [2(70)] – [2(131)+ (205)]
∆Sorxn = J/mole (lower entropy)
(Seems correct, water is more ordered than the two gases)

59. Why is this reaction spontaneous? Disorder is
not increased! What’s the force, here?
-Hrxn
Ssurr =
-(-572 kJ/mol)
298K = T
This equals 1919 Joules/mole (converted to Joules)
Need to calculate the entropy for the universe……..
(OH! A lot bigger than -327!)

60. Suniv = Srxn + Ssurr Suniv = -327 J/mole + 1919 J/mole
Answer: J/mole
+ answer means that entropy for the overall
process, including system AND surroundings
DID, in fact, increase!
Suniv > so IS spontaneous

61. ∆Gorxn ∆Gorxn = ∆Horxn – (T)∆Sorxn
Further study of these calculations and processes lead
to a culmination formula that connects ∆H, heat or enthalpy
and ∆S, entropy or disorder:
This is another way to calculate/prove
that a reaction is spontaneous……
Gibbs Free Energy: The amount of energy that is
available to do work on the UNIVERSE in a chemical
reaction.
∆Gorxn
∆Gorxn = ∆Horxn – (T)∆Sorxn
∆Gorxn < 0 FOR spontaneous reactions
( WORK? What work?)

62. 2H2(g) + O2(g)  2H2O(l) + 572 kJ/mole
(where’s the work?)
2H2(g) + O2(g)  2H2O(l) kJ/mole
Remember negligible work?
There is not much energy available to do work on this
system; three moles of gas on one side, NONE on the
other. Negligible to zero work done on the system.
Most of the energy here is given off as enthalpy, not work.
But according to the Second Law, for this reaction to
proceed as spontaneous, enough energy as work must be given
away to the surroundings so that the increase in entropy of
the surroundings offsets the decrease in the entropy of the
system.
( how much energy is that?)

63. ∆Gorxn = ∆Horxn – (T)∆Sorxn
2H2(g) + O2(g)  2H2O(l) kJ/mole
∆Hosurr > (0.327kJ/mol K given to surroundings) * (298K) =
97kJ/mole available to do work on surroundings
∆Gorxn = ∆Horxn – (T)∆Sorxn
∆Gorxn = needs to be negative to have a
spontaneous reaction
∆Gorxn = kJ/mol – [ (298K)(-0.327kJ/mole)]
∆Gorxn = kJ/mol
Number is – so reaction is spontaneous

64. ∆Gorxn = ∆Horxn – (T)∆Sorxn
For any reaction to be spontaneous the sole condition
is that Gibbs Free Energy needs to be negative!
∆Gorxn = ∆Horxn – (T)∆Sorxn

65. Go is negative for any reaction for which Ho is
Go is negative for any reaction for which  Ho is negative and  So is positive.
 Go is negative for any reaction where both the enthalpy and entropy terms are negative.

66. NO Yes Keys to determining why a reaction happens H Sreaction G +
at high temp
+ at low temp -
- at low temp
+ at high temp NO
Yes

67. Our description of why a reaction occurs
sometimes needs to include HOW
need to describe a step-wise progression
on exactly how a compound is formed.
needs to include all energies along the way
The Born Haber Cycle
(normally associated with an ionically bonded compound)

68. Include these energy changes as needed:
Standard enthalpy of formation
Standard enthalpy of vaporization (substance into gas)
First Ionization energy ( gas losing an electron)
Enthalpy of bond dissociation (molecules to atoms)
First electron affinity (gas gaining an electron)
Enthalpy of Lattice (energy associated with one
mole of solid is formed from its gaseous ions)

69. Formation of LiF Li(s) + ½ F2(g) (pgs. 363 – 367 of text)
Li+ (g) + F (g)
Li+ (g) + ½ F2(g)
77kJ
-328 kJ
Li+ (g) + F- (g)
520kJ
Li (g) + ½ F2(g)
161kJ
Li(s) + ½ F2(g)
-1047 kJ
Overall -617 kJ
LiF