Chapter 5 final(1)

Contents:
1. Nature of energy
2. 1st law of thermodynamics
3. Calorimetry
4. Enthalpy
5. Enthalpy of reaction
6. Hess’s law
7. Enthalpy of formation

The nature of ENERGY
ThermoChemistry

Potential Energy
Energy an object
possesses by virtue
of its position or
chemical composition
Kinetic Energy
Energy an object
possesses by virtue
of its motion
1
2
KE =  mv2
PE = mgh

The ability to do work or transfer heat
Work: Energy used to cause an object that has
mass to move over some distance
Heat: Energy used to cause the temperature of an
object to rise
Heat and work exchange between system
and surroundings

System and Surrounding:
o The system includes the
molecules we want to
study (here, the hydrogen
and oxygen molecules)
o The surroundings are
everything else (here, the
cylinder and piston)

System and Surrounding:
o The system includes the
molecules we want to
study (here, water and ice
molecules)
o The surroundings are
everything else (here, the
cup, air and the whole
universe)

Work: Energy used to cause an object that has
mass to move over some distance
w =- F  d
where
w is work,
F is the force,
d is the distance over
which the force is
exerted.

Work: When a process
occurs in an open
container, commonly
the only work done is a
change in volume of a
gas pushing on the
surroundings (or being
pushed on by the
surroundings) w = - PexV
(+ve means work applied to the system)

Work:
We can measure the work done by the gas if the
reaction is done in a vessel that has been fitted
with a piston
(-ve means work
done by the system;
also known as work
of expansion)
w = − Pex V

Heat: Energy used to cause the temperature of
an object to rise
 Energy can also be
transferred as heat.
 Heat flows from
warmer objects to
cooler objects.

Heat: Energy used to cause the temperature of
an object to rise
 Heat added to the
system is endothermic
(+ve)
 Heat taken from the
system is exothermic
(-ve)

Units of energy:
SI unit of energy is the Joule (J)
1 J = 1kg.m/s2
Note: calorie (cal) is another unit of energy, where:
1 cal = 4.184 J
1 Cal = 1 kcal = 1000 cal
Definition:
1 cal = amount of energy required to raise the
temperature of 1 g of water by 1oC

The 1st law of thermodynamics
ThermoChemistry

“Energy is neither created nor destroyed”
In other words, the total energy of the universe is
a constant;
if the system loses energy, it must be gained by the
surroundings, and vice versa

The Internal Energy (E) of a system is the sum of all
kinetic and potential energies of all components of
the system
By definition, the change in internal energy, E, is
the final energy of the system minus the initial
energy of the system:
E = Efinal − Einitial

The total energy of the universe is a constant;
if the system loses energy, it must be gained by the
surroundings, and vice versa
Exergonic Endergonic

Endergonic
Exergonic
Example:
Draw an energy diagram showing the process of formation of
liquid water from its gaseous components

When energy is exchanged between the system and
the surroundings, it is exchanged as either heat (q)
or work (w). That is,
E = q + w

For q + means system gain heat
(endothermic)
- means system loses heat
(exothermic)
For w + means work done on system
- means work done by system
For E + means net gain of energy by system
(endergonic)
- means net loss of energy by system
(exergonic)
E = q + w

E = q + w
Any energy given to (added)
to the system is positive
Any energy taken from
(subtracted) from the
system is negative
During exchange of heat
and work, E sign depends
on the larger value of
q and w

E = q + w
Exercise:
Calculate E, and determine whether the process is
“endothermic” or “exothermic” for the following cases:
1. A system absorbs 105 kJ of heat from its surroundings
while doing 29 kJ of work on the surroundings
2. q = 1.50 kJ, and w = -657 J
2. A system releases 57.5 kJ of heat while doing 22.5 kJ of
work on the surroundings
Solution Approach

q = + 105 kJ
(Endothermic)
E = q + w
Exercise:
1. A system absorbs 105 kJ of heat from its surroundings
while doing 29 kJ of work on the surroundings
Solution:
w = - 29 kJ
 E = 105-29 = + 76 kJ (Endergonic)

q = + 1.50 kJ
(Endothermic)
E = q + w
Exercise:
2. q = 1.50 kJ, and w = -657 J
Solution:
w = - 657 J
= - 0.657 kJ
 E = 1.50 – 0.657 = + 0.843 kJ (Endergonic)

q = - 57.5 kJ
(Exothermic)
E = q + w
Exercise:
3. A system releases 57.5 kJ of heat while doing 22.5 kJ
of work on the surroundings
Solution:
w = - 22.5 kJ
 E = -57.5 -22.5 = - 80 kJ (Exergonic)

State Function
It depends only on the present state of the system,
not on the path by which the system arrived at that
state (e.g. E)
Example:
Liquid water at 25oC can be reached by 2 different pathways,
but E is the same as it depends on Ei and Ef ONLY

State Function
Example:
Whether the battery is shorted
out or is discharged by running
the fan, its E is the same, as it
depends on Ei and Ef ONLY
Even though q and w are
different in the two cases
It depends only on the present state of the system,
not on the path by which the system arrived at that
state (e.g. E)

Enthalpy
Assuming that the only work done is this pressure-volume
work, then Enthalpy (H) is the internal energy (E) plus the
product of pressure and volume (PV)
It is the amount of heat measured at constant
pressure (= qp)
H = E + PV
When the system changes at constant pressure, the
change in enthalpy, H, is:
H = (E + PV)
Which can be written as:
H = E + PV

Enthalpy
Prove that H = qp
pressure (= qp)
Based on the definition of enthalpy (H)
H = E + PV
Based on the definition of the 1st law of thermodynamics
E = q - PV
Where P = Pex at constant pressure
Therefore,
H = qp - PV + PV
i.e. H = qp

Enthalpy
pressure (= qp)
H is also a state function and could be +ve or –ve

Calorimetry
It is the measure of heat (q) flow during a chemical
reaction
Heat (q) flow takes place at
 constant volume (qv) , or
 constant pressure (qp)
 Constant volume Calorimetry
 Constant pressure Calorimetry

Calorimetry
reaction
Heat Capacity
is the amount of heat required by an object to raise
its temperature by 1 K or 1oC (units: J/oC or J/K)
Molar
Heat Capacity
(per mole)
Specific
Heat Capacity
(per gram)

Calorimetry
Molar Heat Capacity
is the amount of heat required by 1 mol of an
object to raise its temperature by 1 K or 1oC
(units: J/mol oC or J/mol K)
Specific Heat Capacity
is the amount of heat required by 1 gram of an
object to raise its temperature by 1oC
(units: J/g oC or J/g K)
Therefore:
Molar heat capacity =
(Specific heat capacity) x (molecular weight)

Calorimetry
reaction
Constant volume
Calorimetry
Constant pressure
Calorimetry
Also known as Bomb Calorimetry Also known as Coffee Cup Calorimetry

Calorimetry
Constant pressure Calorimetry
Heat measured at constant pressure
(qp) is also known as enthalpy (H)
Heat released (qp or H) is -ve
Heat absorbed (qp or H) is +ve
Recall:
 Endothermic reaction
 Exothermic reaction
Here, A reaction is the system, while
a solution is the surroundings
qrxn = - qsol

Calorimetry
Heat measured at constant pressure
(qp) is also known as enthalpy (H)
Here, A reaction is the system, while
a solution is the surroundings
qrxn = - qsol
= -[(specific heat of solution) x (grams of solution) x T]

Calorimetry
Solution Approach
Example:
When a 9.55 g sample of solid sodium hydroxide dissolves
in 100.0 g of water in a coffee-cup calorimeter, the
temperature rises from 23.6oC to 47.4oC.
Calculate H (in kJ/mol NaOH) for the solution process:
NaOH (s)  Na+ (aq) + OH- (aq)
Assume that the specific heat of the solution is the same as
that of pure water (4.18 J/g.oC)

Calorimetry
Example:
Total mass (m)
= (9.55+100.00)
= 109.55 g
Change in
temperature
T = 47.4-23.6
= 23.8oC
Heat is released
These mean
a Constant P
calorimetry
qp (H) should
be per mole of
NaOH
nNaOH = 9.55/40
=0.2388
Because
water is the
majority

Calorimetry
Solution:
Example:
H = qsol = - [(specific heat of solution) x (grams of solution) x T]
H = - [4.18 x 109.55 x 23.8] = - 10.898 kJ
H = - 10.898 / 0.2388 = - 45.64 kJ/mol of NaOH

Calorimetry
Solution Approach
Example:
A 25 g piece of iron (Fe) at 398 K was placed in 25 mL of
water at 298 K. What will be the final temperature of Fe?
(Specific heat of Fe = 0.449 J/g.K,
Density of water = 1.0 g/mL)

Calorimetry
Example:
A 25 g piece of iron (Fe) at 398 K was
placed in 25 mL of water at 298 K. What
will be the final temperature of Fe?
Fe is at higher temperature
than water. So, heat
transfers from Fe to Water
i.e. qFe = - qH2O
Until both have same Tf
Mass of water
= density x volume
= 1.0 x 25 = 25 g

Solution:
Here, qFe = - qH2O
[0.449 x 25 x (Tf - 398)] = - [4.18 x 25 x (Tf - 298)]
0.449 Tf – 178.7 = - 4.18Tf + 1245.6
 Tf = 307.7 K
Calorimetry
Example:
A 25 g piece of iron (Fe) at 398 K was placed in 25 mL of
water at 298 K. What will be the final temperature of Fe?
q = - [(specific heat) x (mass in grams) x T]

Calorimetry
Example:
The H for the solution process when solid NaOH dissolves
in water is 44.4 kJ/mol. When a 13.9 g sample of NaOH
dissolves in 250.0 g of water in a coffee-cup calorimeter, the
temperature increases. If the initial temperature is 23.0oC,
calculate the final temperature, assuming that the Specific
heat of the solution is that of liquid water as 4.18 J/gK
Solution Approach

Calorimetry
Example:
H is – 44.4 kJ/mol
Because temperature
increases
H is – 44.4 kJ/mol
It has to be made
in kJ or J ONLY

Solution:
Calorimetry
Example:
q = - [(specific heat) x (mass in grams) x T]
 - 15,429 = - 4.18 x (13.9+250) x T
 T = 13.98  Tf = 36.98oC
q = - H = - (44.4 x 103) x (13.9/(23+16+1)) = - 15,429 J

Calorimetry
Constant volume Calorimetry
This is used for combustion reactions,
especially of organic compounds
Most of these reactions release heat,
i.e. exothermic
Equation used to calculate qv here is:
Where:
Ccal is called calorimeter constant
T is the temperature difference
q = - Ccal x T

Calorimetry
Example:
A 1.800 g sample of phenol (C6H5OH) was burned in a bomb
calorimeter whose total heat capacity is 11.66 kJ/oC. The
temperature of the calorimeter plus contents increased
from 21.36oC to 26.37oC.
a) Write a balanced chemical equation for the bomb
calorimeter reaction.
b) What is the heat of combustion:
i) per gram of phenol?
ii)per mole of phenol?
Solution Approach

Calorimetry
Example:
A 1.800 g sample of phenol (C6H5OH) was burned in a
bomb calorimeter whose total heat capacity is 11.66 kJ/oC.
The temperature of the calorimeter plus contents increased
This is the Ccal value
T = 26.37-21.36
= 5.01 oC
Constant vol Calorimeter
i.e. qv = -Ccal x T
nPhenol = 1.8/94
= 0.019

Calorimetry
Example:
Solution:
Burning or combustion usually takes place in air (presence of O2)
C6H5OH + 7O2  6CO2 + 3H2O

Calorimetry
Example:
Solution:
q = - Ccal x T = - 11.66 x 5.01 = - 58.42 kJ
q (per gram of phenol) = - 58.42 / 1.8 = - 32.45 kJ/g
q (per mol of phenol) = - 58.42 / 0.019 = - 3074.74 kJ/mol

Enthalpy of reaction
ThermoChemistry

pressure (= qp)
The change in enthalpy,
H, is the enthalpy of the
products minus the
enthalpy of the reactants:
H = Hproducts − Hreactants
H, is called the enthalpy
of reaction, or the heat of
reaction

Characteristics of H
o Enthalpy (H) is an extensive property
 If a reaction is multiplied or divided by a factor,
H will be multiplied or divided by the same factor
o Enthalpy (H) of a forward reaction has the same value
but different sign of that of the opposite reaction
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = - 890 kJ
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(l) H = -1780kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = - 890 kJ
CO2(g) + 2H2O(g)  CH4(g) + 2O2(l) H = +890 kJ
o Enthalpy (H) of a reaction depends on the state of matter
of all reactants and products
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ

Example:
Consider the following reaction:
2Mg (s) + O2 (g)  2MgO (s) H = -1204 kJ
a) Is this reaction exothermic or endothermic?
b) Calculate the amount of heat transferred when 2.4 g of
Mg(s) reacts at constant pressure.
c) How many grams of MgO are produced during an
enthalpy change of – 96.0 kJ?
d) How many kilojoules of heat are absorbed when 7.50 g
of MgO (s) is decomposed into Mg (s) and O2 (g) at
constant pressure?
Solution Approach

Example:
2Mg (s) + O2 (g)  2MgO (s) H = -1204 kJ
a) Is this reaction exothermic or endothermic?
Solution:
 Based on the sign of H, this reaction is exothermic
b) Calculate the amount of heat transferred when 2.4 g of
Mg(s) reacts at constant pressure.
Solution:
 This H is for the combustion of 2 moles of Mg
 we have 2.4/24 = 0.1 moles of Mg
 i.e. H = (-1204) x (0.1/2) = - 60.2 kJ

Example:
2Mg (s) + O2 (g)  2MgO (s) H = -1204 kJ
c) How many grams of MgO are produced during an
enthalpy change of – 96.0 kJ?
Solution:
 This H is for the production of 2 moles of MgO
 if -96.0 kJ are to be produced, nMgO = (2 x 96) / 1204
= 0.159 moles
 Number of grams of MgO = 0.159 x (24+16)
= 6.36 grams

Example:
2Mg (s) + O2 (g)  2MgO (s) H = -1204 kJ
d) How many kilojoules of heat are absorbed when 7.50 g
of MgO (s) is decomposed into Mg (s) and O2 (g) at
constant pressure?
Solution:
 If MgO is to be decomposed, then an opposite reaction
is the case:
2MgO (s)  2Mg (s) + O2 (g) H = +1204 kJ
 This H (+ 1204 kJ) is for the decomposition of 2 moles
of MgO
 For the decomposition of 7.5 grams of MgO (0.1875
moles), H = (1204) x (0.1875/2) = + 112.875 kJ

Hess’s Law
If a reaction is carried out in a series of steps, H for
the overall reaction will be equal to the sum of the
enthalpy changes for the individual steps
H1 = H2 + H3
It can be understood
in terms of the fact
that H is a state
function

Hess’s Law
Example:
Calculate the enthalpy change for the reaction:
P4O6 (s) + 2O2 (g)  P4O10 (s)
Given the following enthalpies of reaction:
P4 (s) + 3O2 (g)  P4O6 (s) H = -1640.1 kJ
P4 (s) + 5O2 (g)  P4O10 (s) H = -2940.1 kJ
Solution Approach

Hess’s Law
Solution:
Consider the target reaction # 3
(3) P4O6 (s) + 2O2 (g)  P4O10 (s)
Consider the given equations # 1 and # 2:
(1) P4 (s) + 3O2 (g)  P4O6 (s) H1 = -1640.1 kJ
(2) P4 (s) + 5O2 (g)  P4O10 (s) H2 = -2940.1 kJ
Since P4O6 is a reactant in equation #3, equation #1 should
be reversed (the sign of H will be reversed), then add it to
equation # 2
P4O6 (s)  P4 (s) + 3O2 (g) H1 = +1640.1 kJ
P4 (s) + 5O2 (g)  P4O10 (s) H2 = -2940.1 kJ
P4O6 (s) + 2O2 (g)  P4O10 (s) H3 = -1300.0 kJ

Enthalpy of formation
ThermoChemistry

If 1 mol of compound is formed from its constituent
elements, then the enthalpy change for this reaction
is called the “enthalpy (or heat) of formation”, Hf
Example:
C (s) + O2 (g)  CO2 (g) Hf = -395.4 kJ/mol
Enthalpy (heat) of
formation of CO2
If 1 mol of compound is formed from its constituent
elements under standard conditions, then the
enthalpy change for this reaction is called the
“standard enthalpy (or heat) of formation”, Ho
f
Note:
Standard conditions are: 1 atm and 25oC (298 K).

Examples:
Note:
Ho
f is given in kJ/mol

o If there is more than one state for a substance under
standard conditions, the most stable one is used.
o Standard enthalpy of formation (Ho
f) of the most
stable form of an element is zero.
o We use Hess’ Law to calculate enthalpies of a reaction
from enthalpies of formation.
Rules:

Example:
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
What is Ho
rxn ?
Solution
The first step is to confirm the balance of the equation, and
write the Ho
rxn based on Hess’s Law
 The equation is balanced and its Ho
rxn is given as:
Ho
rxn = 3Ho
f (CO2) + 4Ho
f (H2O) – Ho
f (C3H8)
Based on the standard Hf of CO2 (g), H2O (l) and C3H8 (g)
in the table (previous slide),
 Ho
rxn = (3 x -393.5) + (4 x -285.8) - (-103.85)
= - 2,218.65 kJ

Example:
Given the following standard enthalpy of reaction, use the
standard enthalpies of formation of the reactants and
products to calculate the standard enthalpy of formation of
CuO (s)
CuO (s) + H2 (g)  Cu (s) + H2O (l), Ho
rxn = -129.7 kJ
Solution
First, confirm the balance of the equation, then write the
Ho
rxn based on Hess’s Law
 The equation is balanced and its Ho
rxn is given as:
Ho
rxn = Ho
f (H2O) – Ho
f (CuO) (Note: Ho
f of H2 & Cu = 0)
-129.7 = - 285.8 - Ho
f (CuO)
 Ho
f (CuO) = -156.1 kJ

Example:
Use the enthalpies of the following reactions to calculate
H for the reaction of ethylene with F2 (equation # 4)
Solution
Based on the target equation (4), the following steps are needed:
oRev # 3, C2H4 (g)  2C (s) + 2H2 (g) H3 = - 52.3 kJ
oMultiply # 1 by 2 2H2 (g) + 2F2 (g)  4HF H1 = -1074 kJ
oMultiply # 2 by 2 2C (s) + 4F2 (g)  2CF4 (g) H2 = -1360 kJ
o Add  C2H4 (g) + 6F2 (g)  2CF4 (g) + 4HF (g) H4 = - 2486.3 kJ
(1) H2 (g) + F2 (g)  2HF H1 = -537 kJ
(2) C (s) + 2F2 (g)  CF4 (g) H2 = -680 kJ
(3) 2C (s) + 2H2 (g)  C2H4 (g) H3 = + 52.3 kJ
(4) C2H4 (g) + 6F2 (g)  2CF4 (g) + 4HF (g) H4 ????

End of Chapter 5
ThermoChemistry

Chapter 5 final(1)

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