Thermochemistry
By: Ms. Buroker
Thermochemistry is the
study of the relationship
between chemical
reactions and energy
changes.
The Nature of Energy
We define energy as the capacity to do work or
to transfer heat.
HEAT is the form of energy that flows between
2 objects because of their difference in
temperature.
The Law of Conservation of Energy states that
energy can be converted from one form to
another but can be neither created nor
destroyed.
The Nature of Energy
Some Definitions ….
• Thermal Energy: energy associated with the
random motion of molecules
• Chemical Energy: energy “stored” due to the
structure of chemical substances
• Kinetic Energy: energy of motion
• Potential Energy: available energy due to an
object’s position
• Radiant Energy: energy in the form of
electromagnetic radiation
Potential Energy
Potential Energy: energy due to position
or composition. “Stored” energy
resulting from attractions and
repulsions an object experiences in
relation to other objects.
1.) Electrostatic Forces between
charged particles
2.) Chemical Energy
3.) Thermal Energy
Kinetic Energy
Kinetic Energy: energy in motion
Ek = ½ mv2
Mass
Velocity
Potential and Kinetic Enegry
Potential Energy can be converted to kinetic energy and do work in the
process.
Units of Energy:
SI Unit: Joule, (J)
1J = 1kg-m2/s2
Cgs Unit: Erg
1Erg = 1g-cm2/s2
Units of Energy
The joule is not very large, so energies
associated with chemical reactions are
commonly expressed in kJ.
Non- SI Base Unit commonly used …
calorie (cal)
Calorie = the amount of energy required
to raise the temperature of 1g of water 1○C.
1 cal = 4.184J (exactly)
Internal Energy of the System
Kinetic (thermal) energy:
Translational
Rotational
Vibrational
Potential Energy:
Nuclear Forces
Electrostatic attractions (Chemical energy):
Intramolecular - Bonds between atoms
Intermolecular - Attractions between molecules
Temperature vs. Heat
Temperature: a property that
reflects the random motions of the
particles in a particular substance.
Heat: involves the transfer of energy
between two objects due to a
temperature difference … it is NOT a
substance contained in the object!
Temperature vs. Heat Continued…
The more thermal energy a substance has, the
greater the motion of its atoms and molecules.
The total thermal energy in an object is the sum of
the individual energies of all the atoms, molecules,
or ions in that object.
So …
The thermal energy of a given substance depends
not only on temperature but also on the amount
of substance. (E.g. Warm bath vs. hot coffee)
State Function
A State Function or State Property is a
property of a system that depends only on
its present state.
* It does not depend on how the system
arrived at the present state, it depends
only on the characteristics of the present
state.
* A change in this function or property
in going from one state to another state is
independent of the particular pathway
taken between the two states.
An Example of a State Function
Elevation on the earth’s surface vs.
distance between to points.
Chicago (elevation 674 ft)
Denver (elevation 5280 ft)
Elevation = State Function
Distance = Not a State Function
Energy, Pressure, Volume, and Temperature are State
Functions.
Energy is a
state function
while, heat
and work are
NOT!!!!
System and Surrounding
System: the portion we single out for
our study
Surrounding: everything else
• When studying chemical reactions …
the chemicals are the system and the
containers and everything else are
the surroundings.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + energy(heat)
Three Types of Systems
1. Open system is one in which both matter and energy
(usually in the form of heat) are “flowing into and out
of the system from the surroundings, i.e., there is
exchange of mass and energy with its surroundings.
2. Closed system is one in which matter does NOT flow
into or out of the system but energy does flow, i.e.,
there can be an exchange of energy but not of mass.
3. Isolated system is one in which neither mass nor
energy can flow from surroundings to the system or
from the system to the surroundings, i.e., there is no
exchange of either mass or energy .
Thermal Equilibrium
Heat transfer occurs when two objects at different temperatures
are brought into contact.
Heat energy is transferred until the system comes to thermal
equilibrium.
Within a system, the heat lost by the hotter object is equal to the
heat gained by the cooler object.
When heat transfer occurs across the boundary between a
system and the surrounding, the directionality of the heat transfer
is described as:
Exothermic: heat is transferred from a system to the
surroundings … means “out of” (negative)
Endothermic: Heat is transferred from the surroundings to the
system … means “in to” (positive)
The First Law of Thermodynamics
The energy of the universe is
constant.
Internal Energy, E, of a system can be
defined most precisely as the sum of
the kinetic and potential energies of all
the “particles” in the system.
∆E = Eproducts – Ereactants
∆E = q + w
Heat Transfer
work
Specific Heat Capacity and Energy
Transfer
Specific Heat: The amount of energy required to
increase one gram of a substance by 1K (or oC)
∆E = q + w
The sign reflects the system’s point of view.
If energy flows into the system via heat … q is
positive and is thus an endothermic process.
If energy flows out of the system via heat … q is
negative and is thus an exothermic process.
Specific Heat and Energy Transfer
Endothermic vs. Exothermic
Sample Problems
1. In an experiment it was determined that 59.8J was required to
change the temperature of 25.0g of ethylene glycol (a compound
used as antifreeze in automobile engines) by 1.00K. Calculate the
specific heat capacity of ethylene glycol from these data.
2. A 15.5g piece of chromium, heated to 100.0oC, is dropped into
55.5g of water at 16.5oC. The final temperature of the metal and
the water is 18.9oC. What is the specific heat capacity of
chromium? (Assume no heat is lost to the container or to the
surrounding air.)
3. A piece of iron (400.g) is heated in a flame and then dropped
into a beaker containing 1,000.g of water. The original
temperature of the water was 20.0oC, and the final temperature of
the water and iron is 32.8oC after thermal equilibrium has been
attained. What was the original temperature of the hot iron bar?
(Assume no heat is lost to the beaker or to the surrounding air.)
You need the specific heat capacity of iron = 0.449J/gK
Answer: Sample Problem #1
Sample Problems
1. In an experiment it was determined that 59.8J was
required to change the temperature of 25.0g of
ethylene glycol (a compound used as antifreeze in
automobile engines) by 1.00K. Calculate the
specific heat capacity of ethylene glycol from these
data.
Answer:
q= c x m x DT
Specific Heat Capacity Units = J/gK
59.8J / [(25.0g)(1.00K)] = 2.39 J/gK
Answer: Sample Problem #2
Sample Problems
2. A 15.5g piece of chromium, heated to 100.0oC, is dropped into
55.5g of water at 16.5oC. The final temperature of the metal and the
water is 18.9oC. What is the specific heat capacity of chromium?
(Assume no heat is lost to the container or to the surrounding air.)
Answer:
Heat lost by chromium:
-q = C(15.5g)(18.9oC – 100.0oC) = -1257.05(C)
Heat gained by water:
q = (55.5g)(4.184J/goC)(18.9oC – 16.5oC) = 557.309J
1257.05g(C) = 557.309J
C = 557.309 / 1257.05 = .443J/goC
Answer: Sample Problem #3
Sample Problems
3. A piece of iron (400.g) is heated in a flame and then dropped
into a beaker containing 1,000.g of water. The original
temperature of the water was 20.0oC, and the final temperature of
the water and iron is 32.8oC after thermal equilibrium has been
attained. What was the original temperature of the hot iron bar?
(Assume no heat is lost to the beaker or to the surrounding air.)
You need the specific heat capacity of iron = 0.449J/gK
Answer:
Ti = 330.992oC = 331oC
Phase Changes
Solid → Liquid →Gas
Heat of Fusion: Heat required to convert a
solid at its melting point to a liquid
Heat of Vaporization: Heat required to convert
a liquid at its boiling point to a gas.
Heat of Sublimation: Heat required to take a
substance directly from its solid to its gas
state.
Note that the temperature of a substance does not change as it is
going through a phase change.
Example Problem
How much heat must be absorbed to warm 25.0g of liquid
methanol, CH3OH, from 25.0oC to its boiling point (64.6oC) and then
to evaporate the methanol completely at that temperature? The
specific heat capacity of liquid methanol is 2.53J/gK. The heat of
vaporization of methanol is 2.00kJ/g
Answer:
Step 1
q1 =c x m x DT
q1 = (2.53J/gK) x (25.0g) x (64.6oC – 25.0oC)
q1 = 2504.7J
Step 2
q2 = 25.0g (2000.0J/g)
q2= 50,000J
Work Associated with Chemical Processes
Work done by a gas … expansion
Work done to a gas … compression
W = -P∆V
P= external pressure
-∆V= compression
+∆V= expansion
NOTE: For an ideal gas, work can occur ONLY when its
volume changes. Thus, if a gas is heated at constant
volume, the pressure increases but no work occurs.
Enthalpy
Thermodynamics
defines a new
function:
ENTHALPY(H) (q p)
as the heat flow part
of the change in
internal energy when
the pressure is kept
constant as it is in an
open container or in
an expandable
container.
DE = q - P (DV)
DE = q p - P (DV)
(q p is given the name
Enthalpy and the
symbol (H))
DE = DH - P (DV)
or re-arrange to
express it:
DH = DE + P (DV) =
Change in Enthalpy
Enthalpy Continued
At constant pressure,
the change in
enthalpy of the
system is equal to the
energy flow as heat.
DH = qp
DH = Hproducts - Hreatctants
Note: DH = qp ONLY at
constant pressure …
the change in
enthalpy of a system
has no easily
interpreted meaning
except at constant
pressure, where DH =
heat.
At constant pressure, exothermic means DH is negative ;
endothermic means DH is positive.
Enthalpy Changes For Chemical
Reactions
Enthalpy (heat) of Reactions
Enthalpy changes accompany chemical Reactions:
H2O(g) → H2(g) + ½ O2(g) DH = +241.8kJ (Endothermic)
For the opposite reaction …
H2(g) + ½ O2(g) → H2O(g)
H = -241.8kJ (Exothermic)
Calorimetry
The devise used to determine the
heat associated with a chemical
reaction is called a calorimeter.
Calorimetry is the science of
measuring heat, based on the
temperature change a body
undergoes when it absorbs or
discharges energy as heat.
Heat Capacity
The heat capacity of a substance, is defined
as:
C = heat absorbed
increase in temperature
Specific Heat Capacity: the energy required
to raise the temperature of one gram of a
substance by one degree Celsius.
Molar Heat Capacity: the energy required to
raise the temperature of one mole of a
substance by one degree Celsius.
Coffee Cup Calorimetry
Simple way of
measuring heat.
Two Styrofoam
cups are nested
together where the
inner cup holds
the solution in
which the reaction
takes place.
An example of constant-pressure calorimetry. Used in
determining the changes in enthalpy for reactions occurring in
solution.
Remember!! Under the conditions of constant pressure … enthalpy
equals heat!
Bomb Calorimetry
Constant- Volume Calorimetry
qreaction = -Ccal x DT
Calorimeter
Constant
Bomb Calorimetry ….
E = qv = qrxn = -(qbomb + qwater)
The energy transferred to the bomb is given
by the heat capacity (usually in J/K)
The energy transferred to the water is given
by the specific heat equation:
qwater = m x C x T
Example Problems
1. Assume you mix 200.mL of 0.400M HCl with 200.mL of
0.400M NaOH in a coffee-cup calorimeter. The temperature
of the solutions before mixing was 25.10oC; after mixing
and allowing the reaction to occur, the temperature is
27.78oC. What is the molar enthalpy of neutralization of the
acid? (Assume that the densities of all solutions are
1.00g/mL and their specific heat capacities are 4.20 J/gK)
2. A 1.00g sample of ordinary table sugar (sucrose
C12H22O11) is burned in a bomb calorimeter. The
temperature of 1.50x103g of water in the calorimeter rises
from 25.00oC to 27.32oC. The heat capacity of the bomb is
837J/K, and the specific heat of the water is 4.20J/gK.
a) Calculate the heat evolved per gram of sucrose.
b) Calculate the heat evolved per mole of sucrose.
Answer: Sample Problem #1
1. Assume you mix 200. mL of 0.400M HCl with 200. mL of 0.400M
NaOH in a coffee-cup calorimeter. The temperature of the
solutions before mixing was 25.10oC; after mixing and allowing the
reaction to occur, the temperature is 27.78oC. What is the molar
enthalpy of neutralization of the acid? (Assume that the densities
of all solutions are 1.00g/mL and their specific heat capacities are
4.20 J/gK)
HCl + NaOH → NaCl + H2O
Total volume of solutions = 200.mL + 200.mL = 400.mL
Total mass = 400.mL (1g/mL) = 400.g
Moles of HCl and NaOH (each) = .200L (0.400mol/L) = .0800mol
q = (400.g)(4.20J/gK)(27.78oC – 25.10oC) = 4502.4J
Molar enthalpy = 4502.4J / .0800mol = 56,280J/mol = 56.3kJ/mol
Answer: Sample Problem #2
2. A 1.00g sample of ordinary table sugar (sucrose C12H22O11) is
burned in a bomb calorimeter. The temperature of 1.50x103 g of
water in the calorimeter rises from 25.00oC to 27.32oC. The heat
capacity of the bomb is 837J/K, and the specific heat of the water
is 4.20J/gK.
a) Calculate the heat evolved per gram of sucrose.
b) Calculate the heat evolved per mole of sucrose.
a) Temperature change = 27.32oC – 25.00oC = 2.32oC
qwater = (1500g)(4.20J/gK)(2.32oC) = 14,616J
qbomb = 837J/K(2.32oC) = 1,941.84J
qrxn = -(14,616J + 1,941.84J) = -16,557.8J = -16.6kJ/g
b) Moles sucrose = 1.00g / (342.2965g/mol) = 2.921x10-3mol
qrxn = -16,557.8J / 2.921x10-3mol = 5.6685x106J/mol = 5.67x103kJ/mol
Hess’s Law
Since enthalpy is a state function, the
change in enthalpy in going from some
initial state to some final state is
independent of the pathway…..
So, in going from a particular set of
reactants to a particular set of
products, the change in enthalpy is the
same whether the reaction takes place
in one step or in a series of steps.
Characteristics of Hess’s Law
1.) If a reaction is reversed, the sign of
DH is also reversed.
2.) The magnitude of DH is directly
proportional to the quantities of
reactants and products in a reaction.
If the coefficients in a balanced
reaction are multiplied by an
integer, the value of DH is multiplied
by the same integer.
Example: Given the reaction information to the left, determine the
unknown enthalpy of reaction “a”.
Answer: Ha + -283.0kJ = -393.5kJ Ha = -110.5kJ
Example Problem #1
Use Hess’s law to calculate the enthalpy change for the formation
of CS2(l) from C(s) and S(s) from the following enthalpy values:
C(s) + O2(g) → CO2(g)
DH = -393.5kJ
S(s) + O2(g) → SO2(g)
DH = -296.8kJ
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
C(s) + 2 S(s) → CS2(l)
DH = -1103.9kJ
DH = ?
Hint: Remember you can reverse and multiply equations through.
Whatever you do to the reaction equation, you do to the enthalpy.
Answer: Example Problem #1
Use Hess’s law to calculate the enthalpy change for the formation
of CS2(l) from C(s) and S(s) from the following enthalpy values:
C(s) + O2(g) → CO2(g)
DH = -393.5kJ
S(s) + O2(g) → SO2(g)
DH = -296.8kJ
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) DH = -1103.9kJ
C(s) + 2 S(s) → CS2(l)
DH = ?
C(s) + O2(g) → CO2(g)
DH = -393.5kJ
2S(s) + 2O2(g) → 2SO2(g)
DH = -593.6kJ (Double)
CO2(g) + 2 SO2(g) → CS2(l) + 3 O2(g) DH = 1103.9kJ (Reverse)
C(s) + 2 S(s) → CS2(l)
H = -393.5kJ + -593.6kJ + 1103.9kJ = 116.8kJ
Example #2
Diborane (B2H6) is a highly reactive boron
hydride that was once considered as a
possible rocket fuel for the U.S. space
program. Calculate DH for the synthesis
of diborane from its elements, according
to the equation:
2B(s) + 3H2(g)  B2H6(g)
Use the following data:
(a) 2B(s) + 3/2O2(g)  B2O3(s)
DH= -1273kJ
(b) B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g)
DH = -2035kJ
(c) H2(g) + 1/2O2(g)  H2O(l)
DH = -286kJ
(d) H2O(l)  H2O(g)
DH = 44kJ
Hints for using Hess’s Law
Hint #1: Work backward from the required
reaction, using the reactants and
products to decide how to manipulate the
other given reaction at your disposal.
Hint #2: Reverse any reactions needed to
give the required reactants and products.
Hint #3: Multiply reactions to give the
correct number of reactants and
products.
Standard Enthalpies of Formation
Standard enthalpy of
formation (DHºf) for a
compound is the change in
enthalpy that accompanies
the formation of one mole of
a compound from its
elements with all substances
in their standard states.
The degree symbol on a
thermodynamic function, for
example, DHºf, indicates that the
corresponding process has been
carried out under standard
conditions.
The standard state for a
substance is a precisely defined
reference state.
Calculation
DHºreaction = SnpDHºf(products) – SnrDHºf(reactants)
Elements in their standard states are not
included in the DHreaction calculations.
That is, DHºf, for an element in its
standard state is zero.
Sample Problem
Calculate the standard enthalpy of combustion for benzene, C6H6.
C6H6(l) + 7.5 O2(g) → 6 CO2(g) + 3 H2O(l)
Horxn = ?
Hof[C6H6(l)] = +49.0kJ/mol. Other values needed can be found in the
appendix in the back of your book.
Horxn = [6(-393.51) + 3(-285.83)] – [1(49.0) + 7.5(0)]
= -3,267.55 = -3.27x103kJ/mol