Week 8.1 thermodynamic and equilibriaPresentation Transcript
Thermodynamic and Equilibria
First Law of Thermodynamics
Second Law of Thermodynamics
Physical Transformation of Pure Substances
Mrs Faraziehan Senusi
Reference: Chemistry: the Molecular Nature of Matter and Change,
6th ed, 2011, Martin S. Silberberg, McGraw-Hill
• Thermodynamics ~ energy changes that accompany
physical and chemical processes. Usually these
energy changes involve heat.
• Thermochemistry ~ is the study of heat change in
concerned with how we observe, measure, and
predict energy changes for both physical changes and
use energy changes to tell whether or not a given
process can occur under specified conditions to give
predominantly products (or reactants) and how to
make a process more (or less) favorable.
• System :
specific part of the universe that is of interest to us.
The substances involved in the chemical and
physical changes that we are studying
• Surroundings: the rest of the universe
Open system: can exchange
mass and energy (heat) with
• system where mass and energy
can cross the boundary
transfer of heat but not mass
• consists of a fixed amount of
mass and no mass can cross its
• Energy in the form of heat or
work can cross the boundary
Isolated system: does not
allow transfer either mass or
• A system where no mass, heat
and work can cross the boundary.
THE FIRST LAW OF THERMODYNAMICS
• The first law of thermodynamics (the conservation of energy
principle) provides a sound basis for studying the relationships among the
various forms of energy and energy interactions.
The first law states that energy can be neither
created nor destroyed; it can only change forms.
When a rock falls, the
decrease in potential
energy is equals to the
increase in kinetic
The increase in the energy of a potato
in an oven is equal to the amount of
heat transferred to it.
INTERNAL ENERGY, E
Each particle in a system has potential and kinetic energy;
the sum of these energies for all particles in a system is the
internal energy, E.
In a chemical reaction: when reactants are converted to
products, E changes (DE).
DE = Efinal - Einitial = Eproducts - Ereactants
Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings
Heat and Work
Energy transfer outward from the system or inward from
the surroundings can appear in two forms, heat and work .
DE = q + w
where q = heat and w = work
Heat (or thermal energy, symbol q) is the energy transferred
between a system and its surroundings as a result of a
difference in their temperatures only.
All other forms of energy transfer (mechanical, electrical, and
so on) involve some type of work (w), the energy transferred
when an object is moved by a force.
Sign Conventions for q, w and DE
The numerical values of q and w can be either positive or
negative, depending on the change the system undergoes.
Energy coming into the system is positive; energy going out
from the system is negative.
depends on magnitudes of q
depends on magnitudes of q
For q: (+) means system gains heat, (-) means system loses heat.
For w: (+) means work done on system (compression),
(-) means work done by system,(expansion).
Law of Conservation of Energy
(First Law of Thermodynamics)
The energy of the system plus the energy of the surroundings remains constant:
energy is conserved.
DEsystem + DEsurroundings = 0
Units of Energy
1 J = 1 kg m2/s2
1 cal = 4.184 J
British Thermal Unit
1 Btu = 1055 J
Thermodynamic state of a system
• The properties of a system—such as P, V, T—are
called state functions
• The value of a state function depends only on the
state of the system and not on the way in which the
system came to be in that state.
• A change in a state function describes a difference
between the two states. It is independent of the
process or pathway by which the change.
• The most important use of state functions in
thermodynamics is to describe changes.
ΔX = ΔXfinal – ΔXinitial
• When X increases, the final value is greater than
the initial value, so ΔX is positive; a decrease in X
makes ΔX a negative value.
• We can determine the energy change associated with a
chemical or physical process by using an experimental
technique called calorimetry.
• This technique is based on observing the temperature
change when a system absorbs or releases energy in the
form of heat.
• The experiment is carried out in a device called a
calorimeter, in which the temperature change of a known
amount of substance (often water) of known specific heat
• The temperature change is caused by the absorption or
release of heat by the chemical or physical process under
Specific heat, c : amount required to raise temperature of one
gram of the substance by one degree Celsius (J/g.oC)
• Heat capacity or calorimeter constant, C : amount of heat
required to raise the temperature of a given quantity of the
substance by one degree Celsius (J/oC)
C = mc (where m is the mass)
• If specific heat and amount of substance is known, change in
sample temperature ∆T will tell us the amount of heat, q, that
has been released/ absorbed in particular process.
q = mc∆T
• q is positive (endothermic) and negative for exothermic
• Calorimeter can be used to measure the amount of
heat absorbed or released when a reaction takes place
in aqueous solution.
We add 3.358 kJ of heat to a calorimeter that contains 50.00 g of
water. The temperature of the water and the calorimeter, originally
at 22.34°C, increases to 36.74°C. Calculate the heat capacity of the
calorimeter in J/°C. The specific heat of water is 4.184 J/g.°C.
Calculate the amount of heat gained by the water in the
The rest of the heat must have been gained by the
Determine the heat capacity of the calorimeter.
amount of heat gained by the water
amount of heat gained by the calorimeter
A 50.0 mL sample of 0.400 M copper(II) sulfate solution at 23.35°C is
mixed with 50.0 mL of 0.600M sodium hydroxide solution, also at
23.35°C, in the coffee-cup calorimeter. Heat capacity of calorimeter is
24.0 J/ C. After the reaction occurs, the temperature of the resulting
mixture is measured to be 25.23°C. The density of the final solution is
1.02 g/mL. Calculate the amount of heat evolved. Assume that the
specific heat of the solution is the same as that of pure water, 4.184
The amount of heat released by the reaction is absorbed by the
calorimeter and by the solution.
To find the amount of heat absorbed by the solution, we must
know the mass of solution; to find that, we assume that the
volume of the reaction mixture is the sum of volumes of the
The practice of calorimetry
• Two common types are :
A "coffee-cup" calorimeter is often used to
measure the heat transferred (qp) in processes
open to the atmosphere.
One type of constant-volume apparatus is the
bomb calorimeter, designed to measure very
precisely the heat released in a combustion
• One common use is to find the specific heat
capacity of a solid that does not react with or
dissolve in water.
• The solid (system) is weighed, heated to some
known temperature, and added to a sample of
water (surroundings) of known temperature and
mass in the calorimeter.
• With stirring, the final water temperature, which
is also the final temperature of the solid, is
• The heat lost by the system (-qsys, or -qsolid) is
equal in magnitude but opposite in sign to the
heat gained by the surroundings (+qsurn or +qH2O):
- qsolid = q H2O
- (c solid x mass solid x ΔT solid ) = c H2O x mass H2O x ΔTH2O
Determining the Specific Heat Capacity of a Solid
PROBLEM: A 25.64 g sample of a solid was heated in a test tube to 100.00 oC in boiling
water and carefully added to a coffee-cup calorimeter containing 50.00 g of
water. The water temperature increased from 25.10 oC to 28.49 oC.
What is the specific heat capacity of the solid? (Assume all the heat is
gained by the water)
It is helpful to use a table to summarize the data given. Then work the problem
realizing that heat lost by the system must be equal to that gained by the
25.64 g x
csolid = - 50.00 g x
4.184 J/g.K x 3.39 K
25.64 g x
4.184 J/g.K x 3.39 K
= - 50.00 g x
Figure 6.8 depicts the preweighed
combustible sample in a metal-walled
chamber (the bomb), which is filled
with oxygen gas and immersed in an
insulated water bath fitted with
motorized stirrer and thermometer.
A heating coil connected to an
electrical source ignites the sample, and
the heat evolved raises the temperature
of the bomb, water, and other
Because we know the mass of the
sample and the heat capacity of the
entire calorimeter, we can use the
measured ΔT to calculate the heat
Calculating the Heat of Combustion
A manufacturer claims that its new diet 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 heat capacity of the calorimeter =
8.151 kJ/K). The temperature increases by 4.937 oC. Is the
manufacturer’s claim correct?
- qsample = qcalorimeter
= heat capacity x DT
= 8.151 kJ/K x 4.937 K
= 40.24 kJ
40.24 kJ x
= 9.62 kcal < 10 Calories = 10 kcal
The manufacturer’s claim is correct.
• The quantity of heat transferred into or out of a
system as it undergoes a chemical or physical
change at constant pressure.
• Extensive property: magnitude depends on amount
of substance present
• Impossible to determine enthalpy of substance
• Measure change in enthalpy, ∆H
• Enthalpy of reaction, ∆H
∆H = H (product) – H (reactant)
• Exothermic : negative
• Endothermic: positive
• EXOTHERMIC PROCESS – a process that
releases energy in the form of heat into its
surroundings. (Ex: combustion reaction)
• ENDOTHERMIC PROCESS – a process that
absorbs energy from its surroundings
Enthalpy diagrams for exothermic and endothermic processes
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + heat
heat + H2O(s)
Heat is released;
Heat is absorbed;
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
1C4H10(l) + 13/2O2(g)
4CO2(g) + 5H2O(g)
heat of formation (DHf)
K(s) + 1/2Br2(l)
heat of fusion (DHfus)
heat of vaporization (DHvap)
of either reactant
or product: 1 mol
• A balanced chemical equation, together with its
value of ΔH
• The ΔHrxn value shown refers to the amounts (moles)
of substances and their states of matter in that
Combustion of methane :
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O(l)
• 1367 kJ of heat is released when one mole of C2H5OH(l) reacts
with three moles of O2(g) to give two moles of CO2(g) and three
moles of H2O(l).
• We can refer to this amount of reaction as one mole of reaction,
which we abbreviate ―mol rxn.‖
• We can also write the thermochemical equation as
We always interpret ΔH as the enthalpy change for the reaction
as written; as (enthalpy change)/(mole of reaction), where the
denominator means ―for the number of moles of each substance
shown in the balanced equation.‖
– Stoichiometric coefficient refer to number of moles of
– Reverse equation, magnitude of ∆H same but sign
– Multiply equation by factor of n then ∆H must also
change by the same factor
– Must always specify the physical state
Summary of the relationship between
amount (mol) of substance and the heat
(kJ) transferred during a reaction
of compound A
molar ratio from
of compound B
gained or lost
Using the Heat of Reaction (DHrxn) to Find Amounts
The major source of aluminum in the world is bauxite (mostly aluminum
oxide). Its thermal decomposition can be represented by:
2Al(s) + 3/2O2(g)
DHrxn = 1676 kJ
If aluminum is produced this way, how many grams of aluminum can form
when 1.000 x 103 kJ of heat is transferred?
1.000 x 103 kJ x
2 mol Al
1676 kJ = 2 mol Al
mol of Al
x M (g/mol)
g of Al
= 32.20 g Al
26.98 g Al
1 mol Al
Standard heat of reaction
Specifying Standard States
For a gas, the standard state is 1 atm; ideal gas behavior is
For a substance in aqueous solution, the standard state is
1 M concentration (1 mol/liter solution).
For a pure substance (element or compound), the standard state
is usually the most stable form of the substance at 1 atm and the
temperature of interest (usually 25 oC (298 K).
DHorxn = standard heat of reaction
(enthalpy change determined with all
substances in their standard states)
Standard enthalpy of formation
• Standard enthalpy of formation (∆Hof) : heat
change that results when one mole of a compound
is formed from its elements at a pressure of 1 atm
• ∆Hof of any element in its most stable form is
∆Hof (O2) = 0
∆ Hof (O3) ≠ 0
Direct method of measuring ∆Hof
In a formation equation, 1 mol of a compound forms
from its elements. The standard heat of formation
(DHof) is the enthalpy change for the formation
equation when all substances are in their standard
C(graphite) + 2H2(g)
C(graphite) + O2 (g) CO2 (g)
DHof = -74.9 kJ
∆Horxn= - 393.5kJ
An element in its standard state is assigned a DHof of 0.
Indirect method measuring ∆Hof
• Many compounds cannot be directly synthesized from
their elements (proceed too slowly/ side reactions produce
other substance than desired compound)
• Hess’s law: when reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps
• General rule of applying:
– Arrange series of chemical equation in a way that when
added together all species will cancel except reactant
and product that appear in the overall reaction
– Multiply or reverse equation
Hess’s Law of Heat Summation
The enthalpy change of an overall process is the sum of
the enthalpy changes of its individual steps.
Used to predict the enthalpy change (a) of an overall reaction
that cannot be studied directly, and/or (b) of an overall reaction
that can be separated into distinct reactions whose enthalpy
changes can be measured individually.
Using Hess’s Law to Calculate an Unknown DH
Two gaseous pollutants that form auto exhaust are CO and NO. An
environmental chemist is studying ways to convert them into less harmful
gases through the following equation:
CO(g) + NO(g)
CO2(g) + 1/2N2(g)
DH = ?
Given the following information, calculate the unknown DH.
Equation A: CO(g) + 1/2O2(g)
2NO(g) DHB = +180.6 kJ
Equation B: N2(g) + O2(g)
CO2(g) DHA = -283.0 kJ
Equations A and B have to be manipulated by reversal and/or multiplied by
factors in order to sum to the target equation.
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
CO(g) + NO(g)
1/2N2(g) + 1/2O2(g)
CO2(g) + 1/2N2(g)
DHA = -283.0 kJ
DHB = -90.3 kJ
DHrxn = -373.3 kJ
• From ∆Hof values, the standard enthalpy of
reaction ∆Horxn can be calculated
aA + bB cC + dD
∆Horxn = [c∆Hof (C) + d∆Hof (D)] – [a∆Hof (A) + b∆Hof (B)]
∆Horxn = ∑n∆Hof (Product) –∑m∆Hof(Reactant)
The general process for determining DHorxn from DHof values
Calculating the Standard Heat of Reaction from
Standard Heats of Formation
PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kg, is
used to make many products, including fertilizers, dyes and explosives.
The first step in the industrial production process is the oxidation of
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate DHorxn from DHof values.
DHorxn = S nDHof (products) - S mDHof (reactants)
DHorxn = [4DHof NO(g) + 6DHof H2O(g)]
- [4DHof NH3(g) + 5DHof O2(g)]
= [(4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol)] [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
DHorxn = -906 kJ
• Chemical reactions involve the breaking and making of chemical
• Energy is always required to break a chemical bond.
• The bond energy (B.E.) is the amount of energy necessary to
break one mole of bonds in a gaseous covalent substance to form
products in the gaseous state at constant temperature and pressure.
• The greater the bond energy, the more stable (stronger) the bond
is, and the harder it is to break. Thus bond energy is a measure of
• A special case of Hess’s Law involves the use of bond energies to
estimate heats of reaction.
A schematic representation of the relationship between bond energies
and ΔHrxn for gas phase reactions.
(a) For a general reaction (exothermic)
(b) For the gas phase reaction : H2 (g) + Br2 (g) 2HBr (g)
Use the bond energies listed in Table 15-2 to estimate the heat
of reaction at 298 K for the following reaction:
Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K
for the following reaction: