Transcript Chapter 1: Matter and Measurement

General Chemistry
M. R. Naimi-Jamal
Faculty of Chemistry
Iran University of Science & Technology
‫فصل پنجم‪:‬‬
‫شیمی گرمایی‬
Contents
5-1
5-2
5-3
5-4
5-5
5-6
5-7
5-8
5-9
Getting Started: Some Terminology
Heat
Heats of Reaction and Calorimetry
Work
The First Law of Thermodynamics
Heats of Reaction: U and H
The Indirect Determination of H: Hess’s Law
Standard Enthalpies of Formation
Fuels as Sources of Energy
Thermochemistry: Basic Terms
• Thermochemistry is the study of energy changes
that occur during chemical reactions.
• System: the part of the universe being studied.
• Surroundings: the rest of the universe.
Types of Systems
• Open: energy and matter can be
exchanged with the surroundings.
• Closed: energy can be exchanged
with the surroundings, matter
cannot.
• Isolated: neither energy nor matter
can be exchanged with the
surroundings.
A closed system;
energy (not matter)
can be exchanged.
Terminology
• Energy, U
– The capacity to do work.
• Work
– Force acting through a distance.
• Kinetic Energy
– The energy of motion.
Energy
• Kinetic Energy
ek =
1
2
mv2
kg m2
= J
[ek ] =
2
s
• Work
w = Fd
kg m m
= J
[w ] =
2
s
Energy
• Potential Energy
– Energy due to condition, position, or composition.
– Associated with forces of attraction or repulsion
between objects.
• Energy can change from potential to kinetic.
Energy and Temperature
• Thermal Energy
– Kinetic energy associated with random
molecular motion.
– In general proportional to temperature.
– An intensive property.
• Heat and Work
– q and w.
– Energy changes.
Thermal Equilibrium
• Heat transfer ALWAYS occurs from a hotter
object to a cooler object (directionality)
• Transfer of heat continues until both objects are at
the same temperature (thermal equilibrium)
• The quantity of heat lost by a hotter object and the
quantity of heat gained by a cooler object when
they are in contact are numerically equal (required
by the law of conservation of energy)
Units of Heat
• Calorie (cal)
heat required to raise the temperature of 1.00 g of
pure liquid water from 14.5 to 15.5oC.
• Joule (J):
SI unit for heat
1 cal = 4.184 J
Heat Capacity
• The quantity of heat required to change the
temperature of a system by one degree.
– Specific heat capacity, c.
q = mcT
• System is one gram of substance
– Heat capacity, C.
• C = Mass x specific heat (mc).
q = CT
Many metals have
low specific heats.
The specific heat of
water is higher than
that of almost any
other substance.
Example:
How much heat, in joules and in kilojoules, does it
take to raise the temperature of 225 g of water
from 25.0 to 100.0 °C?
Example:
What will be the final temperature if a 5.00-g silver
ring at 37.0 °C gives off 25.0 J of heat to its
surroundings? The specific heat of silver is 0.235
Jg-1°C-1.
Example: An Estimation Example
Without doing detailed calculations, determine which of
the following is a likely approximate final temperature
when 100 g of iron at 100 °C is added to 100 g of 20 °C
water in a calorimeter, the specific heat of iron is 0.449
Jg-1°C-1:
(a) 20 °C
(b) 30 °C
(c) 60 °C
(d) 70 °C.
Conservation of Energy
• In interactions between a system and its
surroundings the total energy remains
constant. energy is neither created nor
destroyed.
qsystem + qsurroundings = 0
qsystem = -qsurroundings
First Law: Sign Convention
• Energy entering a system carries a positive
sign:
– heat absorbed by the system, or
– work done on the system
• Energy leaving a system carries a negative
sign
– heat given off by the system
– work done by the system
Determination of Specific Heat
Example:
Determining Specific Heat from Experimental Data.
Use the data presented on the last slide to calculate the specific
heat of lead.
qlead = -qwater
qwater = mcT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C
qwater = 1.4x103 J
qlead = -1.4x103 J = mcT = (150.0 g)(c)(28.8 - 100.0)°C
clead = 0.13 Jg-1°C-1
Heats of Reaction and Calorimetry
• Chemical energy.
– Contributes to the internal energy of a system.
• Heat of reaction, qrxn.
– The quantity of heat exchanged between a
system and its surroundings when a chemical
reaction occurs within the system, at constant
temperature.
Heats of Reaction (qrxn)
• An exothermic reaction gives off heat
– In an isolated system, the temperature increases.
– The system goes from higher to lower energy;
– qrxn is negative.
• An endothermic reaction absorbs heat
– In an isolated system, the temperature decreases.
– The system goes from lower to higher energy;
– qrxn is positive.
Bomb Calorimeter
qrxn = -qcal
qcal = qbomb + qwater + qwires
+…
Define the heat capacity of the
calorimeter:
qcal = ∑miciT = CT
i
heat
Example:
Using Bomb Calorimetry Data to Determine a Heat of Reaction.
The combustion of 1.010 g sucrose, in a bomb calorimeter,
causes the temperature to rise from 24.92 to 28.33°C. The
heat capacity of the calorimeter assembly is 4.90 kJ/°C.
(a) What is the heat of combustion of sucrose, expressed in
kJ/mol C12H22O11
(b) Verify the claim of sugar producers that one teaspoon of
sugar (about 4.8 g) contains only 19 calories.
Example:
Calculate qcalorimeter:
qcal = CT = (4.90 kJ/°C)(28.33-24.92)°C = (4.90)(3.41) kJ
= 16.7 kJ
Calculate qrxn:
qrxn = -qcal = -16.7 kJ per 1.010 g
Example:
Calculate qrxn in the required units:
-16.7 kJ
qrxn = -qcal =
= -16.5 kJ/g
1.010 g
qrxn
343.3 g
= -16.5 kJ/g
1.00 mol
= -5.65 x 103 kJ/mol
(a)
Calculate qrxn for one teaspoon:
qrxn
4.8 g 1.00 cal
)= -19 kcal/tsp
= (-16.5 kJ/g)(
)(
4.184 J
1 tsp
Note: in food industry we say calorie instead of kcal!
(b)
Internal Energy
• Internal Energy, U.
– Total energy (potential and kinetic) in a system.
• Translational kinetic energy.
• Molecular rotation.
• Bond vibration.
• Intermolecular attractions.
• Chemical bonds.
• Electrons.
First Law of Thermodynamics
• A system contains only internal energy.
– A system does not contain heat or work.
– These only occur during a change in the system.
U = q + w
q = heat gained by system
w = work gained (+) by or done on system (-)
• Law of Conservation of Energy
– The energy of an isolated system is constant
Example:
A gas does 135 J of work while expanding,
and at the same time it absorbs 156 J of heat.
What is the change in internal energy?
U = q + w = 156 – 135 = 21 J
State Functions
• The state of a system: its exact condition at a fixed
instant.
• State is determined by the kinds and amounts of matter
present, the structure of this matter at the molecular
level, and the prevailing pressure and temperature.
• A state function is a property that has a unique value
that depends only on the present state of a system, and
does not depend on how the state was reached (does
not depend on the history of the system).
State Functions
• Water at 293.15 K and 1.00 atm is in a specified
state.
• d = 0.99820 g/mL
• This density is a unique function of the state.
• It does not matter how the state was established.
Functions of State
• U is a function of state.
– Not easily measured.
• U has a unique value between two states.
– Is easily measured.
Path Dependent Functions
For example:
Fuel-Consume to go from one point to the another one is
NOT a function of state, but a path dependent function
4 Litre
B
A
2 Litre
Internal Energy Change at Constant Volume
• For a system where the
reaction is carried out at
constant volume, V = 0
and U = qV.
• All the thermal energy
produced by conversion
from chemical energy is
released as heat; no P-V
work is done.
Work
• In addition to heat effects chemical reactions
may also do work.
• Gas formed pushes against
the atmosphere.
• Volume changes.
• Pressure-volume work.
Heats of Reaction: U and H
Reactants → Products
Ui
Uf
U = Uf - Ui
U = qrxn + w
In a system at constant volume: PV=0
U = qrxn + 0 = qrxn = qv
But we live in a constant pressure world!
How does qp relate to qv?
lnternal Energy Change at Constant Pressure
• For a system where the
reaction is carried out at
constant pressure,
U = qP – PV or
U + PV = qP
• Most of the thermal energy
is released as heat.
• Some work is done to
expand the system against
the surroundings (push
back the atmosphere).
Heats of Reaction
Heats of Reaction
U = qP + w
We know that w = - PV (minus because done by system)
therefore:
U = qP - PV
qP = U + PV
These are all state functions, so define a new function.
Let
H = U + PV
Then
H = Hf – Hi = U + PV= U + PV + VP
If we work at constant pressure and temperature: VP = 0
H = U + PV = qP
Standard States and Standard Enthalpy
Changes
• Define a particular state as a standard state.
• Standard enthalpy of reaction, H°
– The enthalpy change of a reaction in which all
reactants and products are in their standard
states.
• Standard State
– The pure element or compound at a pressure
of 1 atm and at the temperature of interest
(usually 25 °C).
Enthalpy Diagrams
Indirect Determination of H:
Hess’s Law
• H is an extensive property.
– Enthalpy change is directly proportional to the
amount of substance in a system.
N2(g) + O2(g) → 2 NO(g)
½N2(g) + ½O2(g) → NO(g)
H = +180.50 kJ
H = +90.25 kJ
• H changes sign when a process is reversed
NO(g) → ½N2(g) + ½O2(g)
H = -90.25 kJ
Hess’s Law
• Hess’s law of constant heat summation
– If a process occurs in stages or steps (even hypothetically),
the enthalpy change for the overall process is the sum of
the enthalpy changes for the individual steps.
½N2(g) + ½O2(g) → NO(g)
H = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
H = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
H = +33.18 kJ
Example:
Calculate the enthalpy change for reaction (a) given the data
in equations (b), (c), and (d).
(a) 2 C(graphite) + 2 H2(g)  C2H4(g)
H = ?
(b) C(graphite) + O2(g)  CO2(g)
Hb = –393.5 kJ
(c) C2H4(g) + 3 O2  2 CO2(g) + 2 H2O(l)
Hc = –1410.9 kJ
(d) H2(g) + ½ O2  H2O(l)
Hd = –285.8 kJ
(a) = 2 (b) + 2 (d) – (c)
&
Ha = 2Hb + 2Hd - Hc
Standard Enthalpies of Formation
Hf°
• The enthalpy change that occurs in the
formation of one mole of a substance in the
standard state from the reference forms of
the elements in their standard states.
• The standard enthalpy of formation of a
pure element in its reference state is 0.
Standard Enthalpy of Formation
When we say:
“The standard enthalpy of formation of CH3OH(l) is 238.7 kJ”,
we are saying that the reaction:
C(graphite) + 2 H2(g) + ½ O2(g)  CH3OH(l)
has a value of ΔH of –238.7 kJ.
We can treat ΔHf° values as though they were absolute
enthalpies, to determine enthalpy changes for reactions.
Question: What is ΔHf° for an element in its standard state
[such as O2(g)]? Hint: since the reactants are the same as
the products …
Standard Enthalpies of Formation
Enthalpy of Reaction
Hrxn = ∑Hf°products - ∑Hf°reactants
Table 7.3 Enthalpies of Formation of Ions
in Aqueous Solutions
Example:
Synthesis gas is a mixture of carbon monoxide and hydrogen that is
used to synthesize a variety of organic compounds. One reaction for
producing synthesis gas is
3 CH4(g) + 2 H2O(l) + CO2(g)  4 CO(g) + 8 H2(g) ΔH° = ?
Use standard enthalpies of formation from Table 6.2 to calculate the
standard enthalpy change for this reaction.
Example:
The combustion of isopropyl alcohol, common rubbing alcohol, is
represented by the equation
2 (CH3)2CHOH(l) + 9 O2(g)  6 CO2(g) + 8 H2O(l) ΔH° = –4011 kJ
Use this equation and data from Table 6.2 to establish the standard
enthalpy of formation for isopropyl alcohol.
Fuels as Sources of
Energy
• Fossil fuels.
– Combustion is exothermic.
– Non-renewable resource.
– Environmental impact.
Chapter 5 Questions
6, 8, 17, 23, 25, 27
30, 32, 34, 36, 39
47, 53, 60, 63