Transcript Chapter 6 - ETSU.edu

Chapter 6
Thermochemistry
What is Thermodynamics?
• Thermodynamics is the study of the effect of
work, heat and energy on a system. It is
concerned with only large scale observation.
• Thermochemistry is the study of the heat
released or required by chemical reactions.
Energy, Heat and Enthalpy
• The reaction type where energy is obtained
are usually combustion in which a
hydrocarbon fuel burns in oxygen to form
carbon dioxide and water. A typical reaction is
the combustion of natural gas, which is largely
methane.
• CH4(g)+2O2(g) CO2(g)+2H2O (l) +energy
• Since energy is released in this reaction we
include it in the product. Living organisms make
use of a very sophisticated form of combustion to
obtain the energy to move , grow and think.
• C6H12O6(aq) + 6O2(g)--- 6CO2(g) +6H2O(l) + energy.
• The glucose does not burn with a flame inside us.
A series of reaction takes place at a controlled
rate. The field of bioenergetics is the study of the
uses of energy by organisms.
The conservation of energy
• Energy comes in a variety of form. The kinetic
energy which is the energy due to motion and the
potential energy that is the energy due to
position. A body of mass m at a height h above
the surface of the earth has a potential energy Ep
= mgh g= acceleration of free fall on earth
(g=9.8m/s2)
• The total energy of an object such as a planet or a
ball is the sum of its kinetic and potential
energies: E=Ek +Ep
• The law of conservation of energy states that
total energy of an isolated body is constant.
• The internal energy of an object is a measure
of its capacity to do work and supply heat. It is
the total energy of all its atoms and
molecules.
System and Surroundings
• A system is the part of the world we want to
study. In chemistry this is usually a reaction
mixture in a flask.
• A system can be open, closed or isolated.
• An open system can exchange both matter
and energy with the surroundings.Examples
are an automobile engine, the human body,
and an open flask.
• A closed system has a fixed amount of matter,
but can exchange only energy with the
surroundings. Examples are electric batteries,
cold packs used for athletic injuries.
• An isolated system can exchange neither
matter nor energy with its surroundings.
Example a sealed vacuum flask.
Heat and work
• When a system does work, such as an engine
raising a weight, energy is transferred from the
system to the surroundings. Energy is also
transferred when a system releases heat, as in a
combustion. A system with a high internal energy
can do a lot of work or provide a lot of heat to
the surroundings. Example atightly coiled spring,
a fully charged battery and the hot steam of a
turbine. A system with a low internal energy can
do less work example an uncoiled spring, a nearly
exhausted battery, and cold water.
• We can change the internal energy of an open
system by adding or removing matter. An
automobile is an open system, and we
increase its internal energy by putting gasoline
in its tank. A closed system can be prepared
by adding reagents and then sealing the
container. Chemical reactions can occur inside
the system, but all the reactants and products
remain inside.
• A second way to change internal energy of a
system is by heating or cooling it. Heat is the
transfer of energy that occurs as a result of
temperature difference. Energy flows as heat
from a hot region into a cooler region .
• When a hot object is placed in a cool
surroundings energy flows out of the system
and stirs up chaotic motion of the molecules
in the surroundings. Random molecular
motion is called thermal motion, meaning
heat stirs up the thermal motion of molecules
in the system.Heat can raise the temperature
of a system and it can also change the physical
state , such as from liquid to vapor.
• The third way to change the internal energy of
a system is by doing work on the system or by
letting the system do work on the
surroundings.
• Work is a transfer of energy that takes place
when an object is moved against an opposing
force.
The first Law
• Internal energy is the sum of kinetic energy of
all the particles in the system and potential
energy arising from their interactions with one
another. If molecules move rapidly their
kinetic energy is high and so the internal
energy. If the particles are too close example
in a solid the squashing them even closer
together will increase their potential energy
and hence increase the internal energy.
Units
• Kinetic energy=1/2 mv2 mass is measured in
kg and speed in (meter/sec)2
kg.m2/s2
1 kg.m2/s2 = 1J
1kJ =103 J
1 calorie(cal) is the energy needed to raise the
temperature of 1g of water by 1⁰C.
1cal=4.184 J
• Internal energy changes when energy enters
or leaves a system.
• ∆U= Ufinal –Uinitial
• A positive value of ∆U means that the internal
energy of the final state of the system is
higher than the initial state. The negative
value of ∆U means that the final state has a
lower energy than the initial state.
• When we do work on a system for instance,
by winding a spring inside it its internal energy
rises. If we do 100kJ of work; then the system
increases its store of energy by 100kJ and we
write as ∆U=+100kJ
• Changes in internal energy= energy supplied
to system as work
• ∆U=w
• If we transfer 40kJ of energy to a large beaker of
water by standing it on ahot surface; then the
internal energy of the water increases by 40kJ
and we write ∆U=+40kJ.
• Change in internal energy= energy supplied to
system as heat
• ∆U=q
• Change in internal energy=energy supplied to
system as heat + energy transferred to system as
work
• If we supply 40kJ of heat (q=+40kJ)and do
25kJ of work on the system (w=+25kJ), then
the total change in internal energy is
• ∆U= q+w =40 +25 kJ =+65 kJ
• If, in another process ,40kJ of heat leaks out of
the system while we are doing 25kJ of work on
it we would write q= -40kJthe minus sign
indicates that heat has left the system
• ∆U =q+w = (-40) + 25 kJ =-15 kJ
Class Practice
• A battery drives an electric motor in an
aquarium pump, and we consider the battery
plus the motor as the system. During a certain
period, the system does 555kJ of work on the
pump and releases 124kJ of heat into the
surroundings. What is the change in internal
energy of the system?
• Transferring Energy as Work
• How to measure changes in the internal energy
brought about by doing work?
• Work done = distance moved х Opposing force
• Opposing force = area х external pressure
• When we combine these two expressions, we get
• work done= distance moved х area х external pressure
• The distance moved multiplied by the area is the
change in volume that occurs during the expansion,
• work done = change in volume х external pressure
• The above relation can be expressed in
symbols. The change in volume (final – initial)
is ∆V, and the external pressure is Pexternal.
• w= -Pexternal∆V
• The negative sign indicates that w is negative
(there is a decrease in internal energy that is
the energy leaves the system) and the positive
means the system expands.
• 1 L-atm = 101.325 J
Class Practice
• The hot gaseous products of combustion in an
internal combustion engine expand by 2.2 L
against an external pressure of 1.0 atm. How
much work is done by the expansion?
• Transferring Energy as Heat
• Suppose we arrange for a reaction to take place inside
a sealed container:
• At constant volume ∆U=q
• If 75 kJ of heat is supplied into the container full of gas.
If the volume could not change then all the heat
supplied to it would be used to raise the internal
energy of the system, and ∆U=+75kJ
• When the gas is free to change its volume some of the
energy supplied is used to raise the temperature and
the rest leaks back to the surroundings as work. This is
art constant pressure.
• If 10kJ is used to expand the system ( w= 10kJ) in this case though q= +75kJ, the change
in internal energy ∆U = q+w =75-10kJ =+65kJ
• At constant pressure ∆U = q+ w =q - P∆V
Enthalpy
• A quantity called enthalpy is a convenient
procedure that takes into account the expansion
work automatically
• ∆H=∆U + P∆V
• ∆H=q-P∆V+P∆V
• ∆H=q
• A change in internal energy can be identified with
the heat supplied at constant volume. A change
in enthalpy is equal to the heat supplied at
constant pressure
Class practice
• In a certain reaction at constant pressure ,50kJ
of heat left the system and 30kJ of energy left
the system as expansion work to make room
for the products. What are the values of ∆H
and ∆U for the reaction?
• -50kJ
• -80kJ
Exothermic and Endothermic reaction
• An exothermic process is one that releases
heat; when ∆H< 0 signifies exothermic
process.(At constant pressure).
• An endothermic reaction absorbs heat;
• ∆H>0 signifies endothermic process.
Heat Capacity
• Heat capacity, C is the heat required to raise
the temperature of an object through 1K;
• Heat capacity=heat supplied/ temperature
rise
• C= q / ∆T
• Unit J/K if temperature is expressed in Kelvin.
Specific heat capacity
• The heat capacity divided by the mass of the
sample in grams.
• Heat capacity = mass x specific heat capacity
• C= m x Cs
• If we know the temperature rise it undergoes
during an experiment, then the heat transferred
to it is obtained as
• Heat supplied = mass(g) x specific heat
capacity(J/⁰C.g)x change in temperature(⁰ C)
• q = m x Cs x ∆T
Class Practice
• A swimming pool contains 2x104L of water.
How much energy in Joules is required to raise
the temperature of the water from 20⁰ C to
25⁰C?The specific heat capacity of water is
4.184 J/⁰ C.g
• 4.2x 10⁵ kJ
Thermochemistry of Physical Change
• Vaporization: This is an endothermic process
because energy has to be supplied to
overcome the forces that hold molecules
together in a liquid. The enthalpy change per
mole of molecules when a liquid vaporizes is
called the enthalpy of vaporization, ∆Hvap. For
water at 100⁰C
• ∆Hvap= Hvapor – Hliquid = =40.7kJ/mol
Class Practice
• An electric heater is used to raise the
temperature of a sample of water to its boiling
point in a constant pressure calorimeter. We
bring the water to a boiling point and the
continue heating until 35 g of water has
vaporized. The calculated time taken the
vaporization alone requires 79 kJ of heat.
What is the enthalpy of vaporization of water
at 100⁰ C?
Melting and sublimation
• The enthalpy changes that accompanies melting
is called the enthalpy of fusion,
• ∆H fusion= Hliq –Hsolid
• The enthalpy of fusion of water at 0⁰C is
6.0kJ/mol. This means we have to supply 6.0kJ of
energy as heat to melt one mole of water.
• The enthalpy of freezing is the change in enthalpy
per mole when a liquid turns into a solid. For
water at 0⁰C the enthalpy of freezing is 6.0kJ/mol
• ∆H(reverse process)= − ∆H(forward process)
• Sublimation is the direct conversion of a solid
into its vapor. The enthalpy of sublimation,
∆Hsub=Hvapor – Hsolid
• ∆Hsub = ∆Hfusion + ∆Hvap
• The temperature of a sample is constant at its
boiling and melting points, even though heat
is being applied.
Reaction Enthalpies
• A thermochemical equation is a combination
of a chemical equation with the enthalpy
change for the reaction as written. The
reaction enthalpy is the change in enthalpy
per mole for the stoichiometry numbers of
moles of reactants in the chemical equation.
Class practice
• When 2.31 g of phosphorus reacts with
chlorine to form phosphorus trichloride,PCl3,
in a calorimeter of heat capacity 2.16kJ/⁰C, the
temperature of the calorimeter rises by
11.06⁰C. Write the thermochemical equation
for the reaction.
Hess’s Law
• A reaction that is exothermic in one direction is
endothermic in the other direction:
• C₆H12O₆(aq) +6O2(g) 6CO2(g) + 6H2O(l)
•
∆H⁰=−2808kJ
• This is exothermic reaction.
• The reverse of this reaction , the formation of
glucose and oxygen from carbon dioxide and
water, is endothermic:
• 6CO2(g) + 6H2O(l) C₆H12O₆(aq) +6O2(g)
•
∆H⁰=+2808 kJ
• According to Hess’s Law, thermochemical
equations for the individual steps of a reaction
sequence may be combined to obtain the
thermochemical equation for the overall
reaction.
• explanation
Class practice
• Gasoline, which contains octane as one
component, may burn to carbon monoxide if
the air supply is restricted. Derive the
standard reaction enthalpy for the incomplete
combustion of octane:
• 2C8H18(l) +17O2(g) --- 16CO(g) +18H2O(l)
Enthalpies of combustion
• The heat absorbed or given off by a reaction
can be treated like a reactant or product in a
stoichiometry relation.
Class practice
• How much propane should a backpacker carry:
do we really need to carry a kilogram of gas?
Calculate the mass of propane that you would
need to burn to obtain 350kJ of heat, which is
just enough energy to heat 1 L of water from
27⁰C to boiling at sealevel (if we ignore all heat
losses). The thermochemical equation is
• C3H₈(g) + 5O2(g) - 3CO2(g) + 4H2O(l)
∆H⁰ =−2220kJ
• The specific enthalpy tells how much heat can
be obtained per gram of fuel. The enthalpy
density indicates how much heat can be
obtained per liter of fuel.
Homework
• Page 258
• 6.8, 6.18,6.20,6.26,6.32