Transcript Thermochemistry - Chemistry at Winthrop University

Energy
• We depend on energy for our existence, and to maintain our
standard of life.
• It is not something that we can touch, but we experience it in
many ways
• Commonly, we think of
energy as being either
kinetic or potential
(stored), but there are
many variations of
these classifications.
Chemical Energy
• Chemical energy is a kind of potential energy that is stored
within chemical bonds.
• For example, combusting fuel (oil, coal) breaks the C-C and
C-H bonds, releasing their trapped energy as heat.
• Connecting the ends of a battery allows a redox reaction to
occur, forcing electrons to flow from one end to the other
(current)
Introduction
• In this unit, we will discuss thermochemistry
• Thermochemistry is the study of the energy changes associated
with chemical processes, and will lead to a much deeper
discussion on thermodynamics, which investigates the flow of
energy between systems.
• The first law of thermodynamics, introduced in this lecture,
allows us to “keep track” of energy change.
• Two fundamental concepts of thermochemistry and
thermodynamics are heat and work.
Internal Energy Change
• The internal energy of a system, U (referred to as E in the
book), is the sum of all potential and kinetic energies of the
molecules in that system.
– Kinetic energies include the motion of atoms and
subatomic particles in a system; the temperature of a
substance is directly proportional to its total kinetic energy.
• The exact value of U is very difficult to calculate, but changes
in the value of U are easy to calculate.
Energy Basics: Recap
• What is Energy?
– Energy is defined as the capacity to perform “work”
• How do we define work?
• Work is defined as the result of a force acting through
some distance (F x d)
System and Surroundings
• Two important terms to be defined in this unit are system,
surroundings, and universe.
– System: The portion of the universe that we are interested
in. Ex. A beaker filled with a chemical solution
– Surroundings: Literally, everything else in the universe
that is not part of the system
• We know that energy must be conserved. Therefore, if energy
leaves the system, it must enter the surroundings, and if it
enters the system, it must come from the surroundings.
• If we define energy losses as negative change, and energy
gains as positive change:
∆𝑈𝑠𝑦𝑠 = −∆𝑈𝑠𝑢𝑟𝑟
Isolated, Closed, and Open Systems
• There are three types of systems: isolated, open,
and closed.
– Isolated systems are theoretical, and describe
a system that is completely cut off from the
surroundings, such that no energy or mass can
be transferred. Ex. Thermos
– Closed systems are more realistic versions of
isolated systems, where only energy, but not
mass, is allowed to transfer between system
and surroundings. Ex. Sealed soda can
– Open systems allow exchange of both energy
and matter with the surroundings.
Ex. Glass of water
Calculating PV Work
• You can not perform work without energy
– Ex.: You need at least 100 J of energy to do 100 J of work
• Let’s use a gas cylinder equipped with a piston as an example.
The gas inside of the cylinder is our system. Everything else is
our surroundings.
• The application of pressure to the piston causes displacement
of the system by a distance Δh
Calculating Work of A Trapped Gas
w = F x d = F x (∆h)
• Pressure is defined as force divided by area!!
F
P=
→
A
𝐏𝐀 = 𝐅
• Since the pressure acts against the natural expansion
of the gas, the pressure term is negative
w = −𝐏𝐀 (∆h)
• Volume = (LW)H = (Area) x height
w = −P∆V
• Pressure is expressed in units of atmospheres (atm)
and V in Liters.
PΔV Work For Trapped Gas
• Compression: Since the compression is caused by the
surroundings, we say that the surroundings are doing work on
the system
• .
• Ex. The gas is compressed under a constant pressure of 1 atm
from 3.5L to 1.2L. Calculate the work of the system.
𝑤𝒔𝒚𝒔 = −P∆V
= − 1 atm (−𝟐. 𝟑 𝐋)
= +𝟐. 𝟑 𝐋 𝐚𝐭𝐦
expansion
• Work done on the system is ALWAYS POSITIVE. Energy from the
surroundings is transferred into the system, so the system gains
that energy.
A Quick Note
• You may have noticed that the units of the previous answer
were L•atm, not J. You must convert to joules.
𝟏 𝐋 𝐚𝐭𝐦 = 𝟏𝟎𝟏. 𝟑𝟐𝟓 𝐉
• Now, we can complete the previous problem.
𝑤𝒔𝒚𝒔
101.325 J
= 2.3 L atm x
= 𝟐𝟑𝟑 𝐉
L atm
• Of course, this would mean that:
𝑤𝒔𝒖𝒓𝒓 = −𝟐𝟑𝟑 𝐉
PΔV Work For Trapped Gas
• Expansion: If the gas expands against the piston, then work is
done by the system on the surroundings.
• Work done by the system is ALWAYS NEGATIVE because the
system now has that much less energy.
• Ex. A gas expands against a constant pressure of 1 atm from
5L to 9L. Calculate the work of the system.
J
w = − 1 atm (4L)(101.325
)
L atm
w = −𝟒𝟎𝟓. 𝟑 𝐉
Heat
• Energy that is not used to perform work or stored is lost
as heat. Heat is the flow of energy caused by a
temperature gradient.
• For example, only 25% of the thermal energy released
from the combustion of gasoline is actually used to
perform work (move the car).
– The rest is lost as heat to the surroundings (heating up
the car and the air around the car).
– Without this limitation, cars could easily exceed 100
mpg.
• Heat, denoted with the symbol q, will always flow when
there is a temperature difference between the system and
the surroundings (from high temp to low temp).
Heat
• The sign convention for heat is the same as for work.
•
– When heat is lost by the system, it must be transferred to
the surroundings. In such a case, qsys is negative, and qsurr
is positive (-qsys = qsurr)
– A reaction in which qsys < 0 is called an exothermic reaction
Fuel + O2  CO2 + H2O + heat
Combustion Reactions
• Fire is, by definition, a redox reaction involving a fuel, an
ignition, and oxygen. This self-sustaining reaction emits heat
and light.
– Ex. Natural gas (CH4)When insufficient oxygen is available,
incomplete combustion occurs, leading to side products
like C(s) and CO(g)
Complete oxidation of
carbon
Incomplete oxidation. Formation
of carbon monoxide and carbon
particles (emits yellow).
Heat
• When heat is absorbed by the system, it is taken from the
surroundings, and qsys is positive (qsurr is negative).
• When qsys >0, the reaction is endothermic.
H2O(s) + heat -----> H2O (L)
H2O(L) + heat -----> H2O (g)
First Law of Thermodynamics
• We can measure the change in internal energy of a system by:
∆𝐔 = 𝐪 + 𝐰
Energy is never created or destroyed, merely converted
between forms and transferred from place to place as heat
or work. The total energy of the universe is finite
(conservation of energy).
State Functions
• In thermodynamics, we must distinguish between those terms
that are state functions, and those that are not.
• A state function is a property of the system that is based on
the system’s condition, but NOT on the path taken to reach
that condition. Internal energy is a state function.
• Example: Imagine that our system is 50g of water at 25oC. The
value of U is the same regardless of how that state is reached.
Heat and Work are NOT State Functions
• U is a state function, but q and w are not (state functions are
capitalized).
• The amount of heat and work transferred will depend on the path,
although their sum does not.
• Take the system (battery) below. The value of ΔU doesn’t depend
on HOW the energy is used, but q and w do.
Fully Charged
Heat
Heat
Battery is short circuited.
Coil heats up. No work is
done. All energy is lost as
heat.
ΔU
Dead
Work
Some energy is used
to do work (turn fan).
Enthalpy
• We can represent the energy change of a system at constant
pressure as:
∆𝐔 = 𝐪 + 𝐰 = 𝐪 − 𝐏∆𝐕
• Rearranging the equation to solve for q, we obtain:
𝟏 𝐪𝐩 = ∆𝐔 + 𝐏∆𝐕
(𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐩𝐫𝐞𝐬𝐬𝐮𝐫𝐞)
• If a reaction is carried out in a closed, rigid container, the
volume is constant. Thus, no PV work can be done, and the
energy change in the system is entirely heat (ex. phase change)
𝟐 𝐪𝐯 = ∆𝐔
(𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐯𝐨𝐥𝐮𝐦𝐞)
• More commonly, a reaction would be carried out in an open
flask under constant atmospheric pressure, and the energy
change would be due to both heat and work. q would then
follow expression (1).
Enthalpy
• The heat change of a reaction at constant pressure is known
as ENTHALPY (H), or heat of reaction (∆𝐻 = 𝑞𝑝 )
• In most chemical reactions, only ΔH is important. As with ΔU,
ΔH is a state function. Typically, the difference between ΔU
and ΔH is small.
• For a reaction at a constant pressure of 1 atm and 25oC, we
can define the standard enthalpy change (denoted by ‘o’
symbol) of a reaction
∆𝐇 𝐨 𝐫𝐱𝐧 = 𝐇𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 − 𝐇𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
• Enthalpy is a molar quantity, and is usually expressed in
KJ/mol. As with q, positive values of ΔH indicate endothermic
reactions and negative values indicate exothermic reactions.
Phase Changes
• If a reaction is exothermic in one direction, it must be
endothermic in the other. For example, freezing water requires
the removal of heat from the system.
𝐻2 𝑂 𝐿 → 𝐻2 𝑂(𝑠)
∆𝐻𝑓𝑢𝑠 = −6.01
𝑘𝐽
𝑚𝑜𝑙
• Melting water would require the same magnitude of heat to be
absorbed.
𝐻2 𝑂 𝑠 → 𝐻2 𝑂 𝐿
∆𝐻𝑓𝑢𝑠 = +6.01
𝑘𝐽
𝑚𝑜𝑙
• The enthalpy of melting/freezing is called the enthalpy of
fusion. The enthalpy of boiling/condensation is called the
enthalpy of vaporization.
𝐻2 𝑂 𝐿 → 𝐻2 𝑂 𝑔
∆𝐻𝑣𝑎𝑝 = +40.7
𝑘𝐽
𝑚𝑜𝑙
Heating Curves
• Heating curves can be used to calculate the energy required
to raise the temperature of a substance. For example,
consider 1000g water at -18oC that is heated to 200oC.
Calculating Total Energy From a Heating Curve
1.
2.
3.
4.
5.
Heat H2O(s) from -18oC to 0oC
Melt H2O(s) at 0oC, H2O(s)  H2O (L)
Heat H2O(L) from 0oC to 100oC
Boil H2O(L) at 100oC, H2O(L)  H2O(g)
Heat H2O(g) to 200oC
q1
q2
q3
q4
q5
• The total energy of this process is the sum of the individual q
values.
– Note: During a phase change, the temperature remains
constant because all of the added heat is being used to
overcome the intermolecular forces and separate the
molecules (heat converted to work)
Calculating Total Energy From a Heating Curve
• To calculate the heat absorbed/released when a substance is
changing temperature, we define q as:
𝑞 = 𝑛𝐶𝑝 ∆𝑇
where Cp is the heat capacity, which describes the amount of
energy needed to raise the temperature of one mole of a
substance by 1oC.
• The heat absorbed/released during a phase change is given
by:
𝑞 = 𝑛∆𝐻𝑓𝑢𝑠,𝑣𝑎𝑝
ice
liquid
vapor
1.
Heat H2O(s) from -18oC to 0oC
𝑞1 = 𝑛𝐶𝑝,𝑖𝑐𝑒 ∆𝑇 = 55.55 𝑚𝑜𝑙
2.
38
J
mol K
18 K = 𝟑𝟕𝟗𝟗𝟔 J
Melt H2O(s) at 0oC, H2O(s)  H2O (L)
𝑞2 = 𝑛∆𝐻𝑓𝑢𝑠 = 55.55 mol
6010
J
= 𝟑𝟑𝟑𝟖𝟓𝟓 J
mol
3. Heat H2O(L) from 0oC to 100oC
𝑞3 = 𝑛𝐶𝑝,𝑙𝑖𝑞 ∆𝑇 = 55.55 𝑚𝑜𝑙
4.
J
mol K
100 K = 𝟒𝟏𝟖𝟐𝟗𝟏 J
Boil H2O(L) at 100oC, H2O(L)  H2O(g)
𝑞4 = 𝑛∆𝐻𝑣𝑎𝑝 = 55.55 mol
5.
75.3
40700
J
= 𝟐𝟐𝟓𝟖𝟖𝟓𝟎 J
mol
Heat H2O(g) to 200oC
𝑞5 = 𝑛𝐶𝑝,𝑣𝑎𝑝 ∆𝑇 = 55.55 𝑚𝑜𝑙
34.1
J
mol K
100 K = 𝟏𝟖𝟗𝟒𝟐𝟓 J
𝒒𝒕𝒐𝒕𝒂𝒍 = 𝟑. 𝟐𝟒 𝐱 𝟏𝟎𝟔 𝐉
Heat Transfer
• Heat ALWAYS flows from hotter substances to cooler
substances until both substances reach thermal equilibrium
(same temperature)
• In an insulated system containing two masses at different
temperatures in physical contact, the heat lost by one material is
exactly equal to the heat gained by the other.
Example
• 50g of copper metal at 80oC is placed in 100 mL of water at
10oC within an insulated system.
What will the final
temperature of the water and copper be in oC? Set the molar
heat capacities of Cu and H2O(L) as 24.4 J/moloK and 75.3
J/moloK,
What we know:
−𝒒𝑪𝒖 = 𝒒𝒘𝒂𝒕𝒆𝒓
time
𝑇𝐹,𝐶𝑢 = 𝑇𝐹,𝑊𝑎𝑡𝑒𝑟
𝑞 = 𝑛𝐶𝑝 ∆𝑇
− 0.786 𝑚𝑜𝑙 𝐶𝑢
24.4
𝐽
𝑚𝑜𝑙 𝐾
𝑇𝐹 − 353 𝐾 = 5.55 𝑚𝑜𝑙 𝐻2 𝑂
75.3
−19.18 𝑇𝐹 + 6770 = 417.91 𝑇𝐹 − 118270
125040 = 437.1 𝑇𝐹
286 𝐾 = 𝑇𝐹
13𝑜 𝐶 = 𝑇𝐹
𝐽
(𝑇𝐹 − 283 𝐾)
𝑚𝑜𝑙 𝐾
Practice Problem
• You have 72 cans of Bud Light in a cooler. The cans are at
room temperature (25oC). Each can contains 355mL of
(mostly) water, and the cans are made of 12.5g of aluminum.
To cool the beer, you decide to buy bags of ice (10 lb) which
are at -18oC.
• How many bags of ice do you need to cool all of the booze
down to 0oC? *Refer to the heat capacity table on slide #29.
Pg. 379 in the text provides a solution.
Calculating ΔHo
• The enthalpy change associated with a reaction can be
determined by several methods, depending on the information
provided. These methods include:
1.
2.
3.
4.
Stoichiometric calculation, using standard enthalpies
Hess’s Law, using standard reactions of known ΔHorxn
Heats of formation
Bond enthalpy calculations
Stoichiometric Enthalpy Calculations
• Lets take the combustion of hydrogen gas as
an example of an exothermic reaction.
2H2(g) + O2(g)  2H2O(g)
Enthalpy Diagram
2H2(g) + O2(g)
ΔHorxn= -483.6 kJ/mol
1. Enthalpy is an extensive property. The
magnitude of ΔH is proportional to the amount of
reactant consumed. So, 2 mol of H2 will evolve
-483.6 kJ. 4 mol of H2 will evolve double that.
2. Enthalpy of the reverse reaction has the
opposite sign. The decomposition of 2 moles of
H2O into 2 moles of H2 and 1 mole of O2 is +483.6
kJ.
3. The state of the reactants matters. If liquid
water formed instead of steam, ΔH would be
different.
ΔH < 0
exothermic
2H2O(g)
Example
• Calculate the heat released when 30.0 g of iron(III)oxide is
combined with 15.0 g of Al(s) at 1 bar.
Fe2O3(s) + 2Al(s)  Al2O3(s) + 2 Fe(s)
ΔHorxn= -851.5 kJ/mol
• To determine the enthalpy change of the reaction, we must
first determine the moles of the limiting reactant.
mol Fe2 O3
30 g Fe2 O3 x
= .376 mol Fe2 O3
159.7 g Fe2 O3
mol Al
15g Al x
= .556 mol Al
27 g Al
Limiting Reactant
−851.5 kJ heat
.556 mol Al x
= −𝟐𝟑𝟕 𝐤𝐉
2 mol Al reacted
Hess’s Law
• For multi-step processes, enthalpy changes are additive.
• Lets take the following reaction:
Sn s + 2Cl2 g → SnCl4 (L)
• This reaction is two steps, and proceeds through an
intermediate (short-lived species), as shown below.
Sn s + Cl2 g → SnCl2 s
SnCl2 s + Cl2 g → SnCl4 L
∆H o rxn (1) = −325.1 kJ/mol
∆H o rxn (2) = −186.2 kJ/mol
Hess’s Law, continued
𝑆𝑛 𝑠 + 2𝐶𝑙2 𝑔 → 𝑆𝑛𝐶𝑙4 (𝐿)
• We can add these reactions together, canceling intermediates on
opposing sides of the arrow.
• Since we know the enthalpy change in each step, we can
determine the overall enthalpy change of the reaction in question.
Sn s + Cl2 g → SnCl2 s
∆H o rxn (1) = −325.1 kJ/mol
SnCl2 s + Cl2 g → SnCl4 L
∆H o rxn (2) = −186.2 kJ/mol
Sn s + 2Cl2 g → SnCl4 L
∆Ho rxn (3) = −511.3 kJ/mol
• This additive property is known as Hess’s Law.
Hess’s Law
• The usefulness of Hess’s Law is that it allows you to
express the value of ΔHo of an unknown reaction as the
sum of the ΔHo values of known standard reactions.
Example
• Given the following ΔHorxn values, calculate ΔHorxn for the reaction:
2SO2(g) + O2(g)  2SO3(g)
(1) SO2(g) ---> S(s) + O2(g)
(2) 2S(s) + 3O2(g) ---> 2SO3(g)
ΔHorxn (1) = 296.8 kJ/mol
ΔHorxn (2) = -791. 4 kJ/mol
It is necessary to multiply reaction (1) by a factor of 2 in order to
get 2 moles of SO2(g) and to cancel out S(s) and 2 moles of O2(g).
(1) 2 [SO2(g) ---> S(s) + O2(g) ]
ΔHorxn (1) = 2 (296.8 kJ/mol)
(2)
ΔHorxn (2) = -791. 4 kJ/mol
2S(s) + 3O2(g) ---> 2SO3(g)
(3) 2SO2(g) + O2(g) ---> 2SO3(g)
ΔHorxn (3) = -197.8 kJ/mol
Group Example
• Given the following values of ΔHorxn:
1 2P s + 3Cl2 (g) → 2PCl3 (L)
2 2P s + 5Cl2 (g) → 2PCl5 (s)
ΔHorxn= -639.4 kJ/mol
ΔHorxn= -887.0 kJ/mol
• Calculate the value of ΔHorxn for the following reaction:
(3) 𝐏𝐂𝐥𝟑 𝐋 + 𝐂𝐥𝟐 (𝐠) → 𝐏𝐂𝐥𝟓 (𝐬)
Enthalpies of Formation
• The enthalpy change associated with a the formation of a
molecule is called the enthalpy change of formation, ΔHf (also
called heat of formation).
• For a given reaction, if you know the enthalpies of formation
of the reactants and products:
∆𝐇 𝐨 𝐫𝐱𝐧 =
𝐧 ∆𝐇 𝐨 𝐟 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 −
𝐧 ∆𝐇 𝐨 𝐟 (𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬)
Enthalpies of Formation
For pure elements, and for elements in
their most stable form, ΔHof = 0 kJ/mol
The enthalpies of formation
are exactly known for many
elements.
Example
• Calculate the enthalpy change for the combustion of benzene,
C6H6 to form CO2 and H2O using standard heats of formation.
2𝐶6 𝐻6 𝐿 + 15𝑂2 𝑔 → 12𝐶𝑂2 𝑔 + 6𝐻2 𝑂(𝐿)
Using the heats of formation of the reactants and the
products, we can directly calculate ∆𝐻 𝑜 𝑟𝑥𝑛
∆𝐻 𝑜 𝑟𝑥𝑛 =
∆𝐻𝑜 𝑟𝑥𝑛 =
12
𝑛 ∆𝐻 𝑜 𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 −
𝑘𝐽
−393.5 𝑚𝑜𝑙 + 6
𝑘𝐽
−285.8 𝑚𝑜𝑙
−
𝑛 ∆𝐻𝑜 𝑓 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
2
∆𝐻 𝑜 𝑟𝑥𝑛 = −6535 𝑘𝐽/𝑚𝑜𝑙
𝑘𝐽
𝑘𝐽
49.1 𝑚𝑜𝑙 + 15 0 𝑚𝑜𝑙
Molar Bond Enthalpies
• The enthalpy change for the complete decomposition of water:
H2 O L → O g + 2H(g)
is ∆𝐻 𝑜 𝑟𝑥𝑛 = + 925 kJ/mol.
• In a water molecule, there are 2 O-H
bonds. The positive value of enthalpy
indicates that heat (energy) must be
absorbed by the system in order to break
these bonds.
• Therefore, we can estimate that the
molar bond enthalpy, Hbond, of each O-H
bond (energy needed to break the bond)
is 463.5 kJ/mol
Molar Bond Enthalpies
• In general, if we consider a reaction in the most basic sense,
then we would imagine that the formation of a compound
proceeds as follows:
∆𝑯𝒐 𝒓𝒙𝒏 =
energy absorbed to
break reactant bonds
+
energy released when
product bonds form
Table of common bond
enthalpies. These are the
energies to break the bonds.
When you form bonds, the
sign of H will be negative.
Example
• Using molar bond enthalpies, estimate ∆𝑯𝒐 𝒓𝒙𝒏 of the following:
CH4 g + Cl2 g → CH3 Cl g + HCl(g)
• As you can see, this reaction proceeds by breaking one C-H
bond, and a Cl-Cl bond, followed by the formation of a C-Cl
bond and an H-Cl bond
H
H
C
H
H
Cl
Cl
H
Energy absorbed. Bonds broken
∆𝐇 𝐨 𝐫𝐱𝐧 = (414 kJ/mol + 243 kJ/mol)
H
C
Cl
H
Cl
H
Energy released. New bonds.
+ (-331 kJ/mol - 431 kJ/mol)
= -105 kJ/mol
Group Example
• Nitrogen gas reacts with hydrogen gas to produce ammonia,
NH3. Use molar bond enthalpies to calculate the enthalpy of
formation ammonia. Compare this with the known value, -45.9
kJ/mol.