Thermochemistry
Chapter 6
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Energy is the capacity to do work
•
Radiant energy comes from the sun and is
earth’s primary energy source
•
Thermal energy is the energy associated with
the random motion of atoms and molecules
•
Chemical energy is the energy stored within the
bonds of chemical substances
•
Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
•
Potential energy is the energy available by virtue
of an object’s position
•
Kinetic energy is energy associated with the
motion of an object.
6.1
Energy can be converted from one
form of energy to another.
Potential
Energy
Kinetic
Energy
Energy cannot be created or
destroyed.
=> Law of conservation of energy =
total quantity of energy in the
universe is constant
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
2
1 joule = 1 kg m /sec
K
1 calorie = 4.184 joule
(exactly)
°F
900C
2
°C
400C
greater thermal energy
Thermochemistry is the study of heat change in chemical
& physical processes.
=> chemical reactions
=> phase changes
Study the interconversion of energy from one
form to another.
The system is the specific part of the universe that is of
interest in the study.
The surroundings is the rest of the universe.
SURROUNDINGS
SYSTEM
System Types:
open
Exchange: mass & energy
closed
isolated
energy
nothing
6.2
State – specific characteristics of a system
(physical or chemical)
Examples:
Pressure
Temperature
Composition
State Function – a property of a system that
characterizes a state of a system and is
independent of how the system got to that state.
Examples:
Pressure
Volume
Composition
Thermodynamics
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
Energy it took each
hiker to get to the
top is different !!
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
Exothermic
Endothermic
6.2
First law of thermodynamics – energy
can be converted from one form to another,
but cannot be created or destroyed.
DEsystem + DEsurroundings = 0
or
DEsystem = -DEsurroundings
E is a state function.
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
6.3
Another form of the first law for DEsystem
DE = q + w
Energy is expressed in units of joules of kJ.
DE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PDV when a gas expands against a constant external pressure
Note the conditions that define each process !!
Work Done On The System
w = Fd
DV < 0
w = -P DV
PxV=
-PDV > 0
F
x d3 = Fd = w
2
d
wsys > 0
Work is done
by the
surroundings
on the
system
Work is
not a
state
function!
Dw = wfinal - winitial
initial
final
Work Done By The System
w = Fd
DV > 0
w = -P DV
PxV=
F
x d3 = Fd = w
2
d
-PDV < 0
wsys < 0
Work is done
by the
system on
the
surroundings
Work is
not a
state
function!
Dw = wfinal - winitial
initial
final
6.3
A sample of nitrogen gas expands in volume from 1.6
L to 5.4 L at constant temperature. What is the work
in joules if the gas expands (a) against a vacuum and
(b) against a constant pressure of 3.7 atm?
(a)
w = -P DV
DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b)
DV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = -14.1 L•atm
Energy needs to be converted to joules.
Use 1 L . atm = 101.3 J
101.3 J = -1430 J
w = -14.1 L•atm x
1L•atm
The change in volume was the same in both cases, but the
work done was not the same => work is not a state function.
Chemistry in Action: Making Snow
DE = q + w
Inside machine = air + water vapor
@ 20 atm
Outside machine = 1 atm
Air w/ water vapor rapidly expands
q=0
adiabatic process
DE = w
w<0
DE < 0
Kinetic Energy (gas) a T (K)
DE = CDT
DT < 0, SNOW !!
Enthalpy and the First Law of Thermodynamics
DE = q + w
At constant pressure:
q = DH and w = -PDV
DE = DH - PDV
Note the
conditions
that define
each
process !!
DH = DE + PDV
6.4
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
Enthalpy is a state function.
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
6.4
Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
DH = 6.01 kJ
6.4
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH = -890.4 kJ
6.4
Enthalpy is a state function.
We only care about where we start and where
we finish, not the route.
DHnet = DH1 + DH2 + DH3 + DH4
DH2
DH3
DH1
end
start
DH4
=> We can construct a convenient path.
DHnet
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (l)
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
You must pay attention to the physical state and
the quantity of all reactants and products !!!
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
DH = -3013 kJ
3013 kJ
= 6470 kJ
1 mol P4
6.4
“Feeding the Army”
Flameless Ration Heaters
Chemical Equation:
Mg (s) + 2H2O (l)
Mg(OH)2 (s) + H2 (g)
+ 353kJ
Solid magnesium and water yields solid
magnesium hydroxide, hydrogen gas &
353 kilojoules of heat energy
Source- Tocci, Viehland. 1996. Holt Chemistry Visualizing Matter.
Austin, Texas: Holt, Rinehart and Winston, Inc. pg. 271, 303
and
http://chem200.tripod.com/
It takes about 353kJ of heat energy is needed to heat the MRE to 60°C .
Calculate the amount of magnesium needed.
353 kJ
= 1 mol Mg = 24.3 g Mg
Calculate the volume of water needed.
353 kJ
= 2 mol H20
18.01 g / mol H20 H20 density = 1.00 g/mL
(2 mol H20) x (18.01 g / mol H20) x (1 mL H20 / 1.00 g H20) = 36.0 mL H20
Calculate the mass of magnesium hydroxide Mg(OH)2 produced.
24.0g Mg = 1 mol Mg = 1 mol Mg(OH)2
58.0 g / mol Mg(OH)2
58.0 g Mg(OH)2
Calculate the number of moles of hydrogen gas H2 produced.
353 kJ
= 1 mol Mg = 1.00 mol H2
Calculate the volume of hydrogen gas H2 produced at 1.00 atm and 25°C.
PV = nRT
=>
V = nRT/P
V = (1.00 mol)(0.08206 L atm/K mol)(298 K)/1.00 atm = 24.5 L
What would happen if 25 mL of water is added?
What would happen if 360 mL of water is added?
What would happen is the soldier were smoking
while heating dinner?
Is the material in a used flameless ration heater
safe to touch?
Can the heater be reused?
A Comparison of DH and DE
2Na (s) + 2H2O (l)
DE = DH - PDV
2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol
At 25 0C, 1 mole H2 = 24.5 L at 1.0 atm
PDV = 1.0 atm x 24.5 L = 2.5 kJ
DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
6.4
DHnet = DH1 + DH2 + DH3 + DH4 + DH5
5
4
3
2
1
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity (m)
of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = msDt
q = CDt
Dt = tfinal - tinitial
6.5
How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
s of Fe = 0.444 J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.5
How much heat is required to take exactly one mole
of water from a solid at -25.0°C to steam at 125.0°C?
DHnet = DH1 + DH2 + DH3 + DH4 + DH5
DH1 = q1 = msDt1 m = (1 mol ) (18.01 g/mol) = 18.01 g
s = 2.030 J / g °C
Dt1 = (0°C - -25°C) = 25°C
DH1 = (18.01 g) (2.030 J / g °C) (25°C) = 914 J = 0.914 kJ
DH2 = DHfus = (6.01 kJ/mol ) (1 mol) = 6.01 kJ
DH = 6.01 kJ
H2O (s)
H2O (l)
DH3 = q3 = msDt3
s = 4.184 J / g °C
Dt3 = (100.°C -0°C) = 100.0°C
DH3 = (18.01 g)(4.184 J / g °C)(100.0°C) = 7540 J = 7.54 kJ
DH4 = DHvap = (44.0 kJ/mol ) (1 mol) = 44.0 kJ
DH = 44.0 kJ
H2O (l)
H2O (g)
DH5 = q5 = msDt5
s = 2.043 J / g °C
Dt5 = (125.0°C – 100.0°C) = 25.0°C
DH5 = (18.01 g) (2.043 J / g °C) (25.0°C) = 920 J = 0.920 kJ
DHnet = DH1 + DH2 + DH3 + DH4 + DH5
= 0.914 kJ + 6.01 kJ + 7.54 kJ + 44.0 kJ + 0.920 kJ = 56.1 kJ
Constant-Volume Calorimetry
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = msDt
qbomb = CbombDt
Reaction at Constant V
DH = qrxn
DH ~ qrxn
No heat enters or leaves!
6.5
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = msDt
qcal = CcalDt
Reaction at Constant P
DH = qrxn
No heat enters or leaves!
6.5
What happens is we mix two liquids at different
temperatures?
Example: Mix 100 g water at 25°C with 100 g
water at 75°C What is the final temperature?
Since both liquids are the same (water) and both
have the same weight (100 g) => take the average
100g @ 75°C
Loss of thermal energy
200g @ 50°C
Gain of thermal energy
100g @ 25°C
What happens if the substances are different or
the amounts are different?
If 60.56 g of Copper at 100.0°C is added to 150. g
water at 25.0°C What is the final temperature?
| Heat lost by copper | = | Heat gained by water |
qsys = qwater + qcal + qCu = 0
assume qcal is very very small
- qCu = qH2O
-qCu = - mCusCuDtCu = mH2OsH2ODtH2O
Dt = (tfinal – tinitial)
sCu = 0.385 J / g °C
sH2O = 4.184 J / g °C
-(60.56 g)(0.385 J/g°C)(tfinal - 100.0°C) = (150. g)(4.184 J/g°C)(tinitial -25.0°C)
(-23.3 J/g°C) tfinal + 2330 J = (628 J/g°C) tfinal – 15700 J
15700 J + 2330 J = ((628 + 23.3) J/g°C) tfinal
tfinal = (18000/651.3)°C
tfinal = 27.7°C
6.5
Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
A 0.1375 g sample of solid magnesium is burned in a
constant volume bomb calorimeter that has a heat capacity
of 3027 J/°C. The temperature increases by 1.126 °C.
Calculate the heat change by the burning Mg in kJ/g and in
kJ/mol.
| Heat given off by reaction | = | Heat gained by calorimeter |
qCal = CpDt = (3024 J/°C)(1.126 °C) = 3.405 x103 J
qRxn = (3.405 x103 J)(1 kJ / 1000 J) = 24.76 kJ / g Mg
0.1375 g Mg
(24.76 kJ / g Mg)(24.31 g Mg / mol Mg) = 601.9 kJ / mol Mg
Is the reaction exothermic or endothermic?
What would happen to the temperature of the water if it
were the opposite type reaction?
Because there is no way to measure the absolute value of
the enthalpy of a substance, must I measure the enthalpy
change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
formation (DH0f ) as a reference point for all enthalpy
expressions.
Standard enthalpy of formation (DH0f) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero.
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
6.6
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
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.
(Enthalpy is a state function. It doesn’t matter how you
get there, only where you start and end.)
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
6.6
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?
6.7
The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7