Brad Collins
Thermochemistry-Part 1
Chapter 7
Thermochemistry
• Thermodynamics: The study of energy
• Thermochemistry: The study of energy in
chemical reactions
• Energy: The capacity to do work
• Work = force x distance
• Force = mass ÷ area
7.1
Systems and Surroundings
The system is the specific part of the universe that is of
interest in the study.
SURROUNDINGS
SYSTEM
Exchange:
open
closed
isolated
mass & energy
energy
nothing
7.1
An 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
An 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)
7.1
Exo vs. Endothermic
An Illustration
Exothermic
Endothermic
7.1
Types of Energy
• Potential energy is the energy of position, or stored
energy.
• 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 an atom.
• Elastic energy is the energy stored in a compressed
gas or coiled spring (sometimes called mechanical
energy)
7.1
Types of Energy
• Kinetic energy is the energy of motion
• Radiant (electromagnetic) radiation comes from
the sun and is the earth's primary energy source
• Thermal energy is the energy from the random
motion of atoms and molecules
• Sound energy is the transmitted vibrational
energy of atoms and molecules in bulk.
7.1
First Law of Thermodynamics
Energy can be converted from one form to another,
but cannot be created or destroyed.
More Thermochemistry Definitions
Heat is the transfer of thermal energy between two bodies
that are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature X
= Thermal Energy
900C
400C
greater thermal energy
7.1
Units of Heat
• Heat is measured in joules or calories
• A joule (J) is defined as the energy expended in applying a
force of 1 newton over a distance of 1 meter
2
3
2
2
• 1 J = N x m = Pa x m = (kg x m )/s
• A calorie (cal) is defined as the energy required to raise 1
gram of water 1ºC.
• A kilocalorie (kcal, or Cal) is defined as the energy required to
raise 1 kilogram of water 1ºC. Used for foods.
• 1 calorie = 4.184 joules
7.1
Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l)
heat of reaction = –2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
7.1
Specific Heat
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 = m·s
Heat (q) absorbed or released:
q = m·s·Δt
q = C·Δt
Δt = tfinal - tinitial
7.1
Practice Problem
How much heat is required to raise the temperature
of 12.5 g of copper, 57ºC?
q = m·s·∆T
specific heat (s) of copper = 0.365 J/g·ºC
7.1
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
Δt = tfinal – tinitial = 50C – 940C = -890C
q = msΔt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.5
Calorimetry
• The measurement of heat (q)
transferred between a system
and its surroundings
• Uses a calorimeter
• Types of calorimeters:
• Constant volume
• Constant pressure
7.1
Constant-Volume Calorimetry
Bomb Calorimeter
Reaction at Constant V
qsys = qwater + qbomb + qrxn
qsys = 0 (an isolated system)
qrxn = – (qwater + qbomb)
qwater = msΔt
qbomb = CbombΔt
Cbomb is the heat capacity of the calorimeter
No heat enters or leaves!
7.1
Constant-Pressure Calorimetry
Reaction at constant pressure
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = msΔt
qcal = CcalΔt
Ccal is the heat capacity of the calorimeter
No heat enters or leaves!
7.1
Practice Problem
A lead pellet having a mass of 26.47 g was heated to a temperature of 89.98ºC
and placed in a constant-pressure calorimeter containing 100.0 g of water at
22.50ºC. If the calorimeter has a heat capacity of approximately zero and the
final temperature of the water plus lead pellet is 23.17ºC, what is the specific
heat of the lead pellet?
7.1
Solution
Heat lost by the pellet is gained by the water in the calorimeter.
Because the calorimeter has a heat capacity of zero, no heat is
lost to the calorimeter, so:
qpellet + qwater = 0
Heat gained by the water is:
qpellet = – qwater
∆T = Tf(ºC) – Ti(ºC)
qwater = m·s·∆T, where ∆T = 23.17ºC – 22.50ºC
qwater = 100.0 g·4.184 J/g·ºC·0.67ºC = 280.328 J
Heat lost by the pellet is:
qpellet = – qwater = –280.328 J
Specific heat of lead is:
∆T = 23.17ºC – 89.98ºC
slead = qlead/m·∆T = –280.328 J/(26.47 g·(–66.81ºC)) = 0.158 J/g·ºC
7.1
Enthalpy (H)
• Enthalpy (H) is used to measure the heat flow into or out of a system for a
process that occurs at constant pressure.
• Enthalpy change (∆H) is the difference between the enthalpy under initial
conditions and enthalpy under final conditions
• ∆H = Hfinal – Hinitial
• H and ∆H are both state functions
• Enthalpy is a molar quantity, meaning it is expressed in J/mol
• Under normal lab conditions ∆H = q
• Negative values for ∆H indicate exothermic processes.
• Positive values for ∆H indicate endothermic processes.
7.2
Enthalpy Change (∆H)
ΔH = heat given off or absorbed during a reaction at constant pressure
ΔH = H (products) – H (reactants)
Hproducts < Hreactants
ΔH < 0
Hproducts > Hreactants
7.2
ΔH > 0
Enthalpy Change Examples
Is ΔH negative or positive?
System absorbs heat
Endothermic
ΔH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
ΔH = 6.01 kJ
7.2
Enthalpy Change Examples
Is ΔH negative or positive?
System gives off heat
Exothermic
ΔH < 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)
ΔH = -890.4 kJ
7.2
Thermochemical Equations
•
Chemical equations that include the enthalpy change as
part of the balanced chemical reaction
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
ΔH = 6.01 kJ
If you reverse a reaction, the sign of ΔH changes
H2O (l)
•
H2O (l)
H2O (s)
ΔH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then ΔH must change by the same factor n.
2H2O (s)
2H2O (l)
ΔH = 2 x 6.01 = 12.0 kJ
7.2
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
ΔH = 6.01 kJ
H2O (l)
H2O (g)
ΔH = 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
1 mol P4
P4O10 (s)
x
123.9 g P4
3013 kJ
1 mol P4
ΔH = -3013 kJ
= 6470 kJ
7.2
Enthalpy and Phase
Changes
7.2
Enthalpy and Phase
Changes
• As heat is added to a substance it warms, and undergoes phase
changes.
• The heat required to warm a substance is given by the specific heat
for the phase (s, l, g) of the substance times its mass (q=m·s·∆T)
• The heat required to melt a substance is the enthalpy of fusion (∆Hfus)
#
H2O (s)
H2O (l)
ΔHfus = 6.01 kJ
• The heat required to vaporize a substance is the enthalpy of
vaporization (∆Hvap)
H2O (l)
H2O (g)
ΔHvap = 44.0 kJ
7.2
Practice Problem
How much heat is required to raise the
temperature of 10.0 g of water from –10ºC to
+50ºC?
7.2
Hess's Law
• Germaine Henri Hess (Russian chemist)
• Thermochemical equations are additive
C(s) + ½O2(g) —> CO(g)
∆H = –110.5 kJ
+ CO(g) + ½O2(g) —> CO2(g)
∆H = –283.0 kJ
C(s) + O2(g) —> CO2(g)
∆H = –393.5 kJ
• Hess’s Law works because ∆H is a state function
7.3
State Functions
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy (E), enthalpy (H), pressure, volume, temperature
ΔE = Efinal - Einitial
ΔH = Hfinal - Hinitial
ΔP = Pfinal - Pinitial
ΔV = Vfinal - Vinitial
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
ΔT = Tfinal - Tinitial
7.3
Hess’s Law Practice
Calculate the enthalpy change of:
C2H2(g) + 2Cl2(g) —> C2H2Cl4(l)
Using the following thermochemical equations:
2C(s) + H2(g) —> C2H2(g)
2C(s) + 2Cl2(g) —> C2H2Cl4(l)
∆H = 227 kJ
∆H = 130 kJ
7.3
Enthalpy of Formation
Definition: The enthalpy change required to form one mole of a substance from each
of its elements in their standard states.
• Standard state of an element is its state in pure form at 1 atm.
• Standard states of some common elements
• Metals: are monatomic solids,except Hg(l)
• Diatomic elements: are gases, except Br(l) and I(s)
• Carbon: graphite, written as C(graphite)
• Phosphorus: white phosphorus, P(white)
• Sulfur: rhombic sulfur, S(rhombic)
• Enthalpy of formation (∆Hf) is determined experimentally
C(graphite) + O2(g) —> CO2(g)
∆Hf = –393.5 kJ
7.4
7.4
Using ∆Hf to Calculate ∆Hrxn
Use ∆Hº*f of the reactants and products
C6H12O6(s) + 6 O2(g) —> 6 CO2(g) + 6H2O(l)
Calculate using ∑ n·∆Hºf{products} – ∑ m·∆Hºf{reactants}
∆H0rxn = [6·∆Hºf{CO2(g)} + 6·∆Hºf{H2O(l)}] – [∆Hºf{C6H12O6(s)} + 6·∆Hºf{O2(g)}]
From Table, p. 253
∆Hºf{CO2(g)} = –393.5 kJ/mol
∆Hºf{O2(g)} = 0 kJ/mol
∆Hºf{H2O(l)} = –241.8 kJ/mol
∆Hºf{C6H12O6(s)} = –1274.5 kJ/mol
∆H0rxn = [6·(–393.5) + 6·(–241.8)] – [–1274.5 + 6·(0)] = –2537.3 kJ
Try this one: CaCO3(s) —> CaO(s) + CO2(g)
*Indicates ∆Hf was measured at 25ºC and 1 atm
6.5
∆Hºrxn Practice Problem
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)
∆Hºrxn = ∑n·∆Hºf(products) – ∑m·∆Hºf(reactants)
∆Hºrxn = [12·∆Hºf(CO2) + 6·∆Hºf(H2O)] – ∑2·∆Hºf(C6H6)
∆Hºrxn = [12·(–393.5 kJ) + 6·(–285.8 kJ)] – 2·(49.04 kJ)
–6535 kJ
2 mol
= –3267 kJ/mol C6H6