Download Thermochemistry

Advanced Topics in
Chemistry
Thermochemistry
Objectives
1· Define and apply the terms standard state,
standard enthalpy change of formation, and standard
enthalpy change of combustion
2· Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and
combustion
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Standard Enthalpy 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
900C
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400C
greater thermal energy1/2013
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
open
Exchange: mass & energy
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closed
isolated
energy
nothing
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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)
2Hg (l) + O2 (g)
energy + H2O (s)
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H2O (l)
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Thermodynamics is the scientific study of the
interconversion of heat and other kinds of energy.
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
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different
paths.
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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
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
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system
surroundings
Another form of the first law for DEsystem
DE = q + w
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
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Work Done On the System
w=Fxd
w = -P DV
PxV=
DV > 0
F
x d3 = F x d = w
2
d
-PDV < 0
wsys < 0
Work is
not a
state
function!
Dw = wfinal - winitial
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initial
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final
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A sample of nitrogen gas expands in volume from 1.6 L to
5.4 L at constant temperature. What is the work done in
joules if the gas expands (a) against a vacuum and (b)
against a constant pressure of 3.7 atm?
w = -P DV
(a)
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
w = -14.1 L•atm x
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101.3 J = -1430 J
1L•atm
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Enthalpy and the First Law of Thermodynamics
DE = q + w
At constant pressure:
q = DH and w = -PDV
DE = DH - PDV
DH = DE + PDV
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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)
DH = heat given off or absorbed during a reaction at constant pressure
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Hproducts < Hreactants
DH < 0
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Hproducts > Hreactants
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DH > 0
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)
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H2O (l)
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DH = 6.01 kJ
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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)
Dr. Dura
CO2 (g) + 2H2O (l) DH = -890.4 kJ
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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)
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2H2O (l)
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DH = 2 x 6.01 = 12.0 kJ
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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
Dr. Dura
P4O10 (s)
1 mol P4
123.9 g P4
x
16
DH = -3013 kJ
3013 kJ
= 6470 kJ
1 mol P4
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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 atm
PDV = 1 atm x 24.5 L = 2.5 kJ
DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
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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=mxs
Heat (q) absorbed or released:
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial
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Chemistry in Action:
Fuel Values of Foods and Other Substances
6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
C6H12O6 (s) + 6O2 (g)
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
Substance
DHcombustion (kJ/g)
Apple
-2
Beef
-8
Beer
-1.5
Gasoline
-34
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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
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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.)
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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)
22
DH0 rxn
= -393.5 + (2x-296.1)
+ 1072 = 86.3 kJ
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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
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Sample Problem
Given the following thermochemical equations:
B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g)
H2O (l) → H2O (g)
H2 (g) + (1/2) O2 (g) → H2O (l)
2 B (s) + 3 H2 (g) → B2H6 (g)
(ΔH = 2035 kJ/mol)
(ΔH = 44 kJ/mol)
(ΔH = -286 kJ/mol)
(ΔH = 36 kJ/mol)
Determine the ΔHf of:
2 B (s) + (3/2) O2 (g) → B2O3 (s)
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After the multiplication and reversing of the equations (and
their enthalpy changes), the result is:
B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g)
3 H2O (g) → 3 H2O (l)
3 H2O (l) → 3 H2 (g) + (3/2) O2 (g)
2 B (s) + 3 H2 (g) → B2H6 (g)
(ΔH = -2035 kJ/mol)
(ΔH = -132 kJ/mol)
(ΔH = 858 kJ/mol)
(ΔH = 36 kJ/mol)
Adding these equations and canceling out the common terms
on both sides, we get
2 B (s) + (3/2) O2 (g) → B2O3 (s)
(ΔH = -1273 kJ/mol)
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Objectives
3· Define and apply the terms lattice enthalpy and electron affinity
4. Explain how the relative sizes and the charges of ions affect the
lattice enthalpies of different ionic compounds
5. Construct a Born-Haber cycle for group 1 and group 2 oxides and
chlorides, and use it to calculate an enthalpy change (9.1.12.A.1, 9.1.12.B.1)
6· Discuss the difference between theoretical and experimental lattice
enthalpy values of ionic compounds in terms of their covalent character
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Experimental lattice energies cannot be determined directly. An
energy cycle based on Hess’s Law, known as the Born-Haber
cycle is used. The Key Concepts in Born-Haber Cycle are:
Heat of formation Δ Hf
Ionization Energy I.E.
Electron Affinity
E.A.
Atomization (Sublimation) s  g
Bond enthalpy B.E.
Lattice Energy – is the enthalpy change that occurs when one mole of a
solid ionic compound is separated into ions under standard conditions.
Example: NaCl(s)  Na+ (g) + Cl -(g) (endothermic).
Note that in forming ionic compounds, the oppositely charged gaseous
ions come together to form an ionic lattice – this is a very exothermic
process as there is a strong attraction between the ionsi.e. Na+ (g) + Cl (g) NaCl(s)
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Na+(g)
IE
+
Cl-(g)
EA
Na(g)
Cl(g)
DHsub
½
BDE
Lattice E. = 786.5 kJ/mol
DHf
Na(s)
+
½ Cl2(g)
NaCl(s)
Born-Haber Cycle for NaCl
Do NOW
Construct the Born-Haber cycle for KF and using appropriate data,
calculate the electron affinity of fluorine.
Thermodynamic data
K(s) → K(g)
K(g) → K+(g) + e-
ΔH°a = 90 kJ mol-1
IE = 419 kJ mol-1
½ F2(g) → F(g)
K(s) + F(g) → KF(s)
K+(g) + F-(g) → KF(s)
ΔH°a = 79.5 kJ mol-1
ΔH°f = -569 kJ mol-1
ΔH°lattice = -821 kJ mol-1
Choose the correct answer:
a) EA = 336.5 kJ mol-1
b) EA = -336.5 kJ mol-1
c) EA = -840.5 kJ mol-1
Answer: b
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Practice problem
Draw the Born-Haber cycle for potassium sulphide K2S.
Thermodynamic data
Potassium K
Sulphur S
Potassium Sulphide K2S
IEI = 419 kJ mol-1
ΔH°a= 279 kJ mol-1 ΔH°f = -257 kJ mol-1
ΔH°a = 78 kJ mol-1 EAI= -199.5 kJ mol-1 ΔH°lattice = -1979 kJ
mol
EAII - 648.5 kJ mol-1
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Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0 0C and ice melts above 0 0C
• Heat flows from a hotter object to a colder
• object
• A gas expands in an evacuated bulb
spontaneous
• Iron exposed to oxygen and water
forms rust
nonspontaneous
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Does a decrease in enthalpy mean a reaction
proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH0 = -890.4 kJ
H+ (aq) + OH- (aq)
H2O (l) DH0 = -56.2 kJ
H2O (s)
NH4NO3 (s)
H2O (l) DH0 = 6.01 kJ
H2O
NH4+(aq) + NO3- (aq) DH0 = 25 kJ
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Objectives
7. State and explain the factors that increase entropy
in a system
8· Predict whether the entropy change for a given
reaction or process is positive or negative·
9. Calculate the standard entropy change for a
reaction using standard entropy values (5.2.12.D.2,
9.1.12.A.1)
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Entropy (S) is a measure of the randomness or disorder of
a system.
order
disorder
S
S
DS = Sf - Si
If the change from initial to final results in an increase in
randomness
Sf > Si
DS > 0
For any substance, the solid state is more ordered than
the liquid state and the liquid state is more ordered than
gas state
Ssolid < Sliquid << Sgas
H2O (s)
H2O (l)
DS > 0
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Processes that
lead to an
increase in
entropy (DS > 0)
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How does the entropy of a system change for each of
the following processes?
(a) Condensing water vapor
Randomness decreases
Entropy decreases (DS < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases
Entropy decreases (DS < 0)
(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (DS > 0)
(d) Subliming dry ice
Randomness increases
Entropy increases (DS > 0)
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Entropy
State functions are properties that are determined by the
state of the system, regardless of how that condition was
achieved.
energy, enthalpy, pressure,
volume, temperature, entropy
Potential energy of hiker 1 and hiker
2 is the same even though they took
different paths.
1/2013
First Law of Thermodynamics
Energy can be converted from one form to another but
energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous
process and remains unchanged in an equilibrium
process.
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
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Entropy Changes in the System (DSsys)
0 ) is the entropy
The standard entropy of reaction (DS
rxn
change for a reaction carried out at 1 atm and 250C.
aA + bB
DS0rxn =
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
DS0rxn = SnS0(products) - SmS0(reactants)
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
DS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0rxn= 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
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Entropy Changes in the System (DSsys)
When gases are produced (or consumed)
•
If a reaction produces more gas molecules
than it consumes, DS0 > 0.
•
If the total number of gas molecules diminishes,
DS0 < 0.
•
If there is no net change in the total number of
gas molecules, then DS0 may be positive or
negative BUT DS0 will be a small number.
What is the sign of the entropy change for the following reaction?
2Zn (s) + O2 (g)
2ZnO (s)
The total number of gas molecules goes down,
DS is negative.
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Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at
the absolute zero of temperature.
S = k ln W
W=1
where W is the number of
microstates.
There is only one way to
arrange the atoms or
molecules to form a perfect
crystal.
S=0
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Objectives
10.Predict whether a reaction or process will be
spontaneous using the sign of ΔG
11· Calculate ΔG for a reaction using the equation with
enthalpy, temperature, and entropy and by using values of
free energy change of formation, ΔGf (9.1.12.A.1)
12. Predict the effect of a change in temperature on the
spontaneity of a reaction using standard entropy and
enthalpy changes and the equation
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Gibbs Free Energy
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
Gibbs free
energy (G)
DG < 0
For a constant-temperature
process:
DG = DHsys -TDSsys
The reaction is spontaneous in the forward direction.
DG > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = 0
The reaction is at equilibrium.
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The standard free-energy of reaction (DG0rxn) is the freeenergy change for a reaction when it occurs under
standard-state conditions.
aA + bB
cC + dD
0
DGrxn
= [cDG0f (C) + dDG0f (D)] - [aDG0f (A) + bDG0f (B) ]
0
DGrxn
= SnDG0f (products)- SmDG0f (reactants)
Standard free energy of
formation (DG0f ) is the freeenergy change that occurs when
1 mole of the compound is
formed from its elements in their
standard states.
. DGf 0 of any element in its stable
form is zero
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What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
DGrxn
= SnDG0f (products)- SmDG0f (reactants)
0
DGrxn
= [12DG0f (CO2) + 6DG0f (H2O)] - [ 2DG0f (C6H6)]
0
DGrxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
DG0 = -6405 kJ < 0
spontaneous
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DG = DH - TDS
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