# Thermochemistry lecture notes

```Thermochemistry
Heat Transfer in Chemical
Reactions
Energy: Foundational
Definitions
1
• Energy • Work • Forms of energy:
– Kinetic energy – Potential energy -
Forms of Energy, continued
• Chemical Energy • Thermal energy • Radiant energy - solar energy, that is energy that
comes from the sun.
1
Law of the Conservation of
Energy
• The total quantity of energy in the universe is
assumed constant.
•
• Heat • Thermochemistry - the study of heat change in
chemical reactions.
• The heat exchange occurs between:
the system (part of universe of interest) and
the surroundings (the rest of the universe)
Types of Systems
Matter
Heat
Heat
2
Which would represent a can of
green beans on the shelf at a store?
1. Open
2. Closed
3. Isolated
• Exothermic process • Endothermic process CH4 + 2O2
Heat given off by the
system to the surrounding
2Hg + O2
Heat absorbed by the
system from the surroundings
2HgO
CO2 + 2H2O
The process of an ice cube
melting to form water is…
1. Endothermic
2. Exothermic
3
Introduction to
Thermodynamics
2
• Thermodynamics • State of a system - the values of all relevant
macroscopic properties.
– composition
–
– temperature
– pressure
– volume
• State function • For example:
– Volume is a state function. 6 ounces is 6 ounces
regardless of how you got 6 ounces.
• The magnitude of change only depends on the initial
and final values
• State functions are abbreviated with capital letters.
• Other examples of state functions:
– Energy
–Total internal energy (E)
–
–
– Temperature
– Pressure
4
State Functions are “Path
Independent”
• First Law of Thermodynamics -
• It is not possible to determine the total amount
of energy in the universe.
• What we determine is E.
• Let us focus on the E of the system
• There are two ways that energy can leave a system:
1.
2.
• Therefore the 1st law of Thermodynamics looks like
this:
5
Sign convention of q and w :
Heat (q)
neg.(-)
pos.(+)
Work (w) neg.(-)
pos.(+)
E = q + w
Expansion of a Gas Does Work
WORK AND HEAT
You probably already know that w = F &times; d
When working with gases, work can also be
defined as
This is called PV work and is the work involved
when gases expand or are compressed.
6
w = –PV
• When using this equation, what are the
units of work?
•
• This isn’t a very useful energy unit, to convert
to Joules:
A certain gas expands from 0.5 L to 2.5 L against
an external pressure of 0 atm. How much work is
done?
A certain gas expands from 0.5
L to 2.5 L against an external
pressure of 1 atm
1.
2.
3.
4.
2.0 J
-2.0 J
200 J
-200 J
7
An exothermic reaction produces a net
increase of gas when reacting. What is
the sign of q and w?
1.
2.
3.
4.
q &amp; w are both positive
q &amp; w are both negative
q is positive, w is negative
q is negative, w is positive
Can we determine the sign of
E for the above process?
1. Yes, E is positive
2. Yes, E is negative
3. No, you need to know
the size of w and q to
determine this.
3
Enthalpy
• Enthalpy • It is impossible to determine the enthalpy of a
substance.
• We will measure the change in enthalpy (H)
• The Greek letter delta means change.
• So (anything) =
8
• For a chemical reaction H, the enthalpy of
reaction, is given by:
H =
CH4 + 2O2
Heat given off by the
system to the surrounding
2Hg + O2
Heat absorbed by the
system from the surroundings
2HgO
CO2 + 2H2O
Thermite demo: is the reaction
endothermic or exothermic? What is the
sign of H?
1.
2.
3.
4.
Endothermic, positive
Exothermic, positive
Endothermic, negative
Exothermic, negative
Freeze flask to a board demo: is the
reaction endothermic or exothermic? What
is the sign of H?
1.
2.
3.
4.
Endothermic, positive
Exothermic, positive
Endothermic, negative
Exothermic, negative
9
Which statement is true for the reaction:
SO2(g)+ &frac12; O2(g)  SO3(g) H = -99.1 kJ
1. The reaction is endothermic, heat is consumed.
2. The reaction is exothermic and work is positive.
3. The E is positive.
For reactions that take place in
constant pressure environments, the
enthalpy change is synonymous with
the heat change.
4
Thermochemical Equations
• Thermochemical equations - show the enthalpy
changes as well as the mass relationships.
• Here is an example:
• H2O(s)  H2O(l)
H = 6.01 kJ
•
10
H2O(l)
6.01 kJ Heat absorbed by the
system from the surroundings
H2O(s)
• The coefficients always refer to the number of
moles of the substance.
• Reversing an equation, the sign changes for H.
– In our example with water, if we are freezing
1 mol of water, the H = –6.01 kJ. or
H = –6.01 kJ
– H2O(l)  H2O(s)
• If we multiply both sides of the equation by n, must
also multiply H by n.
H = 12.02 kJ.
– 2H2O(s) 2 H2O(l)
• Equations must always specify the physical states of
all the substances in the equation.
Let’s examine the following thermochemical
equation:
CH4(g) + 2O2(g)  CO2(g)+ 2H2O(l) H = –890.4 kJ
11
What would H be for this equation?
2CO2(g) + 4H2O(l)  2CH4(g) + 4O2(g)
1.
2.
3.
4.
– 890.4 kJ
890.4 kJ
–1780.8 kJ
1780.8 kJ
• Let’s continue to examine the equation:
CH4(g) + 2O2(g)  CO2(g) +2 H2O(l)
H = – 890.4 kJ
• When 1 mole of CO2(g) is produced, what is
the enthalpy change?
•
• When 2.5 mol of O2 is reacted with an
excess of CH4, what is the enthalpy change?
• Note the unit factor. In a
thermochemical equation, a unit factor
can be written for any substance in the
equation.
• We now have a new “conversion tool”
and a new road on our map.
•
12
Determine the amount of heat (in kJ) given off when
1.26 &times; 104 g of ammonia are produced according to the
equation:
N2(g) + 3H2(g) 2NH3(g) Horxn = –92.6 kJ
1.
2.
3.
4.
– 6.86  104 kJ
– 5.83 105 kJ
– 3.43 &times; 104 kJ
– 1.17 106 kJ
A reaction is used to propel rockets is
N2O4(l) + 2N2H4(l)  3N2(g) + 4H2O(g)
When 10.0 g of liquid N2O4 is consumed in
this reaction, 124 kJ of heat is released.
(a) The sign of H is
1. Positive
2. Negative
A reaction is used to propel rockets is
N2O4(l) + 2N2H4(l)  3N2(g) + 4H2O(g)
When 10.0 g of liquid N2O4 is consumed in
this reaction, 124 kJ of heat is released.
(b) Is more or less heat released than the
124 kJ for the quantities given in the
chemical reaction equation?
1. More
2. Less
13
A reaction is used to propel rockets is
N2O4(l) + 2N2H4(l)  3N2(g) + 4H2O(g)
When 10.0 g of liquid N2O4 is consumed in
this reaction, 124 kJ of heat is released.
(c) What is the value of H for the chemical
reaction as written? We will do together.
Consider the reaction:
2CH3OH(l) + 3O2(g)  4H2O(l) + 2CO2(g) H = – 1452.8 kJ/mol
If 15.0 grams of methanol (CH3OH) reacted. How much heat would
be produced in kJ?
When 100. g of NO is burned, 191 kJ of heat is released.
Determine H for the reaction 2NO(g) + O2(g)  2NO2(g)
14
When 55.5 g ammonia reacts with oxygen, 734 kJ of heat is
released. Determine H for the following balanced equation.
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
1.
2.
3.
4.
5.
6.
901 kJ
-901 kJ
225 kJ
-225 kJ
13.2 kJ
-13.2 kJ
Enthalpy Change and the First Law of
Thermodynamics.
5
E = q + w at constant pressure becomes
We can now determine the change in internal
energy of a reaction under constant P conditions.
What is qrxn when 1.0 g of Na reacts with H2O
according to this reaction?
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
Horxn = – 368.4 kJ
1.
2.
3.
4.
– 8.0 kJ
– 16 kJ
+ 4.0 kJ
– 180 kJ
15
Na is placed into the water it produces H2(g). If we know:
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) H = – 367.5 kJ
We can use E = H – PV to calc. E
E = H – PV
• If, in the course of a reaction, there is no change
in the number of gas moles, what can we say
• Even when there are moles of gas produced, the
PV term is small so
• We can use PV = nRT to substitute into the top
equation giving:
Calculate the change in internal energy E when 1 mol
of CH4 reacts with two moles of O2 according to the
following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = – 890.4 kJ
16
Consider the reaction
H2(g) + Cl2(g)  2HCl(g) H = ─184.6 kJ
1) What is the heat released by the reaction if 5.0 grams of
hydrogen reacts with 20.0 grams of chlorine? 2) What is the
work done against a pressure of 1.0 atm at 25oC. 3) What is
E for this reaction, for the amounts reacted? 4) What is E
for the equation as written?
Four ways to determine H of
a reaction.
1. Calorimetry. We will experimentally determine the heat
change of the surrounding when the reaction is performed in
the laboratory. (Learning Objective 6)
2. Direct method - using Hof (Learning Objective 7)
3. Indirect method - using Hess’s Law. (Learning Objective 7)
4. Estimating it using bond dissociation energies. (Learning
Objective 7)
17
Calorimetry: Determination of
Enthalpy via Experimentation
6
• is the measurement of heat changes.
• To determine the H of a reaction, we can
measure the heat change of the surroundings.
•
Terms and Equations for
Calorimetry
• Specific heat (s) -
•
• Heat capacity (C) =
•
Calculating heat absorbed or
released by an object:
• q=
• q =heat change
t = change in temperature; tf - ti.
18
Calculate the amount of heat liberated (in kJ)
from 366 g of mercury when it cools from 77.0oC
to 12.0oC. s(of Hg) = 0.139 J/(goC)
What is the heat absorbed by 50.0 g
copper block when it is heated from
25oC to 90oC? (s of Cu is 0.385 J/goC)
1.
2.
3.
4.
0.481 kJ
1.73 kJ
2.21 kJ
1.25 kJ
Constant Volume Calorimetry
•Used for combustion
reactions.
•C,H,O+O2  CO2+H2O
19
Constant Pressure Calorimetry
•Used often for aq. soln.
reactions.
Dissolving 6.00 g of CaCl2 in 300. mL of water causes
the temperature of the solution to increase 3.43oC.
Assuming the specific heat of the solution is 4.18 J/goC
and assuming the calorimeter absorbs no significant
amount of heat, determine H for the reaction.
CaCl2(s)  Ca2+(aq) + 2Cl–(aq)
What is the mass of the surroundings?
1. Not enough
information is given
2. 300 g
3. 306 g
Dissolving 6.00 g of CaCl2 in 300. mL of water causes
the temperature of the solution to increase 3.43oC.
Assuming the specific heat of the solution is 4.18 J/goC
and assuming the calorimeter absorbs no significant
amount of heat, determine H for the reaction.
CaCl2(s)  Ca2+(aq) + 2Cl–(aq)
20
Calculate the number of moles of CH4 (methane) required to heat
200 mL of water (enough to make a cup of coffee or tea) from 20oC
to 100oC. Assume that only 50 percent of the heat generated by the
combustion is used to heat the water; the rest of the heat is lost to
the surroundings.
The reaction for methane is:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H= – 890.4 kJ
7
Calculating Enthalpy
• Once foundational measurements are made to
determine some H values, manipulations can be
made to determine H for other reactions.
• Three other ways
– Using Enthalpy of Formation
– Using Hess’s Law
– Using Bond Energies
Standard Enthalpy of
Formation
• Standard enthalpy of formation Hof -
• Standard state - 1 atm. The o refers to the std.
state.
21
• If we look up the Hof of MgCO3(s) we see
it equals – 1112.9 kJ/mol. What does this
mean?
• It is the enthalpy change for this reaction:
What is the reaction equation
for the Hfo of CH4(g)
1.
2.
3.
4.
5.
CH4(g)  C(s) + 2H2(g)
C(s) + 4H(g)  CH4(g)
C(s) + 2H2(g)  CH4(g)
2C(s) + 4H2(g)  2CH4(g)
Both 3 and 4
Look back at the definition of
standard enthalpy of formation.
What is the value for Hfo of O2?
1.
2.
3.
4.
Must be negative
Must be positive
Must be zero
Cannot be determined from the definition. Need
22
• Standard Enthalpy of Reaction, Horxn • The of ’s are used to determine Horxn
according to the following equation:
Horxn = nHof(products) – mHof(reactants)
• n,m are the coefficients in balanced equation.
• Using this equation to determine Horxn is called
the direct method.
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
Horxn =
1.
2.
3.
4.
-2598.8 kJ
-905.6 kJ
-1928.4 kJ
-1692.4 kJ
Compound
C2H2(g)
CO2(g)
H2O(l)
Hfo
226.6 kJ/mol
-393.5 kJ/mol
-285.8 kJ/mol
What is Hof (in kJ/mol) of C2H6(g) given the following
information?
2 C2H6(g) + 7O2(g)  4 CO2(g) + 6 H2O(l) Horxn= – 3119.4 kJ
o
H f of CO2(g) = –393.5 kJ/mol
Hof of H2O(l)= –285.8 kJ/mol
1.
2.
3.
4.
-84.7 kJ/mol
-169.4 kJ/mol
84.7 kJ/mol
169.4 kJ/mol
23
For the reaction CaCO3(s)  CaO(s) + CO2(g) Ho = 177.8 kJ
Given: Hof(CaO(s)) = – 635.6 kJ/mol
and Hof (CO2(g)) = – 393.5 kJ/mol
What is the standard enthalpy of formation of CaCO3?
1.
2.
3.
4.
– 64.3 kJ/mol
– 419.9 kJ/mol
– 851.3 kJ/mol
– 1206.9 kJ/mol
For the reaction:
2 H2(g) + O2(g)  2 H2O(l)
Horxn =

2.
3.
4.
Hof(H2O(l))
2 Hof(H2O(l))
– Hof(H2O(l))
– &frac12; Hof(H2O(l))
Hess’s Law - The Indirect
Method for cal. Hrxn
• Hess’s Law -
• To calculate the H, you will be given several
thermochemical equations.
• Let’s see it at work by doing an example.
24
From the following data,
C(gr) +O2(g)  CO2(g)
Horxn = – 393.5 kJ
H2(g) + 1/2O2(g)  H2O(l) Horxn = – 285.8 kJ
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)
Horxn = – 3119.6 kJ
Calculate the enthalpy change for the reaction
2C(gr) + 3H2(g)  C2H6(g)
Heat of Solution and Dilution:
An Application of Hess’s Law
Heat of Hydration
Hhydration = -784 kJ/mol
Heat of solution
Hsoln = 4 kJ/mol
25
Estimating Enthalpy Change
using Bond Energies
• Bond energy
• H2(g)  H(g) + H(g) Ho = 436.5 kJ
• Table on the next slide gives some bond energies.
Some of these are “average bond energies”
• We can use the values to estimate the Horxn
Average Bond Energies
*The
Bond
Bond Energy
kJ/mol
Bond
Bond Energy
kJ/mol
H–H
436.4
C–O
351
H–N
393
C=O
745
H–O
460
C=O*
799
H–F
568.2
N–N
193
C–H
414
N=N
418
C–C
347
N≡N
941.4
C=C
620
N–O
176
C≡C
812
N=O
607
C–N
276
O–O
142
C=N
615
O=O
498.7
C≡N
891
F–F
156.9
bond energy from CO2 Bond energies in RED are for diatomic molecules, not an average.
Enthalpy Change Using Bond Energies
Reactant
Molecules
Requires
Energy
Requires
Energy
Product
Molecules
Atoms
Releases
Energy
Releases
Energy
Enthalpy
Atoms
Reactant
Molecules
Product
Molecules
26
• To use this method, must write the Lewis
structures for all substances in the equation
so that you know what bonds are broken and
what bonds are formed.
For the reaction:
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
Predict the enthalpy of reaction from the
average bond energies of Table 9.3
What is the estimated H for the
reaction?
1.
2.
3.
4.
5.
6.
-2812 kJ
-2308 kJ
-803kJ
803 kJ
2308 kJ
2812 kJ
27
If the bond energy for H2 is
436.5 kJ/mol, what is the Hof of H(g)?
1.
2.
3.
4.
5.
6.
436.5 kJ/mol
-436.5 kJ/mol
218.3 kJ/mol
-218.3 kJ/mol
873.0 kJ/mol
-873.0 kJ/mol
Hint: write the reaction
for both bond energy and
heat of formation.
28
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