Ch 6
Thermochemistry: Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy: Heats of Reaction and Chemical Change
6.3 Calorimetry: Laboratory Measurement of Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hess’s Law of Heat Summation
6.6 Standard Heats of Reaction (∆H0rxn)
6.1
Some basic principles
Energy is the capacity to do work.
State A – Fuel in the tank
change in potential energy
EQUALS
kinetic energy
State B - Fuel burned and
exhaust produced
A system of fuel and exhaust. A fuel is higher in chemical potential
energy than the exhaust. As the fuel burns, some of its potential energy is
converted to the kinetic energy of the moving car.
1
A chemical system and its surroundings
When a chemical reaction takes place, we consider the substances involved.
Therefore, the reactants and products are the system.
the surroundings
the system
The system is a part of universe
which attention is focused.
The surrounding exchanges energy with the system
and make up in principle of the rest of the universe.
Thermodynamics is the study of heat and its transformations.
Thermochemistry is a branch of thermodynamics that deals with
the heat involved with chemical and physical changes.
Fundamental premise
When energy is transferred from one object to another, it appears as work
and/or as heat.
For our work we must define a system to study; everything else then
becomes the surroundings.
The system is composed of particles with their own internal energies (E or
U).
Therefore the system has an internal energy. When a change occurs, the
internal energy changes.
State properties
State property of system is described by giving its composition,
temperature, and pressure.
State property depends on the state of the system, not on the way the
system reaches the state.
2
A system transferring energy as heat only.
When q is negative (-), the heat
flows out of the system into
surrounding.
When q is positive (+), the heat
flows into the system from
surrounding.
A system losing energy as work only.
Zn(s) + 2H+(aq) + 2Cl-(aq)
Energy, E
DE<0
work done on
surroundings
H2(g) + Zn2+(aq) + 2Cl-(aq)
When w is negative (-),
work done by system.
When w is positive (+),
work done on system.
3
The Sign Conventions* for q, w and ∆E
q
+
w
=
∆E
+
+
+
+
-
depends on sizes of q and w
-
+
depends on sizes of q and w
-
-
• For q (heat):
-
+ means system gains heat, Endothermic;
- means system loses heat. Exothermic.
• For w (work): + means work done on system;
- means work done by system.
Magnitude of heat
In any process, we are interested in the direction of
heat flow and heat magnitude.
We express heat, q, in the unit of joules (SI unit) and
kilojoules.
The joules is named for James Joule who carried out
the precise thermodynamic measurement.
Traditionally, chemists use the calorie as an energy
unit.
English physicist
Calorie is the amount of heat needed to raise 1.00 g
water 1 °C.
1 cal = 4.184 J
1 kcal = 4.184 kJ
4
6.2
Specific Heat Capacity and Heat Transfer
It is important to discuss the magnitude of heat flow in chemical reactions of phase changes.
The equation, q = C × ∆T, express the relationship between the magnitude of heat flow and
temperature change. ∆T = Tfinal - Tinitial
The quantity C is known as heat capacity of the system, having a unit J/°C.
Finding the Quantity of Heat from Specific Heat Capacity
PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How
much heat is needed to raise the temperature of the copper layer from
25 0C to 300 0C? The specific heat capacity (c) of Cu is 0.387 J/g*K.
PLAN: Given the mass, specific heat capacity and change in temperature, we can use q
= c x mass x DT to find the answer. DT in 0C is the same as for K.
SOLUTION:
q=
0.387 J
g*K
x 125 g x (300-25) 0C = 1.33x104 J
Coffee-cup calorimeter
•
The heat given out by a reaction is
absorbed by water.
•
The mass of water can be determined.
•
The heat capacity of water is 4.18 J/g •
°C.
•
The temperature change can be
measured by the thermometer.
•
Heat flow can be calculated for the
reaction.
•
Equation, q = mass × c × ∆T, express
the relationship of heat flow and
temperature change.
•
The heat flow for the reaction is equal in
magnitude, but opposite in sign to that
measured by calorimeter
5
Determining the Heat of a Reaction
PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00 0C
and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the
final temperature is 27.21 0C.
Calculate qsoln (in J).
(Assume the total volume is the sum of the individual volumes and that the
final solution has the same density and specfic heat capacity as water: d = 1.00
g/mL and c = 4.18 J/g*K)
PLAN:
1. We need to determine the limiting reactant from the net ionic equation.
2. The moles of NaOH and HCl as well as the total volume can be calculated.
3. From the volume we use density to find the mass of the water formed.
4. At this point, qsoln can be calculated using the eqaution, q = mass × c × ∆T.
•
The heat divided by the M of water will give us the heat per mole of water
formed.
Determining the Heat of a Reaction
SOLUTION:
HCl(aq) + NaOH(aq)
H+(aq)
+
OH-(aq)
NaCl(aq) + H2O(l)
H2O(l)
For NaOH
0.500 M × 0.0500 L = 0.0250 mol OH-
For HCl
0.500 M × 0.0250 L = 0.0125 mol H+
HCl is the limiting reactant.
0.0125 mol of H2O will form during the rxn.
total volume after mixing = 0.0750 L
0.0750 L x 103 mL/L × 1.00 g/mL
= 75.0 g of water
q = mass x specific heat x DT
= 75.0 g × 4.18 J/g* 0C × (27.21-25.00) °C
= 693 J
6
Take-home message: The schematic diagram of A bomb calorimeter
6.6
Calorimetery
Insulated outer
container
Sample
dish
Burning
sample
Steel
bomb
•
The heat given out by a
reaction is absorbed by
water.
•
The mass of water can be
determined.
•
The heat capacity of water
is 4.18 J/g • °C.
•
The temperature change
can be measured by the
thermometer.
•
Heat flow can be calculated
for the reaction.
•
Equation, q = mass × c ×
∆T, express the relationship
of heat flow and
temperature change.
•
The heat flow for the
reaction is equal in
magnitude, but opposite in
sign to that measured by
calorimeter
Calculating the Heat of Combustion
PROBLEM:
A manufacturer claims that its new dietetic dessert has “fewer than 10
KiloCalories per serving.”
To test the claim, a chemist at the Department of Consumer Affairs
places one serving in a bomb calorimeter and burns it in O2 (the heat
capacity of the calorimeter = 8.15 kJ/K).
The temperature increases 4.937 0C. Is the manufacturer’s claim
correct?
PLAN:
- q sample = qcalorimeter
SOLUTION:
qcalorimeter
= heat capacity × ∆T
= 8.151 kJ/K × 4.937 K
= 40.24 kJ
40.24 kJ ×
kcal
= 9.63 Kilocalories
4.18 kJ
The manufacturer’s claim is true.
7
6.3
Energy and changes of state
A cooling curve for the conversion of gaseous water to ice.
Five stages – vapor cools, vapor condenses (constant temperature), liquid water cools,
liquid water freezes (constant tempterature),
tempterature), solid water cools
Quantitative Aspects of Phase Changes
Within a phase, a change in heat is accompanied by a change in
temperature which is associated with a change in average Ek as the
most probable speed of the molecules changes.
q = (amount)(molar heat capacity)(∆T)
During a phase change, a change in heat occurs at a constant
temperature, which is associated with a change in Ep, as the
average distance between molecules changes.
q = (amount)(enthalpy of phase change)
8
6.4
The first law of thermodynamics
The Meaning of Enthalpy
w = - P∆V
∆H
≈ ∆ E in
1. Reactions that do not involve gases.
H = E + PV
where H is enthalpy
∆H
2. Reactions in which the number of moles
of gas does not change.
= DE + P ∆ V
3. Reactions in which the number of moles
of gas does change but q is >>> P ∆ V.
qp = ∆ E + P ∆ V = ∆ H
If a chemical reaction occurs under a constant pressure, the difference in
enthalpy between product and reactant equals the heat flow for the reaction.
Qreaction at a constant pressure = ∆E + P∆V = ∆H
Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
H2O(l)
CH4 + 2O2
H2O(g)
heat out
Enthalpy, H
Enthalpy, H
Hinitial
∆H < 0
H2O(l)
Hfinal
Exothermic process
q < 0,
Hproduct < H reactant
Hfinal
∆H > 0
CO2 + 2H2O
A
H2O(g)
B
heat in
Hinitial
Endothermic process
q > 0, Hproduct > H reactant
9
Drawing Enthalpy Diagrams and Determining the Sign of ∆H
PROBLEM:
PLAN:
In each of the following cases, determine the sign of ∆H, state
whether the reaction is exothermic or endothermic, and draw
and enthalpy diagram.
(a) H2(g) + 1/2O2(g)
H2O(l) + 285.8kJ
(b) 40.7kJ + H2O(l)
H2O(g)
Determine whether heat is a reactant or a product. As a reactant,
the products are at a higher energy and the reaction is
endothermic. The opposite is true for an exothermic reaction
SOLUTION:
(a) The reaction is exothermic.
(b) The reaction is endothermic.
H2(g) + 1/2O2(g) (reactants)
EXOTHERMIC
∆H = -285.8kJ
H2O(l)
6.5
H2O(g)
(products)
∆H = +40.7kJ
ENDOTHERMIC
(products)
H2O(l)
(reactants)
Enthalpy change for chemical reactions
Thermal chemical reaction – shows the enthalpy relationship between reactants and products
•
•
•
•
Rules of thermochemistry
The magnitude of ∆H is directly proportional to the amount of reactants or products.
∆H for a reaction is equal in magnitude but opposite in sign for the reverse reaction.
The value of for a reaction is the same whether it occurs in one step or multi-steps.
Hess law ∆H = ∆H1 + ∆H2 + ∙∙∙∙
Some Important Types of Enthalpy Change
heat of combustion (∆Hcomb)
C4H10(l) + 13/2O2(g)
heat of formation (∆Hf)
K(s) + 1/2Br2(l)
heat of fusion (∆Hfus)
NaCl(s)
heat of vaporization (∆Hvap)
C6H6(l)
4CO2(g) + 5H2O(g)
KBr(s)
NaCl(l)
C6H6(g)
10
Using the Heat of Reaction (∆Hrxn) to Find Amounts
PROBLEM:
The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented by
Al2O3(s)
∆Hrxn = 1676 kJ
2Al(s) + 3/2O2(g)
If aluminum is produced this way, how many grams of aluminum can
form when 1.000x103 kJ of heat is transferred?
PLAN:
SOLUTION:
Heat (kJ)
1676 kJ = 2 mol Al
1.000x103 kJ x
mol of Al
2 mol Al
26.98 g Al
1676 kJ
1 mol Al
X M
= 32.20 g Al
g of Al
6.7
Hess’s Law
to Calculate an Unknown ∆H
PROBLEM:
Two gaseous pollutants that form in auto exhaust are CO and NO. An
environmental chemist is studying ways to convert them to less
harmful gases through the following equation:
CO(g) + NO(g)
CO2(g) + 1/2N2(g) ∆ H = ?
Given the following information, calculate the unknown ∆ H:
Equation A: CO(g) + 1/2O2(g)
2NO(g) ∆ HB = 180.6 kJ
Equation B: N2(g) + O2(g)
PLAN:
CO2(g) ∆ HA = -283.0 kJ
Equations A and B have to be manipulated by reversal and/or multiplication by
factors in order to sum to the first, or target, equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g)
CO(g) + NO(g)
CO2(g) ∆ HA = -283.0 kJ
1/2N2(g) + 1/2O2(g)
∆ HB = -90.3 kJ
CO2(g) + 1/2N2(g)
∆ Hrxn = -373.3 kJ
11
6.8
Standard enthalpies of Formation
PROBLEM:
Write balanced equations for the formation of 1 mol of the following
compounds from their elements in their standard states and include
∆H 0f.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
PLAN:
Use the table of heats of formation for values.
SOLUTION:
(a) Ag(s) + 1/2Cl2(g)
∆H 0f = -127.0 kJ
AgCl(s)
(b) Ca(s) + C(graphite) + 3/2O2(g)
CaCO3(s)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
∆H 0f = -1206.9 kJ
HCN(g)
∆H 0f = 135 kJ
Elements
-∆H0f
formation
Reactants
decomposition
Enthalpy, H
The general process for determining ∆H 0rxn from ∆H0f values.
∆H0f
Hinitial
∆H0rxn
Products
Hfinal
∆H0rxn = Σ m∆H0f(products) - Σ n∆H0f(reactants)
12
Calculating the Heat of Reaction from Heats of Formation
PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion
kilograms, is used to make many products, including fertilizer, dyes,
and explosives. The first step in the industrial production process is
the oxidation of ammonia:
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate ∆H0rxn from ∆H 0f values.
Look up the ∆H0f values and use Hess’s Law to find ∆Hrxn.
PLAN:
∆Hrxn = Σ m∆H0f (products) - Σ n∆H0f (reactants)
SOLUTION:
∆Hrxn = [4(∆H0f NO(g) + 6(∆H0f H2O(g)] - [4(∆H0f NH3(g) + 5(∆H0f O2(g)]
= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
∆Hrxn = -906 kJ
If a chemical reaction occurs under a constant pressure, the difference in
enthalpy between product and reactant equals the heat flow for the reaction.
Qreaction at a constant pressure = ∆E + P∆V = ∆H
Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
H2O(l)
CH4 + 2O2
H2O(g)
heat out
Enthalpy, H
Enthalpy, H
Hinitial
∆H < 0
H2O(l)
Hfinal
Exothermic process
q < 0,
Hproduct < H reactant
Hfinal
∆H > 0
CO2 + 2H2O
A
H2O(g)
B
heat in
Hinitial
Endothermic process
q > 0, Hproduct > H reactant
13
Drawing Enthalpy Diagrams and Determining the Sign of ∆H
PROBLEM:
PLAN:
In each of the following cases, determine the sign of ∆H, state
whether the reaction is exothermic or endothermic, and draw
and enthalpy diagram.
(a) H2(g) + 1/2O2(g)
H2O(l) + 285.8kJ
(b) 40.7kJ + H2O(l)
H2O(g)
Determine whether heat is a reactant or a product. As a reactant,
the products are at a higher energy and the reaction is
endothermic. The opposite is true for an exothermic reaction
SOLUTION:
(a) The reaction is exothermic.
(b) The reaction is endothermic.
H2(g) + 1/2O2(g) (reactants)
EXOTHERMIC
∆H = -285.8kJ
H2O(l)
6.5
H2O(g)
(products)
∆H = +40.7kJ
ENDOTHERMIC
(products)
H2O(l)
(reactants)
Enthalpy change for chemical reactions
Thermal chemical reaction – shows the enthalpy relationship between reactants and products
•
•
•
•
Rules of thermochemistry
The magnitude of ∆H is directly proportional to the amount of reactants or products.
∆H for a reaction is equal in magnitude but opposite in sign for the reverse reaction.
The value of for a reaction is the same whether it occurs in one step or multi-steps.
Hess law ∆H = ∆H1 + ∆H2 + ∙∙∙∙
Some Important Types of Enthalpy Change
heat of combustion (∆Hcomb)
C4H10(l) + 13/2O2(g)
heat of formation (∆Hf)
K(s) + 1/2Br2(l)
heat of fusion (∆Hfus)
NaCl(s)
heat of vaporization (∆Hvap)
C6H6(l)
4CO2(g) + 5H2O(g)
KBr(s)
NaCl(l)
C6H6(g)
14
Using the Heat of Reaction (∆Hrxn) to Find Amounts
PROBLEM:
The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented by
Al2O3(s)
∆Hrxn = 1676 kJ
2Al(s) + 3/2O2(g)
If aluminum is produced this way, how many grams of aluminum can
form when 1.000x103 kJ of heat is transferred?
PLAN:
SOLUTION:
Heat (kJ)
1676 kJ = 2 mol Al
1.000x103 kJ x
mol of Al
2 mol Al
26.98 g Al
1676 kJ
1 mol Al
X M
= 32.20 g Al
g of Al
6.7
Hess’s Law
to Calculate an Unknown ∆H
PROBLEM:
Two gaseous pollutants that form in auto exhaust are CO and NO. An
environmental chemist is studying ways to convert them to less
harmful gases through the following equation:
CO(g) + NO(g)
CO2(g) + 1/2N2(g) ∆ H = ?
Given the following information, calculate the unknown ∆ H:
Equation A: CO(g) + 1/2O2(g)
2NO(g) ∆ HB = 180.6 kJ
Equation B: N2(g) + O2(g)
PLAN:
CO2(g) ∆ HA = -283.0 kJ
Equations A and B have to be manipulated by reversal and/or multiplication by
factors in order to sum to the first, or target, equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g)
CO(g) + NO(g)
CO2(g) ∆ HA = -283.0 kJ
1/2N2(g) + 1/2O2(g)
∆ HB = -90.3 kJ
CO2(g) + 1/2N2(g)
∆ Hrxn = -373.3 kJ
15
6.8
Standard enthalpies of Formation
PROBLEM:
Write balanced equations for the formation of 1 mol of the following
compounds from their elements in their standard states and include
∆H 0f.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
PLAN:
Use the table of heats of formation for values.
SOLUTION:
(a) Ag(s) + 1/2Cl2(g)
∆H 0f = -127.0 kJ
AgCl(s)
(b) Ca(s) + C(graphite) + 3/2O2(g)
CaCO3(s)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
∆H 0f = -1206.9 kJ
HCN(g)
∆H 0f = 135 kJ
Elements
-∆H0f
formation
Reactants
decomposition
Enthalpy, H
The general process for determining ∆H 0rxn from ∆H0f values.
∆H0f
Hinitial
∆H0rxn
Products
Hfinal
∆H0rxn = Σ m∆H0f(products) - Σ n∆H0f(reactants)
16
Calculating the Heat of Reaction from Heats of Formation
PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion
kilograms, is used to make many products, including fertilizer, dyes,
and explosives. The first step in the industrial production process is
the oxidation of ammonia:
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate ∆H0rxn from ∆H 0f values.
PLAN:
Look up the ∆H0f values and use Hess’s Law to find ∆Hrxn.
∆Hrxn = Σ m∆H0f (products) - Σ n∆H0f (reactants)
SOLUTION:
∆Hrxn = [4(∆H0f NO(g) + 6(∆H0f H2O(g)] - [4(∆H0f NH3(g) + 5(∆H0f O2(g)]
= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
∆Hrxn = -906 kJ
Specific Heat Capacities of Some Elements, Compounds, and
Materials
Substance
Specific Heat
Capacity (J/g*K)
Elements
Substance
Specific Heat
Capacity (J/g*K)
Materials
aluminum, Al
0.900
wood
1.76
graphite,C
0.711
cement
0.88
iron, Fe
0.450
glass
0.84
copper, Cu
0.387
granite
0.79
gold, Au
0.129
steel
0.45
Compounds
water, H2O(l)
4.184
ethyl alcohol, C2H5OH(l)
2.46
ethylene glycol, (CH2OH)2(l)
2.42
carbon tetrachloride, CCl4(l)
0.864
17
Selected Standard Heats of Formation at 250C(298K)
Formula
calcium
Ca(s)
CaO(s)
CaCO3(s)
∆H0f(kJ/mol)
0
-635.1
-1206.9
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
chlorine
Cl(g)
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
Formula ∆H0f(kJ/mol)
Formula ∆H0f(kJ/mol)
0
-92.3
silver
Ag(s)
AgCl(s)
hydrogen
H(g)
H2(g)
218
0
sodium
nitrogen
N2(g)
NH3(g)
NO(g)
0
-45.9
90.3
oxygen
O2(g)
O3(g)
H2O(g)
0
143
-241.8
H2O(l)
-285.8
Cl2(g)
HCl(g)
Na(s)
Na(g)
NaCl(s)
0
-127.0
0
107.8
-411.1
sulfur
S8(rhombic) 0
S8(monoclinic) 2
SO2(g)
-296.8
SO3(g)
-396.0
18