This document discusses several key concepts in thermochemistry and thermodynamics: - The first law of thermodynamics relates heat, work, and changes in internal energy. Enthalpy is a useful state function for chemical reactions. - Calorimetry can be used to measure heat flows and determine enthalpy changes during chemical and physical processes. Bomb calorimetry and coffee cup calorimetry are described. - Enthalpy changes accompany chemical reactions and phase changes. The sign and magnitude of enthalpy changes can indicate whether a process is exothermic or endothermic. - Hess's law allows calculation of enthalpy changes from thermochemical data using stepwise processes

1. 6.1 Basic principles
6.2 measurement of heat flow– Calorimetry
(Laboratory Measurement of Heats of Reaction)
6.3 Energy and Changes of State
6.4 The first law of thermodynamics
6.5 Enthalpy change for chemical reaction
6.6 Hess’s Law
6.7 Standard enthalpy of formation
6.8 Product or reactant-favored reaction and thermochemistry

2. 6.1 Some basic principles
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.

3. 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.

4. 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.

5. A system transferring energy as heat only
When q is negative (-), the heat When q is positive (+), the heat
flows out of the system into flows into the system from
surrounding. surrounding.

6. A system losing energy as work only
Zn(s) + 2H+(aq) + 2Cl-(aq)
Energy, E
work done on
surroundings
DE<0
H2(g) + Zn2+(aq) + 2Cl-(aq)
When w is negative (-), When w is positive (+),
work done by system. work done on system.

7. 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.

8. 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.
Calorie is the amount of heat needed to raise 1.00 g
English physicist
water 1 °C.
1 cal = 4.184 J
1 kcal = 4.184 kJ

9. 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: 0.387 J
q= x 125 g x (300-25) 0C = 1.33x104 J
g*K

10. 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

11. 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.

12. Determining the Heat of a Reaction
SOLUTION:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) 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

13. Take-home message: The schematic diagram of A bomb calorimeter
6.6 Calorimetery
• 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

14. 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: = heat capacity × ∆T
qcalorimeter
= 8.151 kJ/K × 4.937 K
= 40.24 kJ
kcal
40.24 kJ × = 9.63 Kilocalories
4.18 kJ
The manufacturer’s claim is true.

15. 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), solid water cools

16. 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)

17. 6.4 The first law of thermodynamics
The Meaning of
Enthalpy
w = - P∆V ∆H ≈ ∆ E in
H = E + PV 1. Reactions that do not involve gases.
Where H is enthalpy
2. Reactions in which the number of moles of
gas does not change.
∆H = DE + P ∆ V
3. Reactions in which the number of moles of
qp = ∆ E + P ∆ V = ∆ H gas does change but q is >>> P ∆ V.

18. 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) H2O(g)
CH4 + 2O2 H2O(g)
Hinitial Hfinal
Enthalpy, H
Enthalpy, H
∆H < 0 heat out ∆H > 0 heat in
CO2 + 2H2O H2O(l)
Hfinal Hinitial
A Exothermic process B Endothermic process
q < 0, Hproduct < H reactant q > 0, Hproduct > H reactant

19. Drawing Enthalpy Diagrams and Determining the Sign of ∆H
PROBLEM: In each of the following cases, determine the sign of DH, 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)
PLAN: 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) H2O(g) (products)
EXOTHERMIC ∆H = -285.8kJ ENDOTHERMIC ∆H = +40.7kJ
H2O(l) (products) H2O(l) (reactants)

20. 6.5 Enthalpy change for chemical reactions
Thermal chemical reaction – shows the enthalpy relationship between reactants and products
Rules of thermochemistry
v The magnitude of ∆H is directly proportional to the amount of reactants or products.
v ∆H for a reaction is equal in magnitude but opposite in sign for the reverse reaction.
v The value of for a reaction is the same whether it occurs in one step or multi-steps.
v Hess law ∆H = ∆H1 + ∆H2 + ∙∙∙∙
Some Important Types of Enthalpy Change
Heat of combustion (∆Hcomb) C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)
Heat of formation (∆Hf) K(s) + 1/2Br2(l) KBr(s)
Heat of fusion (∆Hfus) NaCl(s) NaCl(l)
Heat of vaporization (∆Hvap) C6H6(l) C6H6(g)

21. 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) 2Al(s) + 3/2O2(g) ∆Hrxn = 1676 kJ
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)
1.000x103 kJ x 2 mol Al 26.98 g Al
1676 kJ = 2 mol Al
1676 kJ 1 mol Al
mol of Al
X M = 32.20 g Al
g of Al

22. 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) CO2(g) ∆ HA = -283.0 kJ
Equation B: N2(g) + O2(g) 2NO(g) ∆ HB = 180.6 kJ
PLAN: 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) CO2(g) ∆ HA = -283.0 kJ
NO(g) 1/2N2(g) + 1/2O2(g) ∆ HB = -90.3 kJ
CO(g) + NO(g) CO2(g) + 1/2N2(g) ∆ Hrxn = -373.3 kJ

23. 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
DH 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) AgCl(s) ∆H 0f = -127.0 kJ
(b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s) ∆H 0f = -1206.9 kJ
(c) 1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g) ∆H 0f = 135 kJ

24. The general process for determining ∆H 0rxn from ∆H0f values.
Elements
decomposition
Enthalpy, H
-∆H0f ∆H0f
formation
Reactants
Hinitial
∆H0rxn
Products
Hfinal
∆H0rxn = Σ m∆H0f(products) - Σ n∆H0f(reactants)

25. 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 DH0rxn from DH 0f values.
PLAN: Look up the DH0f values and use Hess’s Law to find DHrxn.
SOLUTION:
DHrxn = S mDH0f (products) - S nDH0f (reactants)
DHrxn = [4(DH0f NO(g) + 6(DH0f H2O(g)] - [4(DH0f NH3(g) + 5(DH0f 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
Lectrue 3

26. Specific Heat Capacities of Some Elements, Compounds, and Materials
Substance Specific Heat Substance Specific Heat
Capacity (J/g*K) Capacity (J/g*K)
Elements 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

27. Selected Standard Heats of Formation at 250C(298K)
Formula ∆H0f(kJ/mol) Formula ∆H0f(kJ/mol) Formula ∆H0f(kJ/mol)
calcium silver
Cl2(g) 0
Ca(s) 0 Ag(s) 0
HCl(g) -92.3
CaO(s) -635.1 AgCl(s) -127.0
CaCO3(s) -1206.9 hydrogen
H(g) 218 sodium
carbon
H2(g) 0 Na(s) 0
C(graphite) 0
C(diamond) 1.9 nitrogen Na(g) 107.8
CO(g) -110.5 N2(g) 0 NaCl(s) -411.1
CO2(g) -393.5 NH3(g) -45.9 sulfur
CH4(g) -74.9 NO(g) 90.3 S8(rhombic) 0
CH3OH(l) -238.6 S8(monoclinic) 2
oxygen
HCN(g) 135 SO2(g) -296.8
O2(g) 0
CSs(l) 87.9
O3(g) 143 SO3(g) -396.0
chlorine H2O(g) -241.8
Cl(g) 121.0 H2O(l) -285.8