1. Chapter 6 Thermochemistry
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2. © 2012 Pearson Education, Inc.
Energy
Energy is the ability to do work or transfer heat.
Energy used to cause an object that has mass to move is called work.
w = F  d
where w is work, F is the force, and d is the distance over which the force is exerted.
Energy used to cause the temperature of an object to rise is called heat.(Heat flows from warmer objects to cooler objects)
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3. Potential and Kinetic Energy
Potential energy is energy an object possesses by virtue of its position or chemical composition.
Kinetic energy is energy an object possesses by virtue of its motion:
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Units of Energy
The SI unit of energy is the joule (J):
An older, non-SI unit is still in widespread use: the calorie (cal):
1 cal = J
1 J = 1 
kg m2 s2
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5. Definitions: System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).
The surroundings are everything else (here, the cylinder and piston).
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6. First Law of Thermodynamics
Energy is neither created nor destroyed.
In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.
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Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.
By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:
E = Efinal − Einitial
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8. DE = Efinal - Einitial = Eproducts - Ereactants
Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.
DE = Efinal - Einitial = Eproducts - Ereactants

9. Changes in Internal Energy
If E < 0, Efinal < Einitial
Therefore, the system released energy to the surroundings.
This energy change is called exergonic.
The reaction is exothermic
If E > 0, Efinal > Einitial
Therefore, the system absorbed energy from the surroundings.
This energy change is called endergonic.
The reaction is endothermic
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10. Two forms of energy transfer
Heat (q): Energy transferred between a system and its surroundings as a result of a difference in their temperatures only.
Work (w): Energy transferred when a force acts on an object.
Total change in a system’s internal energy
∆E = q + w

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E, q, w, and Their Signs
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12. Determining the Change in Internal Energy of a System
PROBLEM:
When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal.
PLAN:
Define system and surroundings, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal.
SOLUTION:
q = J
w = J
DE = q + w =
-325 J + (-451 J) =
-776 J
-776J
103J kJ
-0.776kJ
4.18kJ
kcal
= kJ
= kcal

13. The Meaning of Enthalpy
Enthalpy is a thermodynamic function of a system, equivalent to the sum of the internal energy of the system plus the product of its volume multiplied by the pressure exerted on it by its surroundings.
H = E + PV
where H is enthalpy (greek: to warm)
we cannot know the exact enthalpy of the reactants and products, we then measure H
DH = DE + PDV
qp = DE + PDV = DH
at constant pressure

14. DHrxn = Hfinal - Hinitial = Hproducts - Hreactants
Heat of Reaction
DHrxn = Hfinal - Hinitial = Hproducts - Hreactants
Exothermic process Hfinal < Hinitial so ∆H < 0
Heat is released
Decrease in enthalpy
Endothermic process Hfinal > Hinitial so ∆H > 0
Heat is absorbed
Increase in enthalpy

15. Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
H2O(l) H2O(g)
CH4 + 2O2
H2O(g)
Hfinal
Enthalpy, H
Hinitial
Enthalpy, H
heat out
heat in
DH < 0
DH > 0
CO2 + 2H2O
H2O(l)
Hfinal
Hinitial
A Exothermic process
B Endothermic process

16. Sample Problem
Drawing Enthalpy Diagrams and Determining the Sign of DH
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) kJ
(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)
(products)
H2O(g)
EXOTHERMIC
DH = kJ
ENDOTHERMIC
DH = +40.7kJ
H2O(l)
(products)
(reactants)
H2O(l)

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Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.
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18. Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C).
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19. Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
heat transferred
mass  temperature change
Cs = q
m  T
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20. The Practice of Calorimetry At constant Pressure
Heat lost by the system(sample) is equal in magnitude but opposite in sign to the heat gained by the surroundings(H2O)
-qsample = qH2O
Substitute the equation q = Cs x mass x ∆T
-(Cssample x masssample x ∆Tsample) = (CsH20 x massH2O x ∆TH2O)
Cssample = CsH20 x massH2O x ∆TH2O
masssample x ∆Tsample -

21. Sample Problem
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 250C to 300.0C? The specific heat capacity (c) of Cu is 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
g*K
q = x
125 g x
(300-25)0C
= 1.33x104 J

22. The Practice of Calorimetry At constant Volume
Called bomb calorimetry
Compound reacts completely with O2
example of a combustion reaction
Withstands high pressures
q rxn = -Ccal x ∆T

23. Sample Problem
Calculating the Heat of Combustion
PROBLEM:
A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories 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 C. Is the manufacturer’s claim correct?
PLAN:
- q sample = qcalorimeter
SOLUTION:
qcalorimeter
= heat capacity x DT
= kJ/K x K
= kJ
40.24 kJ
kcal
4.18 kJ
= 9.63 kcal or Calories
The manufacturer’s claim is true.

24. Molar Heat Capacity (Cm)
The quantity of heat required to change the temperature of 1 mole of a substance by 1 K
Molar heat capacity (Cm) = q/(moles x ∆T) [ units of J/mol·K]
Liquid H2O has a Cm of J/g·K
So Cm of H2O(l) = J/g·K x g/1 mole = J/mol·K

25. Sample Problem
Determining the Heat of a Reaction
PROBLEM:
You place 50.0 mL of M NaOH in a coffee-cup calorimeter at C and carefully add 25.0 mL of M HCl, also at C. After stirring, the final temperature is C. Calculate qsoln (in J) and DHrxn (in kJ/mol). (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 Cs = 4.18 J/g*K)

26. HCl is the limiting reactant.
Sample Problem
Determining the Heat of a Reaction
continued
SOLUTION:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) H2O(l)
For NaOH
0.500 M x L = mol NaOH
For HCl
0.500 M x L = mol HCl
HCl is the limiting reactant.
mol of H2O will form during the rxn.
total volume after mixing = L
L x
103 mL/L x
1.00 g/mL
= 75.0 g of water
q = mass x specific heat x DT
= 75.0 g x 4.18 J/g*0C x ( )
= 693 J
(-693 J/ mol H2O)(kJ/103 J) = kJ/ mol H2O formed

27. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ
6.4

28. 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) P4O10 (s) DH = kJ
1 mol P4
123.9 g P4 x
-3013 kJ
1 mol P4 x
266 g P4
= kJ
6.4

29. Hess’s Law of Heat Summation
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.
Many reactions cannot be carried out separately
They are often occur as a series of reactions
The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.

30. Sample Problem 6.7
Using Hess’s Law to Calculate an Unknown DH
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) DH = ?
Given the following information, calculate the unknown DH:
Equation A: CO(g) + 1/2O2(g) CO2(g) DHA = kJ
Equation B: N2(g) + O2(g) NO(g) DHB = kJ
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g) CO2(g) DHA = kJ
NO(g) /2N2(g) + 1/2O2(g)
DHB = kJ
CO(g) + NO(g) CO2(g) + 1/2N2(g)
DHrxn = kJ

31. Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.
Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure) page 184 or table 5.3
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (C, graphite) = 0 f
DH0 (O2) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f

32. ∆H0f Standard Enthalpies found on page 184 or Table 5.3.
The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
∆H0f Standard Enthalpies found on page 184 or Table 5.3.

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Calculation of H
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(− kJ) + 5(0 kJ)]
= [(− kJ) + (− kJ)] – [(− kJ) + (0 kJ)]
= (− kJ) – (− kJ) = − kJ
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