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Chapter 6
Thermochemistry: Energy Flow and Chemical Change
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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 (DH0rxn)
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Figure 6.1
A chemical system and its surroundings.
the surroundings
the system
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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.
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Figure 6.2
Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.
DE = Efinal - Einitial = Eproducts - Ereactants
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Figure 6.3
A system transferring energy as heat only.
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Figure 6.4
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)
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Table 6.1 The Sign Conventions* for q, w and DE
q
+
w
=
DE
+ +
-
+ + -
+
depends on sizes of q and w
depends on sizes of q and w
-
* For q: + means system gains heat; - means system loses heat. * For w: + means word done on system; - means work done by system.
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DEuniverse = DEsystem + DEsurroundings
Units of Energy Joule (J) 1 J = 1 kg*m2/s2
Calorie (cal)
1 cal = 4.18J
1 Btu = 1055 J
British Thermal Unit
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Figure 6.5
Some interesting quantities of energy.
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Sample Problem 6.1 PROBLEM:
Determining the Change in Internal Energy of a
System 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. q = - 325 J DE = q + w = -776J kJ 103J w = - 451 J -325 J + (-451 J) = -776 J -0.776kJ kcal 4.18kJ = -0.185 kcal
SOLUTION:
= -0.776kJ
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Figure 6.6
Two different paths for the energy change of a system.
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Figure 6.7
Pressure-volume work.
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The Meaning of Enthalpy
w = - PDV
H = E + PV
where H is enthalpy
DH DE in
1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change.
DH = DE + PDV
qp = DE + PDV = DH
3. Reactions in which the number of moles of gas does change but q is >>> PDV.
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Figure 6.8
Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)
CH4 + 2O2
Hinitial Enthalpy, H Enthalpy, H
H2O(g)
Hfinal
DH < 0
heat out
DH > 0
heat in
CO2 + 2H2O Hfinal
H2O(l)
Hinitial
A
Exothermic process
B
Endothermic process
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Sample Problem 6.2 PROBLEM:
Drawing Enthalpy Diagrams and Determining
the Sign of DH 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)
PLAN:
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 (b) The reaction is endothermic. H2O(g) ENDOTHERMIC (products) DH = +40.7kJ
SOLUTION: (a) The reaction is exothermic. H2(g) + 1/2O2(g) (reactants) EXOTHERMIC DH = -285.8kJ
H2O(l)
(products)
H2O(l)
(reactants)
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Some Important Types of Enthalpy Change
heat of combustion (DHcomb) C4H10(l) + 13/2O2(g) heat of formation (DHf) K(s) + 1/2Br2(l) heat of fusion (DHfus) NaCl(s) NaCl(l) KBr(s) 4CO2(g) + 5H2O(g)
heat of vaporization (DHvap) C6H6(l) C6H6(g)
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Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials Substance Elements aluminum, Al graphite,C iron, Fe copper, Cu gold, Au 0.900 0.711 0.450 0.387 0.129 Compounds Specific Heat Capacity (J/g*K) Substance Materials wood cement glass granite steel 1.76 0.88 0.84 0.79 0.45 Specific Heat Capacity (J/g*K)
water, H2O(l)
ethyl alcohol, C2H5OH(l)
4.184
2.46
ethylene glycol, (CH2OH)2(l)
carbon tetrachloride, CCl4(l)
2.42
0.864
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Sample Problem 6.3
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 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 x 125 g x (300-25)0C = 1.33x104 J
q=
g*K
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Figure 6.9
Coffee-cup calorimeter.
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Sample Problem 6.4
Determining the Heat of a Reaction
PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.00 0C. After stirring, the final temperature is 27.21 0C. 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 c = 4.18 J/g*K)
PLAN:
We need to determine the limiting reactant from the net ionic equation. The moles of NaOH and HCl as well as the total volume can be calculated. From the volume we use density to find the mass of the water formed. At this point qsoln can be calculated using the mass, c, and DT. The heat divided by the M of water will give us the heat per mole of water formed.
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Sample Problem 6.4 continued
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 x 0.0500 L = 0.0250 mol OH0.500 M x 0.0250 L = 0.0125 mol H+
HCl is the limiting reactant. 0.0125 mol of H2O will form during the rxn.
For HCl
total volume after mixing = 0.0750 L 0.0750 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 (27.21-25.00)0C
= 693 J
(693 J/0.0125 mol H2O)(kJ/103 J) = 55.4 kJ/ mol H2O formed
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Figure 6.10
A bomb calorimeter
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Sample Problem 6.5 PROBLEM:
Calculating the Heat of Combustion
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 4.937 0C. Is the manufacturer's claim correct?
PLAN:
- q sample = qcalorimeter qcalorimeter = heat capacity x DT
= 8.151 kJ/K x 4.937 K = 40.24 kJ 40.24 kJ kcal = 9.63 kcal or Calories
SOLUTION:
4.18 kJ
The manufacturer's claim is true.
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Figure 6.11
AMOUNT (mol)
of compound A
Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction.
AMOUNT (mol)
of compound B molar ratio from balanced equation
HEAT (kJ)
DHrxn (kJ/mol)
gained or lost
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Sample Problem 6.6 PROBLEM:
Using the Heat of Reaction (DHrxn) to Find
Amounts 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) DHrxn = 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)
1676kJ=2mol Al mol of Al xM
1.000x103 kJ x
2 mol Al
1676 kJ
26.98 g Al
1 mol Al = 32.20 g Al
g of Al
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Sample Problem 6.7 PROBLEM:
Using Hess's Law to Calculate an Unknown DH
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 = -283.0 kJ
Equation B: N2(g) + O2(g)
PLAN:
2NO(g) DHB = 180.6 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. Multiply Equation B by 1/2 and reverse it. CO(g) + 1/2O2(g) CO2(g) DHA = -283.0 kJ
SOLUTION:
NO(g)
CO(g) + NO(g)
1/2N2(g) + 1/2O2(g)
DHB = -90.3 kJ
DHrxn = -373.3 kJ
CO2(g) + 1/2N2(g)
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Table 6.5 Selected Standard Heats of Formation at 25 0C(298K) Formula calcium Ca(s) CaO(s) CaCO3(s) DH0f(kJ/mol) 0 -635.1 -1206.9 Formula DH0f(kJ/mol) Formula DH0f(kJ/mol)
Cl2(g)
HCl(g) hydrogen H(g) H2(g) nitrogen N2(g) NH3(g) NO(g)
0 -92.3
218 0 0 -45.9 90.3 0 143 -241.8 -285.8
silver Ag(s) AgCl(s)
sodium Na(s) Na(g) NaCl(s)
0 -127.0
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
0 107.8 -411.1
oxygen O2(g) O3(g) H2O(g)
H2O(l)
sulfur S8(rhombic) 0 S8(monoclinic) 2 SO2(g) -296.8 SO3(g) -396.0
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Sample Problem 6.8 PROBLEM:
Writing Formation Equations
Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH0f. (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) DH0f = -127.0 kJ
(b) Ca(s) + C(graphite) + 3/2O2(g)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
CaCO3(s)
HCN(g)
DH0f = -1206.9 kJ
DH0f = 135 kJ
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Figure 6.13
The general process for determining DH0rxn from DH0f values.
Elements
Enthalpy, H
decomposition
Reactants
formation
-DH0f
DH0f Hinitial DH0rxn
Products DH0rxn = S mDH0f(products) - S nDH0f(reactants)
Hfinal
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Sample Problem 6.9 PROBLEM:
Calculating the Heat of Reaction from Heats of
Formation 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 DH0f values.
PLAN: Look up the DH0f values and use Hess's Law to find DHrxn. DHrxn = S mDH0f (products) - S nDH0f (reactants)
SOLUTION:
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)]
DHrxn = -906 kJ
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Figure B6.1
The trapping of heat by the atmosphere.
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Figure B6.2
A Atmospheric CO2concentrations
The accumulating evidence for the greenhouse effect.
B Average global temperatures
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