thermochemistry unit 13 thermochemistry dr. jorge l. alonso miami-dade college – kendall campus...
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![Page 1: Thermochemistry Unit 13 Thermochemistry Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL Textbook Reference: Chapter # 15 (sec. 1-11)](https://reader035.vdocuments.us/reader035/viewer/2022062304/56649e665503460f94b60cbe/html5/thumbnails/1.jpg)
Thermochemistry
Unit 13Thermochemistry
Dr. Jorge L. Alonso
Miami-Dade College – Kendall Campus
Miami, FL
Textbook Reference: •Chapter # 15 (sec. 1-11)
•Module # 3 (sec. I-VIII)
CHM 1046: General Chemistry and Qualitative Analysis
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Thermochemistry
EnergyPotential Energy an object possesses by virtue of its position or chemical composition (bonds).
Kinetic Energy an object possesses by virtue of its motion.
12
KE = mv2
2 C8H18 (l) + 25 O2 (g)
16 CO2(g) + 18 H2O(g)
Energy (-E): work + heat
Chemical bond
energy
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Thermochemistry
Energy
Work (w): Energy (force) used to cause an object that has mass to move (a distance) = kinetic energy. W = F x d
2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g)
H = - 5.5 x 106 kJ/
Energy produced by chemical reactions = heat or work.
gas=2534= +9
E =
Heat (q): spontaneous transfer (flow) of energy (energy in transit) from one object to another; causes molecular motion & vibrations (kinetic energy); measured as temperature.
9 x 22.4L/ = 202 L
q + wLaw of Conservation of Energy:
Chemical bond
energy
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Thermochemistry
Units of Energy: Joule & calorie• The joule (J) = SI unit of energy (work)
• The calorie (cal) = heat required to raise 1 g of H2O 10C
1 cal = 4.184 J
{Nutritional Calorie = 1000 calories (1kcal)=4,184 J}.
= kg m2
s2Joule (J) = N·m =
kg m
s2m
The Newton (N) is the amount of force that is required to accelerate a kilogram of mass at a rate of one meter per second squared.
Energy (Work) = F x d
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Thermochemistry
System and Surroundings
• The system includes the molecules we want to study.
• The surroundings are everything else (here, the cylinder and piston).
System
Surroundings
Work
Heat-Eq(heat)
-Ework
+Eq
+Ew
E = q + w
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Thermochemistry
• Energy is neither created nor destroyed.
First Law of Thermodynamics
• 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.
• Also known as the Law of Conservation of Energy
System
Surroundings
Work
Heat-Eq
-Ew
+Eq
+Ew
Esys + Esurr = 0
Esys = -Esurr
orE sys = (q + w)surr
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Thermochemistry
Changes in Internal Energy (E)
E = q + w
E, q, w, and Their Signs
> q
> w
+ q
+ w
- q
- w
Syst gains heat
Syst looses heat
: energy released/absorbed as either work or heat, is looked at from the perspective of the
system (the chemicals)
Work done on Syst
Work done by Syst
+ q + w- q - w- E =
+ E =
± q ± w± E =
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Thermochemistry
Why consider Energy Change (ΔE) and not the total Internal Energy (E)?
• But when a change occurs we can measure the change in internal energy:
E = Efinal − Einitial
• Total Internal Energy (E) = the sum of all kinetic and potential energies of all components of the system. How can it be determined?
• Usually we have no way of knowing the total internal energy of a system; finding that value is simply too complex a problem.
Esys + Esurr = 0 Esys + Esurr = ETotal (constant)
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Thermochemistry
Energy as Work (w) of Gas Expansion
{WorkGasExpansion}
Work = - (Force x distance)
w = - (F x h)
w = - (P x A x h)
w = - P V(@ constant P)
F = P A
w = - nRT(@ constant V)
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Thermochemistry
Exchange of Heat (H) by chemical systems
• When heat is released by the system to the surroundings, the process is exothermic.
• When heat is absorbed by the system from the surroundings, the process is endothermic.
- H
+ H
q
q
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Thermochemistry
Enthalpies of Reaction (H)The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.
Notice it is the same value, but of opposite sign for the reverse reaction
Reactants:
Products
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Thermochemistry
State Functions
A state function depends only on the present state of the system, not on the path by which the system arrived at that state.
• Other state functions are P, V and T.
∆E=q+w
• Are E, q and w all state functions?
• However, q and w are not state functions.
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Thermochemistry
Calorimetry• The experimental methods of measuring of
heats (H) involved in chemical reactions
• Methods utilized:1. Constant Pressure
Calorimetry (open container in contact with atmosphere)
2. Constant Volume Calorimetry: closed container (bomb)
1 atm
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Thermochemistry
Thermochemical Equations
2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g)
H = - 5.5 x 106 kJ/ C8H18
1. H is per mole of reactant, at the indicated states (s, l, g).
2. Direct quantitative relationship between coefficients of balanced equation and H.
3. Reverse equation has H of opposite sign.
Problem: Calculate H when 25g of C8H18 (l) (MM= 114) are burned in excess oxygen?
HC 25g kJ? 188
g114
1 kJ 10x 1.2- 6
1
kJ 10 x 5.5
6
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Thermochemistry
E = q + w
E + PV = qp
E = qp - PV
qp = Heat of Reaction
=
or Enthalpy of Reaction
H = E + PV
SOLVE FOR
1 atm
• Most chemical reactions are carried out in beakers or flasks that are open to the atmosphere (i.e., at constant pressure, isobaric).
(1) Calorimetry at Constant Pressure
H = Epv and when the volume is constant
q = m c T
E + PV = H Big Problem:
calorimeter also
absorbs heat
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Thermochemistry
(a) Determining the Heat Capacity of a Calorimeter
50 mL of H2O @ 650C are added…….
…..to 50 mL of H2O
@ 250C inside the calorimeter
The final temperature of the 100 mL of H2O inside the calorimeter is 420C
Heat lost by = heat gained by + heat gained by hot water cold water calorimeter
qhot = qcold + qcalorimeter
50 mL H2O
50 mL H2O
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Thermochemistry
qhot = qcold + qcalorimeter
Determining the Heat Capacity of a Calorimeter (Ccal)
(mcT)H2O = (mcT)H2O + (Ccal T)
(mcT)H2O - (mcT)H2O
(T)cal
Ccal =
2542
2542/184.450 - 6542/184.450g
0
0000
C
CCgJgCCgJCcal
C/J492
C)17(
J)4.3556 - (J6.4811-C 0
0cal
Includes both the mass and heat capacity
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Thermochemistry
(b) Calculating the Heats of Reactions (H)
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
qrxn = (mc*T)NaClsoln + (CcalT)
qrxn = - Hrxn
= heat of neutralization rxn* How do you determine the cNaCl soln?
50 mL@25°C + 50 mL@25°C 100 mL of NaCl(aq) @ 30°C
H(qrxn) = qNaCl soln + qcal
Heat of = heat gained by + heat gained by reaction NaCl solution calorimeter
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Thermochemistry
How do you determine the cNaCl soln?
50 mL of H2O @ 650C are added…….
…..to 50 mL of NaCl solution @ 250C inside the calorimeter
The final temperature of the 100 mL of solution inside the calorimeter is 410C
50 mL H2O
50 mL NaCl soln
Heat lost by = heat gained by + heat gained by hot water salt water soln calorimeter
qh = qNaClsoln + qcal
(mcT)H2O = (mc*T)NaClsoln + (CcalT)
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Thermochemistry
(2) Calorimetry at Constant Volume (Bomb Cal.)
E = q + w
Bomb calorimeter
w = P V V =0 w = 0
E = qv
q = m c T
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Thermochemistry
• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.
• For most reactions, the difference is very small.
Bomb Calorimetry
E = qv
H = E + nRT
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Thermochemistry
Review of Thermochemical Equations
E = q + w
E = qv
E = qp - nRT since ∆H= qp
1 atm
E = qp - PV
E = H - PV
H = E + PV
E = q + w
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Thermochemistry
Comparing H and E
H = E + PV
2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g)
E = H - PV (useful at const P)
E = H - nRT (useful at const V)
Problem: calculate E @ 25°C for above equation @ const. V
R= 8.314J/K. gas=34-25 = +9 = 4.5 C8H18
/kJ10 x 5.5/kJ11/kJ10 x 5.5E
)K298)(J10/kJ1)(K/J314.8)(5.4(/kJ10 x 5.5E66
36
H = - 5.5 x 106 kJ/
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Thermochemistry
Constant-Volume CalorimetryProblem: When one mole of CH4 (g) was combusted at 250C in a bomb
calorimeter, 886 kJ of energy were released. What is the enthalpy change for this reaction?
)K ( J
kJK) J/η.η) ( (- kJ H 27325
1000
13182886
CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(l)
E = qv = - 886 kJ
R = 8.31 J/ .K
n = 1 – (1+2) = -2
H = E + PV
= E + nRT
kJ891 kJ 95.4 kJ - 886 - ΔH
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Thermochemistry
Energy in FoodsMost of the fuel in the food we eat comes from carbohydrates & fats.
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Thermochemistry
Methods of determining H1. Calorimetry (experimental)
2. Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method).
3. Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method)
4. Bond Energies used with Hess’s Law
Experimental data combined with theoretical concepts
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Thermochemistry
(2) Determination of H using Hess’s Law
Hrxn is well known for many reactions, but it is inconvenient to measure Hrxn for every reaction.
The Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps are added to lead to reaction of interest (indirect method).
However, we can estimate Hrxn for a reaction of interest by using Hrxn values that are published for other more common reactions.
Standard conditions (25°C and 1.00 atm pressure).
(STP for gases T= 0°C)
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Thermochemistry
Hess’s Law
“If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”
- 1840, Germain Henri Hess (1802–50), Swiss
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Thermochemistry
Calculation of H by Hess’s Law
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
3 C(graphite) + 4 H2 (g) C3H8 (g) H= -104
3 C(graphite) + 3 O2 (g) 3 CO2 (g) H=-1181
4 H2 (g) + 2 O2 (g) 4 H2O (l) H=-1143
C3H8 (g) 3 C(graphite) + 4 H2 (g) H= +104
• Appropriate set of Equations with their H values are obtained (or given), which containing chemicals in common with equation whose H is desired.
• These Equations are all added to give you the desired equation.
• These Equations may be reversed to give you the desired results (changing the sign of H).
• You may have to multiply the equations by a factor that makes them balanced in relation to each other.
• Elimination of common terms that appear on both sides of the equation .
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Thermochemistry
Calculation of H by Hess’s Law
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
3 C(graphite) + 3 O2 (g) 3 CO2 (g) H=-1181
4 H2 (g) + 2 O2 (g) 4 H2O (l) H=-1143 Hrxn =
+ 104 kJ
-1181 kJ
- 1143 kJ
- 2220 kJ
C3H8 (g) 3 C(graphite) + 4 H2 (g) H= +104
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Thermochemistry
Calculate heat of reaction
W + C (graphite) WC (s) ΔH = ?
Given data:
2 W(s) + 3 O2 (g) 2 WO3 (s) ΔH = -1680.6 kJ
C (graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJ
2 WC (s) + 5 O2 (g) 2 WO3 (s) + CO2 (g) ΔH = -2391.6 kJ
Calculation of H by Hess’s Law
C (graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJ
½(2 W(s) + 3 O2 (g) 2 WO3 (s) ) ½(ΔH = -1680.6 kJ)
½(2 WO3 (s) + CO2 (g) 2 WC (s) + 5 O2 (g)) ½ (ΔH = +2391.6 kJ)
W + C (graphite) WC (s) ΔH = - 38.0
W(s) + 3/2 O2 (g) WO3 (s) ) ΔH = -840.3 kJ
WO3 (s) + CO2 (g) WC (s) + 5/2 O2 (g) ΔH = + 1195.8 kJ)
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Thermochemistry
Problem: Chloroform, CHCl3, is formed by the following reaction: Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g)
Determine the enthalpy change for this reaction (ΔH°rxn), using the following:
2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g)ΔH°f = – 103.1 kJ/mol CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g) ΔH°rxn = – 890.4 kJ/mol2 HCl (g) → H2 (g) + Cl2 (g) ΔH°rxn = + 184.6 kJ/molC (graphite) + O2 (g) → CO2(g) ΔH°rxn = – 393.5 kJ/molH2 (g) + ½ O2 (g) → H2O (l) ΔH°rxn = – 285.8 kJ/mol
answers: a) –103.1 kJ b) + 145.4 kJ c) – 145.4 kJ d) + 305.2 kJ e) – 305.2 kJ f) +103.1 kJ
This is a hard question. To make is easer give:C (graphite) + ½ H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔH°f = – 103.1 kJ/mol
Hess’s Law
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Thermochemistry
Methods of determining H1. Calorimetry (experimental)
2. Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method).
3. Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method)
4. Bond Energies used with Hess’s Law
Experimental data combined with theoretical concepts
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Thermochemistry
(3) Determination of H using Standard Enthalpies of Formation (Hf )
Standard Enthalpy of formation Hf are measured under standard conditions (25°C and 1.00 atm pressure).
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.
C + O2 CO2 ∆Hf = -393.5 kJ/
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Thermochemistry
Calculation of H
We can use Hess’s law in this way:
H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients.
CH4(g) + O2(g) CO2(g) + H2O(g)
C + 2H2(g) CH4(g) ΔHf = -74.8 kJ/ŋC(g) + O2(g) CO2(g) ΔHf = -393.5 kJ/ŋ
2H2(g) + O2(g) 2H2O(g) ΔHf = -241.8 kJ/ŋ
H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)]
= - 560.5 kJ
n CO2(g) + n H2O(g)n CH4(g) + n O2(g) -
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Thermochemistry
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
Calculation of H
H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) - (-103.85 kJ)
= -2219.9 kJ
H = nHf(products) - mHf(reactants)
Table of Standard Enthalpy of formation, Hf
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Thermochemistry
• Most simply, the strength of a bond is measured by determining how much energy is required to break the bond.
• This is the bond enthalpy.• The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is measured to be 242 kJ/mol.
(4) Determination of H using Bond Energies
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Thermochemistry
Average Bond Enthalpies (H)
NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than theC—H bond in chloroform, CHCl3.
• Average bond enthalpies are positive, because bond breaking is an endothermic process.
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Thermochemistry
Enthalpies of Reaction (H )
• Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.
• In other words, Hrxn = (bond enthalpies of bonds broken)
(bond enthalpies of bonds formed)
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Thermochemistry
Hess’s Law: Hrxn = (bonds broken) (bonds formed)
Hrxn = [D(C—H) + D(Cl—Cl) [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)] [(328 kJ) + (431 kJ)]
= (655 kJ) (759 kJ)
= 104 kJ
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
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Thermochemistry
Bond Enthalpy and Bond Length
• We can also measure an average bond length for different bond types.
• As the number of bonds between two atoms increases, the bond length decreases.
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Thermochemistry
2003 B Q3
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Thermochemistry
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Thermochemistry
2005 B
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Thermochemistry
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Thermochemistry
2002
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Thermochemistry
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Thermochemistry
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Thermochemistry
2002 #8
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Thermochemistry
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Thermochemistry
2003 A