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CHM 1046 : General Chemistry and Qualitative Analysis. Unit 13 Thermochemistry. Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL. Textbook Reference : Chapter # 15 (sec. 1-11) Module # 3 (sec. I-VIII). 1. KE = mv 2. 2. Energy. - PowerPoint PPT Presentation
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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
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
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
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
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
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
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 =
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)
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)
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
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
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.
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
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
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
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
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
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
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)
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
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
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
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/
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
Thermochemistry
Energy in FoodsMost of the fuel in the food we eat comes from carbohydrates & fats.
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
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)
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
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 .
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
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)
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
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
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/
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) -
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
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
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.
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)
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)
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.
Thermochemistry
2003 B Q3
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2005 B
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2003 A
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