Download Chapter 8 Thermochemistry: Chemical Energy

Thermodynamics
Chapter 8
Thermochemistry:
Chemical Energy
01
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Thermodynamics: study of energy and it’s transformations
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Energy: capacity to do work, or supply heat
Energy = Work + Heat
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Kinetic Energy: energy of motion
EK = 1/2 mv2
(1 Joule = 1 kg⋅m2/s2)
(1 calorie = 4.184 J)
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Thermodynamics
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Conservation of energy law: Energy cannot be
created or destroyed; it can only be converted from
one form into another.
Potential Energy: stored energy
Thermodynamics
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Thermal Energy: kinetic energy of molecular
motion (translational, rotational, and vibrational)
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Heat: the amount of thermal energy transferred
between two objects at different temperatures
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Chemical Energy: potential energy stored in
chemical bonds released in the form of heat or light
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Thermodynamics
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First Law of Thermodynamics: energy of an
isolated system must be kept constant
Thermodynamics
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System ⇒ reactants + products
Surroundings ⇒ everything else
Energy changes are measured from the point of
view of the system!
∆E is negative ⇒ energy flows out of the system
• ∆E is positive ⇒ energy flows into the system
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Thermodynamics
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Work
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w = –PΔV
2
Sign of w
negative
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Work Units
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w = -PΔV (J or kJ)
positive
w = L x atm =
w = -PΔV ⇒ expansion
positive
m2
1m3
1L x 1000mL x 1cm3 x
1L
negative
1mL
101 x 103 kg
(100cm)3
ms2
1000
w = -PΔV ⇒ contraction
= 101 kgm2 = 101J
s2
Energy and Heat
Energy = Work + Heat
ΔE = w + q = q - PΔV
When a person does work, energy diminishes
w = negative
ΔE = negative
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Heat and Enthalpy
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The amount of heat exchanged between the
system and the surroundings is given the symbol q.
q = ΔE + PΔV
At constant volume (ΔV = 0): qv = ΔE
At constant pressure: qp = ΔE + PΔV
enthalpy
= ΔH
q = ΔE + PΔV
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State Functions
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State Function:
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State Functions
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value depends only on the present state of the
system
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State and Nonstate Properties: The two paths
below give the same final state:
N2H4(g) + H2(g) → 2 NH3(g) + heat (188 kJ)
N2(g) + 3 H2(g) → 2 NH3(g) + heat (92 kJ)
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path independent
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temperature, total energy, pressure, density,
volume, and enthalpy (∆H) ⇒ state properties
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when returned to its original position, overall
change is zero
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nonstate properties include heat and work
Enthalpy
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• Enthalpy or heat of reaction:
Standard Enthalpy of Reaction
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Thermodynamic standard state: P = 1atm, [ ] = 1M,
•ΔH = H(products) - H(reactants)
T = 298.15K (25ºC)
• States of the reactants and products are important!
⇒ (g, l, s, aq)
Standard enthalpy of reaction (Hº)
N2(g) + 3H2(g) → 2NH3(g)
ΔHº = -92.2kJ
• Thermodynamic standard state: P = 1atm, [ ] = 1M,
T = 298.15K (25ºC)
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Enthalpy Changes
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Physical Changes
Most changes in a system involve a
gain or loss in enthalpy
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Enthalpies of Physical Change:
• Physical (melting of ice in a cooler)
• Chemical (burning of gas in your car)
Chemical Changes
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Enthalpies of Chemical Change: Often called
heats of reaction (ΔHreaction).
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Enthalpy Changes
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Reversing a reaction changes the sign of ΔH for a
reaction.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH = –2219 kJ
Endothermic: Heat flows into the system from the
3 CO2(g) + 4 H2O(l) → C3H8(g) + 5 O2(g) ΔH = +2219 kJ
surroundings ⇒ ΔH is positive
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Exothermic: Heat flows out of the system into the
surroundings ⇒ ΔH is negative
Multiplying a reaction increases ΔH by the same factor.
3 x [C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ]
ΔH =(-2219kJ x 3) = –6657 kJ
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Example
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How much work is done (in kilojoules), and in which
direction, as a result of the following reaction?
Example
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w = -0.25kJ
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The following reaction has ΔE = –186 kJ/mol.
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Is the sign of PΔV positive or negative?
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What is the sign and approximate magnitude of ΔH?
Expansion, system
loses -0.25kJ
Contraction, PΔV is negative,
w is positive
ΔH = ΔE + PΔV
ΔH = (-186kJ) + (1atm) (-1mole)
ΔH = negative (slightly more
than ΔE)
Example
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The reaction between hydrogen and oxygen to yield
water vapor has ΔH° = –484 kJ. How much PV
work is done, and what is the value of ΔE (kJ) for
the reaction of 0.50 mol of H2 with 0.25 mol of O2 at
atmospheric pressure if the volume change is
–5.6 L?
PΔV = -0.57kJ
Example
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The explosion of 2.00 mol of solid TNT with a
volume of approximately 0.274 L produces gases
with a volume of 448 L at room temperature. How
much PV (kJ) work is done during the explosion?
Assume P = 1 atm, T = 25°C.
2 C7H5N3O6(s) → 12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)
Contraction, so w is positive
PΔV = 45.2kJ
ΔE = -120.43kJ
Expansion, so w = -45.2kJ
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Example
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How much heat (kJ) is evolved or absorbed in each
of the following reactions?
Hess’s Law
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1.) Burning of 15.5 g of propane:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
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Hess’s Law: The overall enthalpy change for a
reaction is equal to the sum of the enthalpy
changes for the individual steps in the reaction.
ΔHº = –2219 kJ
3 H2(g) + N2(g) → 2 NH3(g) ΔH° = –92.2 kJ
-780kJ (exothermic)
2.) Reaction of 4.88 g of barium hydroxide octahydrate with ammonium
chloride:
Ba(OH)2·8 H2O(s) + 2 NH4Cl(s) → BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)
ΔHº = +80.3 kJ
+1.24kJ (endothermic)
Hess’s Law
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(a)
2 H2(g) + N2(g)
N2H4(g)
ΔH°1 = ?
(b)
N2H4(g) + H2(g)
2 NH3(g)
ΔH°2 = –187.6 kJ
(c)
3 H2(g) + N2(g)
2 NH3(g)
ΔH°3 = – 92.2 kJ
Standard Heats of Formation
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Where do ΔH° values come from?
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Standard Heats of Formation (ΔH°f): enthalpy
change for the formation of 1 mole of substance in
its standard state
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ΔH°f = 0 for an element in its standard state!
ΔH°1 = ΔH°3 – ΔH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ
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Standard Heats of Formation
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ΔH°f = –286 kJ/mol
H2(g) + 1/2 O2(g) → H2O(l)
Standard Heats of Formation
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Calculating ΔH° for a reaction:
ΔH° = ΔH°f (products) – ΔH°f (reactants)
3/
2 H2(g)
ΔH°f = –46 kJ/mol
+ 1/2 N2(g) → NH3(g)
Heat of formation must be multiplied by the coefficient of the
reaction
ΔH°f = +227 kJ/mol
2 C(s) + H2(g) → C2H2(g)
2 C(s) + 3 H2(g) + 1/2 O2(g) → C2H5OH(g) ΔH°f = –235 kJ/mol
Standard Heats of Formation
C6H12O6 (s)
ΔH° = [2ΔH°f(ethanol) + 2ΔH°f(CO2)] - ΔH°f (glucose)
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Some Heats of Formation, ΔHf° (kJ/mol)
Bond Dissociation Energy
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CO(g)
-111
C2H2(g)
227
Ag+(aq)
106
CO2(g)
-394
C2H4(g)
52
Na+(aq)
-240
H2O(l)
-286
C2H6(g)
-85
NO3-(aq)
-207
NH3(g)
-46
CH3OH(g)
-201
Cl-(aq)
-167
N2H4(g)
95.4
C2H5OH(g)
-235
AgCl(s)
-127
HCl(g)
-92
C6H6(l)
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Na2CO3(s)
-1131
2C2H5OH (l) + 2CO2 (g)
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Bond Dissociation Energy (D): Amount of energy needed to
break a chemical bond in gaseous state
D = Approximate ΔHº
ΔH° = D(reactant bonds broken) – D(product bonds formed)
H2 + Cl2
2HCl
ΔH° = (DCl-Cl + DH-H) - (2 D H-Cl)
= [(1 mol)(243 kJ/mol) + (1)(436 kJ/mol] - (2)(432 kJ/mol)
= -185 kJ
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Bond Dissociation Energy
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Calorimetry and Heat Capacity
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Calorimetry and Heat Capacity
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Calorimetry: measurement of heat changes (q)
for chemical reactions
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Constant Pressure Calorimetry: measures the
heat change at constant pressure q = ΔH
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Bomb Calorimetry: measures the heat change
at constant volume such that q = ΔE
Calorimetry and Heat Capacity
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Heat capacity {C}: amount of heat required to raise the
temperature of an object or substance a given amount
C =
q
ΔT
Specific Heat: amount of heat required to raise the
temperature of 1.00 g of substance by 1.00°C
Constant Pressure
Bomb
Molar Heat: amount of heat required to raise the
temperature of 1.00 mole of substance by 1.00°C
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Calorimetry and Heat Capacity
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Example
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The industrial degreasing solvent methylene chloride
(CH2Cl2, dichloromethane) is prepared from methane by
reaction with chlorine:
CH4(g) + 2 Cl2(g)
CH2Cl2(g) + 2 HCl(g)
Calculate ΔH° (kJ)
CH4(g) + Cl2(g)
CH3Cl(g) + HCl(g) ΔH° = –98.3 kJ
CH3Cl(g) + Cl2(g)
CH2Cl2(g) + HCl(g) ΔH° = –104 kJ
ΔH° = -98.3 + -104 = -202kJ
Example
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Example
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Calculate ΔH° (kJ) for the reaction of ammonia with O2 to
Calculate ΔH° (kJ) for the photosynthesis of glucose from
yield nitric oxide (NO) and H2O(g), a step in the Ostwald
process for the commercial production of nitric acid.
CO2 and liquid water, a reaction carried out by all green
plants.
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g)
= [(4)(90.2kJ/mol) + (6)(-241.8)] - [(4)(-46.1) + (5)(0)]
= [(1 mole)(-1260kJ/mol) + (6)(0)] - [(6)(-393.5) + (6)(-285.8)]
= -905.6kJ
= 2816kJ
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Example
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Calculate an approximate ΔH° (kJ) for the synthesis
Example
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What is the specific heat of lead if it takes 96 J to
raise the temperature of a 75 g block by 10.0°C?
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When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL
of 1.0 M NaOH at 25.0°C in a calorimeter, the
temperature of the solution increases to 33.9°C.
Assume specific heat of solution is 4.184 J/(g–1·°C–1),
and the density is 1.00 g/mL–1, calculate ΔH for the
reaction.
of ethyl alcohol from ethylene:
C2H4(g) + H2O(g) → C2H5OH(g)
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Calculate an approximate ΔH° (kJ) for the synthesis
of hydrazine from ammonia:
2 NH3(g) + Cl2(g) → N2H4(g) + 2 HCl(g)
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