Download Ch. 6- Energetics

The Nature of Energy
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
http://academic.cengage.com/chemistry/moore
Chapter 6
Energy and Chemical Reactions
Energy (E
(E) = the capacity to do work.
Work (w) occurs when an object moves against a
resisting force:
w = −(resisting force) x (distance traveled)
w = −F d
All energy is either Kinetic or Potential energy.
Stephen C. Foster • Mississippi State University
The Nature of Energy
Kinetic energy (E
(Ek) - Energy of motion
Energy Units
joule (J) - SI unit (1 J = 1 kg m2s-2)
 macroscale = mechanical energy
 random nanoscale = thermal energy
2.0 kg mass moving at 1.0 m/s (~2 mph):
 periodic nanoscale = acoustic energy
Ek = ½mv2
(m = mass, v = velocity of object)
Potential energy (E
(Ep) – Energy of position. Stored E.
It may arise from:
 gravity: Ep = m g h (mass x gravity x height).
 charges held apart.
 bond energy.
Energy Units
Ek
= ½ mv2 = ½ (2.0 kg)(1.0 m/s)2
= 1.0 kg m2 s-2
= 1.0 J
1 J is a relatively small amount of energy.
1 kJ (1000 J) is more common in chemical problems.
Conservation of Energy
calorie (cal)
Originally: “The energy needed to heat of 1g of
water from 14.5 to 15.5 °C.”
It can only change form.
Now:
Total E of the universe is constant.
1 cal = 4.184 J (exactly)
Dietary Calorie (Cal) - the “big C” calorie
Used on food products.
1 Cal = 1000 cal
= 1 kcal
“Energy can neither be created nor destroyed”.
Also called the 1st Law of Thermodynamics.
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Conservation of Energy
Energy and Working
If an object moves against a force, work is done.
• Lift a book
 you do work against gravity. The book’s Ep increases.
• Drop the book:
 Ep converts into Ek
 The book does work pushing the air aside.
A diver:
a) Has Ep due to macroscale position.
b) Converts Ep to macroscale Ek.
c) Converts Ek,macro to Ek,nano (motion of water, heat)
• The book hits the floor
Energy and Working
Energy, Temperature, and Heating
In a chemical process, work occurs whenever
something expands or contracts.
Expansion pushes back the surrounding air.
On heating a balloon:
 The gas heats up
 The gas expands; the balloon swells.
 The gas does work pushing back the rubber and
the air outside it.
Energy, Temperature, and Heating
 no work is done on the floor (it does not move).
 Ek converts to a sound wave and T of the book and floor
increase (Ek converts to heat).
Temperature is a measure of the
thermal energy of a sample.
Thermal energy
• E of motion of atoms, molecules, and
ions.
• Atoms of all materials are always in
motion.
• Higher T = faster motion.
Energy, Temperature, and Heating
Heat
• Thermal E transfer caused by a T difference.
• Heat flows from hotter to cooler objects until
they reach thermal equilibrium (have equal T ).
Consider a thermometer. As T increases:
 Atoms move faster; on average get farther apart.
 V of the material increases.
 Length of liquid column increases.
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Systems, Surroundings, and Internal Energy
Systems, Surroundings, and Internal Energy
Internal energy = E within the system because of
nanoscale position or motion
System = the part of the universe under study
 chemicals in a flask.
 my textbook.
Einternal= sum of all nanoscale Ek and Ep
Surroundings = rest of the universe (or as much as
needed…)
 the flask.
• nanoscale Ek = thermal energy
• nanoscale Ep
 ion/ion attraction or repulsion
 perhaps the flask and this classroom.
 nucleus/electron attraction
 perhaps the flask and all of the building, etc.
 proton/proton repulsion …..
Universe = System + Surroundings
Systems, Surroundings, and Internal Energy
Internal energy depends on
• Temperature
Calculating Thermodynamic Changes
Energy change = final E – initial E
ΔE = Efinal – Einitial
 higher T = larger Ek for the nanoscale particles.
A system can gain or lose E
• Type of material
 nanoscale Ek depends upon the particle mass.
SURROUNDINGS
SYSTEM
 nanoscale Ep depends upon the type(s) of particle.
SURROUNDINGS
SYSTEM
Efinal
ΔE > 0
• Amount of material
E in
Einitial
 number of particles.
 double sample size, double Einternal, etc.
ΔE positive: internal energy increases
Calculating Thermodynamic Changes
Einitial
ΔE < 0
E out
Efinal
ΔE negative: internal energy decreases
Heat Capacity
• No subscript? Refers to the system: E = Esystem
• E is transferred by heat or by work.
• Conservation of energy becomes: ΔE = q + w
heat
SURROUNDINGS
SYSTEM
Heat transfer out
q<0
Heat transfer in
q>0
Heat capacity = E required to raise the T of an object
by 1°C. Varies from material to material.
work
Specific heat capacity (c
(c )
• E needed to heat 1 g of substance by 1°C.
ΔE = q + w
Work transfer in
w>0
Work transfer out
w<0
Molar heat capacity (c
(cm)
• E needed to heat 1 mole of substance by 1°C.
Note the same sign convention for q and w
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Heat Capacity
Heat Capacity
c (J g-1 °C-1) cm (J mol-1 °C-1)
Substance
For other amounts or for other T changes:
Heat required = mass x specific heat x ΔT
q = m c ΔT
or…
Heat required = moles x molar heat capacity x ΔT
q = n cm ΔT
Heat Capacity
How much energy will be used to heat 500.0 g of iron
from 22°C to 55°C? cFe = 0.451 J g-1 °C-1.
Heat required = q = m c ΔT
q = 500.0 g (0.451 J g-1 °C-1)(55−22)°C
q = 7442 J = +7.4 kJ
Elements
C (graphite)
Al(s)
Fe(s)
Cu(s)
Au(s)
Compounds
NH3(ℓ)
H2O(ℓ)
H2O(s)
CCl4(ℓ)
CCl2F2(ℓ)
Common solids
wood
concrete
glass
granite
0.720
0.902
0.451
0.385
0.129
8.65
24.3
25.1
24.4
25.4
4.70
4.184
2.06
0.861
0.598
80.1
75.3
37.1
132.
72.3
1.76
0.88
0.84
0.79
Heat Capacity
24.1 kJ of energy is lost by a 250. g Al block. If the
block is initially at 125.0°C what will be its final T?
(cAl = 0.902 J g-1 °C-1)
q = m c ΔT
ΔT = q / (m c)
Cooling, q is negative:
negative
ΔT =
−24.1 x 103 J
250. g(0.902 J g-1 °C-1)
ΔT = Tfinal – Tinital = −107 °C
+ sign, E added to the system (the iron)
Heat Capacity
A 215 g block of Cu at 505.0°C is plunged into 1.000 kg
of water (T = 23.4 °C) in an insulated container. What
will be the final equilibrium T of the water and the Cu?
(cCu = 0.385 J g-1 °C-1)
q = m c ΔT
qCu = (215. g)(0.385 J g-1 °C-1)(Tfinal− 505.0)
qH2O = (1000. g)(4.184 J g-1 °C-1)(Tfinal− 23.4)
qCu + qH2O = 0
Thus Tfinal = ΔT + Tinital = −107 + 125°C = 18°C
Heat Capacity
215 g Cu (505.0°C) + 1000. g H2O (23.4 °C). Final T ?
qH2O = -qCu
4184(Tfinal – 23.4) = -82.78(Tfinal – 505.0)
(4184 + 82.78)Tfinal = 41804 + 97906
Tfinal = 32.7°C
(conservation of E)
qH2O = -qCu
(Note: Tfinal must be between Thot and Tcold)
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Conservation of Energy and Changes of State
Conservation of Energy and Changes of State
When heat is: Added to a system
A liquid cools from 45°C to 30°C, transferring 911 J to
the surroundings. No work is done on or by the
liquid. What is ΔEliquid?
 q is positive
 the change is endothermic
Removed from a system
 q is negative
ΔEliquid = qliquid + wliquid
 the change is exothermic
exothermic.
H2O(ℓ)  H2O(g)
Water Boils:
Steam Condenses: H2O(g)  H2O(ℓ)
endothermic
here wliquid = 0
Heat transfers from the liquid to the surroundings:
qliquid = -911 J
(qsurroundings = +911 J)
exothermic
Work occurs as the sample expands or contracts.
Overall:
ΔE = q + w
ΔEliquid = -911J
Enthalpy: Heat Transfer at Constant P
Conservation of Energy and Changes of State
A system does 50.2 J of work on its surroundings and
there is a simultaneous 90.1 J heat transfer from the
surroundings to the system. What is ΔEsystem?
Work done on the surroundings by the system
Heat transfers from the surroundings to the system
wsystem = -50.2 J
qsystem = +90.1 J
Because ΔE = q + w:
 At Constant V:
ΔE = qV
• subscript V shows fixed V
• work requires motion against an opposing force.
• constant V = no motion, so w = 0.
 At Constant P:
ΔE = qP + watm= ΔH + watm
• Subscript P shows fixed P.
• watm = work done to push back the atmosphere
• H = enthalpy
enthalpy. ΔH = qp
ΔEsystem = qsystem + wsystem
ΔEsystem = -50.2 J +90.1 J = +39.9 J
Freezing and Melting (Fusion)
During freezing (or melting)
Vaporization and Condensation
ΔHfusion = qP = heat to melt a solid.
• Substance loses (or gains) E, but…
Change
-50
Example:
Convert 1 g of
ice at -50°C to
water at +50°C
Temperature (°C)
-25
0
25
50
• T remains constant.
Ice is melting. T
remains at 0°C
Name
enthalpy of fusion
liq → gas
enthalpy of vaporization
liq → solid
enthalpy of freezing
gas → liq
enthalpy of condensation
333
2260
−333
−2260
Water warms
from 0 to 50°C
Note: ΔHfusion = − ΔHfreezing
100
200
300
400
500
Quantity of energy transferred (J)
etc.
qfusion = −qfreezing
Ice warms from -50
to 0°C
0
value for H2O (J/g)
solid → liq
600
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State Functions and Path Independence
Thermochemical Expressions
State functions
Always have the same value whenever the system is
in the same state.
ΔH = qP can be added to a balanced equation.
Two equal mass samples of water produced by:
1. Heating one from 20°C to 50°C.
State
functions
2. Cooling the other from 100°C to 50°C.
H
E
have identical final H (and V, P, E…).
P
V
T
etc.
State function changes are path independent.
independent
ΔH = Hfinal – Hinitial is constant.
Thermochemical Expressions
The thermite reaction produces tremendous heat:
2 Al + Fe2O3 → Al2O3 + 2 Fe
ΔH° = – 851.5 kJ
How much heat is released when 10.0 g of Al reacts
with excess Fe2O3 at constant P ?
= 10.0 g / 26.982 g/mol
= 0.3706 mol Al
ΔH° = −890.36 kJ
ΔH° is the standard enthalpy change
 P = 1 bar.
 T must be stated (if it isn’t, assume 25°C).
 ΔH° is a molar value. Burn 1 mol of CH4 with 2 mol O2 to
form 2 mol of liquid water and release 890 kJ of heat
 Change a physical state, change ΔH° : H2O(ℓ) vs. H2O(g)
Where Does the Energy Come From?
Bond Enthalpy (bond energy)
• Equals the strength of 1 mole of bonds
• Always positive
 It takes E to break a bond
 Separated parts are less stable than the molecule.
 Less stable = higher E
• E is always released when a bond forms
2 Al ≡ 1 ΔH° = – 851.5. kJ
 Product is more stable than the separated parts.
– 851.5 kJ
qp = 0.3706 mol Al
= –158 kJ
2 mol Al
 More stable = lower E
Bond Enthalpies
During a chemical reaction:
Old bonds break: requires E (endothermic)
New bonds form: releases E (exothermic)
Overall, heat may be absorbed or released:
Exothermic reactions (ΔH < 0)
 E is released.
 New bonds are more stable than the old,
or
 More bonds are formed than broken.
Both typically occur:
2 H(g) + 2 Cl(g)
reactants
products
2 HCl(g)
Endothermic reactions (ΔH > 0)
 E is absorbed.
endothermic
ΔH= +678 kJ/mol
energy
less stability
Bond Enthalpies
H2(g) + Cl2(g)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ)
exothermic
ΔH= -862 kJ/mol
 New bonds are less stable than the old,
or
 Fewer bonds are formed than broken
energy
less stability
nAl
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH° = −803.05 kJ
products
reactants
6
Measuring Enthalpy Changes
Heat transfers are measured with a calorimeter.
calorimeter
Common types:
• Bomb calorimeter.
 rigid steel container.
 filled with O2(g) and a small sample to be burnt.
 constant V, so qV = ΔE
• Flame calorimeter.
 samples burnt in an open flame.
 constant P, so qp = ΔH
• Coffee-cup calorimeter in lab (constant P).
Measuring Enthalpy Changes
Octane (0.600 g) was burned in a bomb calorimeter containing
751 g of water. T increased from 22.15°C to 29.12°C. Calculate
the heat evolved per mole of octane burned. Ccal = 895 J°C-1.
2 C8H18(ℓ) + 25 O2(g)
16 CO2(g) + 18 H2O(ℓ)
qreaction + qbomb + qwater = 0
qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C
= +6238 J
qwater = m c ΔT
= 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C
= +2.190 x 104 J
Measuring Enthalpy Changes
Bomb Calorimeter
Measure ΔT of the water. Constant V:
qV = ΔE
Conservation of E:
qreaction + qbomb + qwater = 0
or
−qreaction = qbomb + qwater
with
qbomb = mcalccalΔT = CcalΔT
A constant for a calorimeter
Bomb Calorimetry
T = 22.15 → 29.12°C. Heat per mol octane (0.600 g) burned.
2 C8H18(ℓ) + 25 O2(g) →16 CO2(g) + 18 H2O(ℓ)
−qreaction = qbomb + qwater
−qreaction = +6238 J + 2.190 x 104 J
= 2.81 x 104 J
= 28.1 kJ
qreaction = −28.1 kJ
Measuring Enthalpy Changes
Measuring Enthalpy Changes
Octane (0.600 g)… Calculate the heat evolved per mole of octane burned
CoffeeCoffee-cup calorimeter
Nested styrofoam cups prevent heat transfer
with the surroundings.
Molar mass of C8H18 = 114.23 g/mol.
nC8H18 = (0.600 g) / (114.23 g/mol)
= 0.00525 mol C8H18
Heat evolved /mol octane =
−28.1 kJ
0.00525 mol
= −5.35 x 103 kJ/mol
= −5.35 MJ/mol
Constant P. ΔT measured.
q = qp = ΔH
Assume the cups do not absorb heat.
(Ccal = 0 and qcal = 0)
7
Measuring Enthalpy Changes
1.02 g of Mg was reacted with excess 1 M HCl(aq)
(255.0 g) in a coffee-cup calorimeter. Tsoln rose from
22.0 to 41.6°C. cHCl = 3.90 J g-1°C-1 . Complete:
Mg(s) + 2 HCl(aq)
H2(g) + MgCl2(aq) ΔH = ?
qsoln
Measuring Enthalpy Changes
1.02 g Mg + 255.0 g of acid. T: 22.0 → 42.5°C. Molar ΔH = ?
qrxn = –19.57 kJ = ΔH for 1.00 g
= msoln c ΔT
(msoln = macid + mMg )
= 256.0 g (3.90 J g-1 °C-1)(41.6 − 22.0)°C
= 1.957 x 104 J
nMg = 1.02 g
So
Conservation of E:
qrxn = –qsoln
qsoln + qrxn = 0
1 mol = 0.04196 mol
24.31 g
ΔH = –19.57 kJ
0.04196mol
1 mol Mg
ΔH = – 466 kJ
or
exothermic
qrxn = –19.57 kJ = ΔH
Hess’s Law
Hess’s Law
“If the equation for a reaction is the sum of the
equations for two or more other reactions, then ΔH°
for the 1st reaction must be the sum of the ΔH°
values of the other reactions.”
Another version:
“ΔH° for a reaction is the same whether it takes place
in a single step or several steps.”
H is a state function
Multiply a reaction, multiply ΔH.
Reverse a reaction, change the sign of ΔH.
2 CO(g) + O2(g) → 2 CO2 (g)
ΔH = −566.0 kJ
Then
2 CO2(g) → 2 CO(g) + O2(g)
ΔH = –1(–566.0 kJ)
= + 566.0 kJ
4 CO2(g) → 4 CO(g) + 2 O2(g)
Hess’s Law
ΔH = –2(–566.0 kJ)
= +1132.0 kJ
Hess’s Law
Use Hess’s Law to find ΔH for unmeasured reactions.
Example
It is difficult to measure ΔH for:
2 C(graphite) + O2(g)
2 CO(g)
Some CO2 always forms. Calculate ΔH given:
C(graphite) + O2(g)
2 CO(g) + O2(g)
CO2(g)
2 CO2(g)
No phase confusion…
drop phases
ΔH = −393.5 kJ
ΔH = −566.0 kJ
Want:
Have: A
B
2 C + O2 →
C
+ O2 →
2CO + O2 →
2 CO
CO2
2 CO2
+2 x A
2C + 2O2 → 2CO2
−1 x B
2CO2 → 2CO + O2
???
−393.5 kJ
−566.0 kJ
2(−393.5) = −787.0
−1(−566.0) = +566.0
2C + 2O2 + 2CO2 → 2CO2 + 2CO + O2
2 C + O2 → 2 CO
−221.0
ΔH° = −221.0 kJ
8
Hess’s Law
Hess’s Law
Want:
Have: A
B
C
Determine ΔH° for the production of coal gas:
2 C(s) + 2 H2O(g)
CH4(g) + CO2(g)
Using:
C(s) + H2O(g)
CO(g) + H2(g)
ΔH° = 131.3 kJ
A
CO(g) + H2O(g)
CO2(g) + H2(g)
ΔH° = −41.2 kJ
B
CH4(g) + H2O(g)
CO(g) + 3 H2(g) ΔH° = 206.1 kJ
C
No phase confusion…
drop phases
Hess’s law problems often use a combustion or …
Formation reaction
Make 1 mol of compound from its elements in their
standard states.
H2 combustion:
CH4 + CO2
CO + H2
CO2 + H2
CO + 3 H2
???
131.3 kJ
−41.2 kJ
206.1 kJ
+2 x A
2 C + 2 H2O → 2 CO + 2 H2
+262.6
−1 x C
CO + 3 H2 → CH4 + H2O
−206.1
+1 x B
CO + H2O → CO2 + H2
−41.2
2C + 2H2O → CH4 + CO2
15.3 kJ
Standard Molar Enthalpy of Formation
Standard state = most stable form of the pure
element at P = 1 bar.
 e.g. C standard state = graphite (not diamond)
ΔHf° for any element in its standard state is zero.
(take 1 mol of the element and make… 1 mol of element)
2 H2O(ℓ)
ΔH° = −571.66 kJ
but the formation reaction is:
H2(g) + ½ O2(g)
→
→
→
→
(After cancelling and adding)
Standard Molar Enthalpy of Formation
2 H2(g) + O2(g)
2 C + 2 H2O
C
+ H2O
CO + H2O
CH4 + H2O
ΔHf° = −285.83 kJ
1 H2O(ℓ)
ΔHf° (Br2(ℓ) ) = 0
ΔHf° (Br2(g) ) ≠ 0
at 298 K
at 298 K
f = formation
Standard Molar Enthalpy of Formation
Appendix J (25°C)
Notes
• Most are negative
(formation releases E),
but can be positive.
• If the physical state
changes, ΔHf° changes.
Compound
ΔHf°, kJ/mol
Al2O3(s)
aluminum oxide
−1675.7
CaO(s)
calcium oxide
−635.09
CH4(g)
methane
−74.81
C2H2(g)
acetylene
+226.73
C2H4(g)
ethylene
+52.26
C2H6(g)
ethane
−84.68
C2H5OH(l) ethanol
−277.69
H2O(g)
water vapor
−241.818
H2O(l)
liquid water
−285.830
NaF(s)
sodium fluoride
−573.647
Standard Molar Enthalpy of Formation
ΔH° ={(nproducts)(ΔHf° products)}
– {(nreactants)(ΔHf° reactants)}
Example
Calculate ΔH° for:
CH4(g) + NH3(g)  HCN(g) + 3 H2(g)
ΔH° = ΔHf°(HCN) + 3ΔHf°(H2) − ΔHf°(NH3) – ΔHf°(CH4)
= +134 + 3(0) − (−46.11) – (−74.85) = 255 kJ
9