Energy and
Thermochemistry
1
Energy
The ability to do work
2 types
Potential:
stored energy
Kinetic: energy in motion
2
Thermochemistry
Changes of heat content and heat transfer
Follow Law of Conservation of Energy
Or, 1st Law of Thermodynamics
Energy
can neither be created nor destroyed
3
Temperature & Heat
Heat not same as temperature
Heat = energy transferred to one system by another
due to temperature difference
Temperature = measure of heat energy content &
ability to transfer heat
Thermometer
Higher thermal energy, greater motion of
constituents
Sum of individual energies of constituents = total
thermal energy
4
Systems and Surroundings
System = the object in question
Surrounding(s) = everything outside the system
When both system and surrounding at same
temperature ⇒ thermal equilibrium
When not
Heat transfer to surrounding = exothermic
(you feel the heat) ⇒ hot metal!
Heat transfer to system = endothermic
(you feel cold) ⇒ cold metal!
5
Math!
Joules (J) used for
energy quantities
But usually kJ (1000 J)
used
Ye Royal Olde School
used calorie (cal)
cal = amt of heat required
to raise the temperature of
1.00 g of water by 1°C
1 kg × m
Joule (J) =
2
s
1 cal = 4.184 J (SI-unit)
But…Calorie (Cal) = 1000
cal
Used in nutrition science
and on food labels
6
2
Heat Capacity
Specific heat capacity
Quantity of heat
required to raise the
temp of 1 gram of any
substance by 1 K
J
C=
g×K
Molar heat capacity
Quantity of heat
required to raise the
temp of 1 mole of any
substance by 1 K
J
c=
mol × K
4.184 J
specific heat capacity of water =
75.4 J
g × K molar heat capacity of water =
mol × K
7
Calculating heat transfer
Q = C × m × ∆T
Q = transferred heat, m = mass of substance, T = temperature change
FYI
Specific heat capacity
of metals is very low
≈ < 1.000 J/(g•K)
What does this tell us
about heat transfer in
metals?
8
Let’s do an example
In your backyard, you have a swimming
pool that contains 5.19 x 103 kg of water.
How many kJ are required to raise the
temperature of this water from 7.2 °C to
25.0 °C?
9
Example solved
Q = C × m × ∆T = (4.184
J
) × (5.19 x 106 g) × (298.2 K - 280.4 K) = 3.87 × 108 J = 3.87 × 105 kJ
g×K
Trick: ∆ T in K = ∆T in °C
10
Practice
How many kJ are required to raise the
temperature of 25.8 g of quicksilver from
22.5 °C to 28.0 °C? CHg = 0.1395 J/(g•K)
11
Solution
∆T = 28.0C-22.5C = 5.5C
J
kJ
= 20. ×10−3 kJ
Q = C × m × ∆T = (0.1395
) × 25.8g × 5.5 C = 20.J ×
g×K
1000J
12
But what if there’s a change of
state?
Temperature constant throughout
change of state
Added
energy overcomes inter-molecular
forces
13
Change of state
What do the flat areas represent?
14
Change of state
qtot = qs + qs⇒l + ql + ql⇒g + qg
qs⇒l = heat of fusion
Heat required to convert solid at melting pt. to liq
Ice = 333 J/g
ql⇒g = heat of vaporization
Heat required to convert liq. at boiling pt. to gas
Water = 2256 J/g
15
Practice
How much heat is required to vaporize
250.0 g of ice at -25.0 °C to 110.0 °C?
Given:
Specific
heat capacity of ice = 2.06 J/g•K
Specific heat capacity of water = 4.184 J/g•K
Specific heat capacity of steam = 1.92 J/g•K
Let’s do this
16
Solution
J
× (0.0 C − −25.0 C)=12875 ≈ 1.3 ×104 J
g×K
J
Qs→l =250.0g × 333 = 83, 250J
g
J
Ql =250.0g × 4.184
× (100.0 C-0.0C)=104,600J
g×K
J
Ql→g =250.0g × 2256 = 564000J ≈ 5.640 ×105 J
g
J
Qg =250.0g ×1.92
× (110.0 C-100.0 C)=4800 ≈ 4.80 ×103 J
g×K
Qs =250.0g × 2.06
Q tot = 1.3 ×104 J + 83, 250J + 104,600J + 5.640 ×105 J + 4.80 ×103 J = 769,650J
17
Calorimetry
The process of measuring heat transfer in
chemical/physical process
qrxn + qsoln = 0
qrxn = -qsoln
Rxn = system
Soln = surrounding
What you’ll do in lab
Heat given off by rxn
Measured by thermometer
Figure out qrxn indirectly
18
Enthalpy
= heat content at constant pressure
If ∆H = “+”, process endothermic
If ∆H = “-”, process exothermic
Enthalpy change dependent on states of
matter and molar quantities
For example:
Is
vaporizing ice an exothermic or
endothermic process?
Thus, will ∆H be “+” or “-”?
19
Hess’s Law
If a rxn is the sum of 2 or more other
reactions, ∆H = sum of ∆H’s for those rxns
So, ∆Htot = ∆H1 + ∆H2 + ∆H3 + … + ∆Hn
20
Let’s solve a problem
C(s) + 2S(s) → CS2(l); ∆H = ?
Given:
C(s) + O2(g) → CO2(g); ∆H = -393.5 kJ/mol
S(s) + O2(g) → SO2(g); ∆H = -296.8 kJ/mol
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ∆H = -1103.9 kJ/mol
How do we manipulate the 3 rxns to achieve the
necessary net rxn?
Does ∆H change if the rxns are reversed and/or
their mole ratios are changed?
Let’s talk about this on the next slide
21
Let’s work it out
1. Switch this rxn: CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ∆H = -1103.9 kJ
Thus, CO2(g) + 2SO2(g) → CS2(l) + 3O2(g) ; ∆H = + 1103.9 kJ
Thus, -∆Hfwd = +∆Hrev
2. Double this rxn: S(s)
(s) + O2(g) → SO2(g); ∆H = -296.8 kJ
Thus, 2S(s) + 2O2(g) → 2SO2(g); ∆H = (-296.8 kJ) x 2 = -593.6 kJ
Since ∆H is per mole, changing the stoichiometric ratios entails an
equivalent change in ∆H
3. Keep this rxn: C(s)
(s) + O2(g) → CO2(g); ∆H = -393.5 kJ
4. Add those on same side of rxns/eliminate those on opposite sides of rxn:
CO2, 2SO2, 3O2
5. Net rxn: C(s)
(s) + 2S(s) → CS2(l)
∆H = 1103.9 kJ - 593.6 kJ – 393.5 kJ = 116.8 kJ
Is it an exo- or endothermic rxn?
22
Practice
Given:
CH4(g) ⇒ C(s) + 2H2(g); ∆H = 74.6 kJ/mol
C(s) + O2(g) ⇒ CO2(g); ∆H = -393.5 kJ/mol
H2(g) + ½ O2(g) ⇒ H2O(g); ∆H= -241.8
kJ/mol
CH4(g) + 2O2(g) ⇒ CO2(g) + 2H2O(g); ∆Hrxn = ?
23
Standard Energies of Formation
Standard molar enthalpies of formation = ∆Hf° = enthalpy
change for formation of 1 mol of cmpd directly from
component elements in standard states
Standard state = most stable form of substance in physical state
that exists @ 1 bar pressure & a specific temp., usually 0°C
(273K)
1 bar = 100kPa
101.325 kPa = 1 atm
So 1 bar ≈ 1 atm (an SI unit)
Example
C(s) + O2(g) → CO2(g); ∆H = ∆Hf° = -393.5 kJ
∆Hf° = 0 for elements in standard state
24
Enthalpy Change for a Rxn
Must know all standard molar enthalpies
∆H°rxn = ∆Hf°prods - ∆Hf°reactants
Given to you in a table in the back of the
book
Again, keep in mind the mole ratios for
each species involved!
25
Practice
Determine ∆H°rxn for:
4NH3(g) + 5O2(g) ⇒ 4NO(g) + 6H2O(g)
Given:
NH3(g) = -45.9 kJ/mol
NO(g) = 91.3
H2O(g) = -241.8
26
Solution
Determine ∆H°rxn for:
4NH 3(g) + 5O 2(g) ⇒ 4NO(g) + 6H 2 O(g)
∆H°rxn = [4mol × 91.3
kJ
kJ
kJ
kJ
+ 6mol × -241.8
] − [4mol × −45.9
+ 5mol × 0
] = −902kJ
mol
mol
mol
mol
27
Example
∆H°rxn for 10.0 g of nitroglycerin?
2C3H5(NO3)3(l) → 3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g)
C3H5(NO3)3(l) = -364 kJ/mol
CO2(g) = -393.5 kJ/mol
H2O(g) = -241.8 kJ/mol
28
Solution
1)∆H rxn = [3 mol × 0
kJ
kJ
kJ
kJ
+ 1 mol × 0
+ 6 mol × -393.5
+ 5 mol × -241.8
]
2
mol
mol
mol
mol
kJ
] = -2842 kJ
mol
1mol -2842kJ
2)10.0 g ×
×
= −62.6kJ
227.2g
2mol
- [2 mol × -364
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