Download Hess's Law

AP CHEM NOTES - Chapter 6 - THERMO
(THERMOCHEMISTRY OR THERMODYNAMICS)
Chemical Reactivity
What drives chemical reactions? THERMODYNAMICS
How do they occur? KINETICS
We have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED.
•
formation of a precipitate
•
gas formation
•
H2O formation (acid-base reaction)
•
electron transfer in a battery
ENERGY is the capacity to do work or transfer heat.
HEAT is the form of energy that flows between 2 samples because of their difference in
temperature.
Other forms of energy —
 Light
 Electrical
 nuclear
 kinetic
 potential
Law of Conservation of Energy:
Energy can be converted from one form to another but can neither be created nor destroyed.
(E
universe
is constant)
The Two Types of Energy:
1.
Potential: due to position or composition - can be converted to work
PE = mgh
(m = mass, g = acceleration of gravity, and h = height)
2. Kinetic: due to motion of the object
KE =
mv2
(m = mass, v = velocity)
Potential energy — energy a motionless body has by virtue of its position.
Kinetic energy — energy of motion.
Translation
Rotation
Vibration
Units of Energy:
1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food “calorie”)
But we use the unit called the JOULE
1 cal = 4.184 joules
James Joule
1818-1889
Temperature vs. Heat
Temperature reflects random motions of particles, therefore related to kinetic energy of
the system.
Heat involves a transfer of energy between 2 objects due to a temperature difference
Extensive & Intensive Properties
Extensive properties depend directly on the amount of substance present.
• mass
• volume
• heat
• heat capacity (C)
Intensive properties are not related to the amount of substance.
• temperature
• concentration
• pressure
• specific heat (s)
State Function
Depends only on the present state of the system - not how it arrived there. It is
independent of pathway.
Energy change is independent of the pathway (and, therefore, a state function), while work
and heat are dependent on the pathway.
System and Surroundings
System: That on which we focus attention
Surroundings: Everything else in the universe
Universe = System + Surroundings
Exothermic and Endothermic
** Heat exchange accompanies chemical reactions.
Exothermic: Heat flows out of the system (to the surroundings).
Endothermic: Heat flows into the system (from the surroundings).
ENERGY DIAGRAMS
Exothermic:
Endothermic
First Law of Thermodynamics: The energy of the universe is constant.
Enthalpy
ΔH = H
final
-H
initial
If H final > H initial then ΔH is positive
Process is ENDOTHERMIC
If H final < H initial then ΔH is negative
Process is EXOTHERMIC
First Law ΔE = q + w
ΔE = change in system’s internal energy
q = heat
w = work
06_73
P=
P =
F
A
F
A
Area = A
A
h
(a) Initial
state
h
V
(b) Final
state
Piston moving a distance against a pressure does work.
Work (see previous diagram)
P=
F
A
F = PA
w = F h
w = P A h
w = P V
w & V must have opposite signs, since work is being done by the system to expand the gas
wsystem = -P V
1 L · atm = 101.3 J
1J=
kgm 2
s2
Enthalpy = H = E + PV
ΔE = ΔH - PΔV
ΔH = ΔE + PΔV
At constant pressure,
q = ΔE + PΔV,
P
where q = ΔH
P
at constant pressure
ΔH = energy flow as heat (at constant pressure)
Some Heat Exchange Terms
specific heat capacity (s)
heat capacity per gram =
J
J
or
°C g
Kg
molar heat capacity (s)
heat capacity per mole =
J
°C mol
or
J
K mol
Specific Heat Capacity
Substance
Spec. Heat
H 2O
4.184
Al
0.9 02
glass
0.8 4
J
Kg
Aluminum
CALORIMETER PROBLEMS
For calorimeter problems, the basic premise is:
Q
lost
=-Q
gained
Heat lost by the system (often what is being burned) is heat gained by the surroundings –usually
the water in the calorimeter or the bomb.
06_74
Thermometer
H° = - qp
qp = m T s
Styrofoam
cover
q = heat (J)
m = mass (g)
s
Styrofoam
cups
Stirrer
Simple Calorimeter
= specific heat
J
°C g
T = “change in” temperature (°C)
H° = “change in” enthalpy (kJ)
Specific Heat Capacity: Example Problem:
If 25.0 g of Al cool from 310 °C to 37 °C, how many joules of heat energy are lost by the Al?
heat gain/lost = q = m T s
where ΔT = T
final
-T
initial
Q will be negative. The negative sign on q signals heat “lost by” or transferred out of Al.
06_75
Stirrer
Ignition
wires
Thermometer
Insulating
container
Steel
bomb
Reactants
in sample
cup
Water
Bomb Calorimeter
Heat Capacity
C=
heat absorbed
J
J
 or
increase in temperature C K
E = qv & qv = -(C T + m T s)
 E = “change in” internal energy (J)
qv = heat at constant volume (J)
C = heat capacity
J
°C g
T = “change”in temperature (°C)
REMEMBER!!!
In regular calorimetry pressure is constant, but the volume will change so:
qp = -  H
qp = E + p  V
In bomb calorimetry, volume is constant so:
qv =  E
since p  V = zero.
Measuring Heats of Reaction
CALORIMETRY
Calculate heat of combustion of octane.
C8H18 +
25
2
O → 8 CO2 + 9 H2O
2

Burn 1.00 g of octane




Temp rises from 25.00 to 33.20 C
Calorimeter contains 1200 g water
Heat capacity of bomb = 837 J/K
Hc m = -(Ct + m ΔT s) where Hc is heat of combustion.
o
Step 1
Calc. heat transferred from reaction to water.
q = m T s
Step 2
Calc. heat transferred from reaction to bomb.
q = C T
Step 3
Total heat evolved
Heat of combustion of 1.00 g of octane =
Hess’s Law
Reactants → Products
**The change in enthalpy is the same whether the reaction takes place in one step or a series of
steps.
Standard States
Compound
- For a gas, pressure is exactly 1 atmosphere.
- For a solution, concentration is exactly 1 molar.
- Pure substance (liquid or solid), it is the pure liquid or solid.
Element
- The form [N2 (g) K(s)] in which it exists at 1 atm and 25°C.
Calculations via Hess’s Law
1.
2.
If a reaction is reversed, ΔH is also reversed.
N2 (g) + O2 (g) → 2NO(g)
ΔH = 180 kJ
2NO(g) → N2 (g) + O2 (g)
ΔH = -180 kJ
If the coefficients of a reaction are multiplied by an integer, DH is multiplied by that
same integer.
6NO(g) → 3N2 (g) + 3O2 (g) ΔH = -540 kJ
Using Enthalpy
Consider the decomposition of water:
1
H2O (g) + 243 kJ → H2 (g) + O2 (g)
2
Endothermic reaction — heat is a “reactant”
ΔH = + 2 4 3 k J
Making H from H O involves two steps.
2
2
H 2O ( l ) + 4 4 k J → H 2O ( g )
1
H2O (g) + 242 kJ → H2 (g) + O2 (g)
2
----------------------------------------------------------------1
H2O (l) + 286 kJ → H2 (g) + O2 (g)
2
Example of HESS’S LAW
If a rxn. is the sum of 2 or more others, the net ΔH is the sum of the ΔH’s of the other rxns.
Calc. ΔH for S (s) + O2 (g) → SO3 (g)
S(s) + O2 (g) → SO2 (g)
SO2 (g) +
O2 (g) → SO3 (g)
-320.5 kJ
-75.2 kJ
_______________________________________
ΔH along one path = ΔH along another path
This equation is valid because ΔH is a STATE FUNCTION
These depend only on the state of the system and not how it got there.
Unlike V, T, and P, one cannot measure absolute H. Can only measure ΔH.
Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of
ΔH°f = standard molar enthalpy of formation
This is the enthalpy change when 1 mol of compound is formed from elements under standard
conditions.
ΔH°f is always stated in terms of moles of product formed.
See Appendix A21-A24.
ΔH°f , standard molar enthalpy of formation H2 (g) + O2 (g) → H2O (g)
ΔH°f = -241.8 kJ/mol
By definition, ΔH°f = 0 for elements in their standard states.
Using Standard Enthalpy Values Use ΔH°’s to calculate enthalpy change for
H2O (g) + C (graphite)
H2 (g ) +
→
CO(g)
(product is called “water gas”)
From reference books we find
H2 (g) + O2 (g) → H2O (g)
ΔH°f of H2O vapor = - 242 kJ/mol
C (s) + O2 (g) → CO (g)
ΔH°f of CO = - 111 kJ/mol
Using Standard Enthalpy Values H2O(g) → H2 (g) + O2 (g)
C(s) +
ΔH ° = + 2 4 2 k J
o
O (g) → CO(g) DH = -111 kJ
2
----------------------------------------------------------------H O(g) + C(graphite) → H (g) + CO(g)
2
2
o
DH
net
= +131 kJ
To convert 1 mol of water to 1 mol each of H and CO requires 131 kJ of energy.
2
The “water gas” reaction is ENDOthermic.
Change in Enthalpy Can be calculated from enthalpies of formation of reactants and products.
DH
° = Sn DH °(products) - Sn DH °(reactants)
rx n
p
f
r
f
H is an extensive property--kJ/mol
For the reaction: 2H
2 (g )
+O
2 (g )
---> 2H O
2
(g )
Enthalpy would be twice the H value for the combustion of hydrogen.
o
Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., DH
for
rx n
CH OH(g) +
O (g) → CO (g) + 2 H O(g)
3
2
o
2
2
o
DH
o
= S DH (prod) - S DH (react)
rx n
f
f
Using Standard Enthalpy Values CH OH(g) +
3
o
DH
o
rx n
o
DH
o
= S DH (prod) - S DH (react)
f
f
o
rx n
o
= DH (CO ) + 2 DH (H O)
f
2
f
o
2
o
- { DH (O ) + DH (CH OH)}
f
2
f
3
= (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
o
DH
rx n
o
DH
rx n
= -675.6 kJ per mol of methanol
is always in terms of moles of reactant.
O (g) --> CO (g) + 2 H O(g)
2
2
2
06_77
C(s)
CH4(g)
(a)
(c)
CO2(g)
2H2(g)
(d)
2O2(g)
(b)
Reactants
Pathway for the Combustion of Methane
2O2(g)
Elements
Products
2H2O(l)
06_1551
Reactants
Elements
0 kJ
C(s)
2O2(g)
2O2(g)
75 kJ
2H2(g)
- 394 kJ
- 572 kJ
Energy
CH4(g)
CO2(g)
= Reactants
= Elements
= Products
Schematic diagram of the energy changes for the
combustion of methane.
Greenhouse Effect
-- a warming effect exerted by the earth’s atmosphere due to
thermal energy retained by absorption of infrared radiation.
Greenhouse Gases:
CO
H O
2
2
2H2O(l)
Products
CH
NO
4
2
Infrared
radiated by
the earth
06_80
Visible light
from the sun
CO2
and H2O
molecules
Earth
Earth’s
atmosphere