Transcript Chapter 6 Energy Transfer

Chapter 6
Energy Transfer
CHM 108
SUROVIEC
SPRING 2014
I. Principles of Heat Flow
 Energy is the capacity to do work
 Work is the result of a force acting through a
distance
A. Different Types of Energy
 Kinetic energy – energy associated with motion of an object
 Thermal energy – energy associated with temperature of an object
 Potential energy – energy associated with position of an object
 Chemical energy – energy associated with position of an electron
around an atom
B. Law of Conservation of Energy
 Energy cannot be created or destroyed
 Energy can be transferred from one kind to another
II. First Law of Thermodynamics
 Total energy of the universe is constant and energy is
neither created or destroyed
A. Internal Energy (IE)
 Sum of kinetic and potential energies of all the particles that
compose the system
 Is a state function
A. Internal Energy
 Internal energy change = DE
B. Exchanging Energy
 A system can exchange energy with the surroundings
through heat and work
B. Exchanging Energy
III. Quantifying Heat and Work
A. Heat
 Heat = exchange of thermal energy between a system
and its suroundings
B. Heat Capacity
 When a system absorbs
 Heat capacity = quantity
heat (q) its temperature
changes by DT.
of heat required to raise
T by 1oC. (J/oC)
 Specific heat capacity =
quantity of heat required
to raise T of 1 gram of
material by 1oC. (J/goC)
Example
 Suppose that you have the idea to start making
stained glass. Given that glass turns to a pliable
liquid around 1500oC and you are starting at 25oC,
how much heat does a 2.50 g piece of glass absorb?
C. Pressure – Volume work
 How does work change with the volume changes?
Example
 Inflating a balloon requires P-V work
1. If I inflate a balloon from 0.200L to 0.985L at an
external pressure of 1.00 atm, how much work in
Latm is done?
2. Given that 101.3 J = 1 Latm, what is this value of
work in Joules?
IV. Measuring ΔE
 The way to measure this is with constant volume
calorimetry
A. Calorimetry
Calorimetry measures change in thermal energy
between a system and its surroundings.
At constant volume the change in temperature is
related to heat absorbed by entire calorimeter
Example
When 2.09g of sucrose (C12H22O11) is combusted the
temperature rose from 23.89 oC to 28.45 oC. What is
the DErxn for this in kJ and then in kJ/mole?
V. Enthalpy
 In most cases we cannot hold the volume constant
 We are interested with how much energy is given off
at constant pressure
V. Enthalpy
 The signs of DH and DE are similar in meaning
Example
 If we burn 1 mole of fuel and constant pressure it
produces 3452 kJ of heat and does 11 kJ or work.
What are the values of DH and DE?
A. Stoichiometry involving DH
 Enthalpy for chemical reactions = DHrxn
For the reaction below, if you start with 13.2 kg of C3H8
what is q in kJ?
C3H8 (l) + 5O2 (g)  3CO2 (g) + 4H2O (l)
DHrxn = -2044 kJ
VI. Measuring DHrxn
 Coffee cup calorimetry is a way to measure DHrxn.
 Knowing the mass of solution, the heat evolved will
cause a DT. Combing that with Cs of solution you can
determine q.
Example
 We want to know the DHrxn of the following reaction:
Ca(s) + 2HCl (aq)  CaCl2 (aq) + H2 (g)
Given 0.204 g of Ca(s) in 100.0 mL of HCl (aq), the
temperature rose from 24.8 °C to 33.9 °C. The
density of the solution is 1.00 g/mL and the Cs,soln is
4.18 J/g°C
•
IV. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of
a substance
H2O (s)
•
•
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
DH = -6.01 kJ
H2O (l)
H2O (s)
If you multiply both sides of the equation by a factor n, then DH must
change by the same factor n.
2H2O (s)
•
H2O (l)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
The physical states of all reactants and products must be specified in
thermochemical equations.
H2O (s)
H2O (l)
H2O (l)
H2O (g)
DH = 6.01 kJ
DH = 44.0 kJ
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g)
CO2 (g)
0 = -393.5 kJ
DHrxn
S(rhombic) + O2 (g)
SO2 (g)
0 = -296.1 kJ
DHrxn
CS2(l) + 3O2 (g)
0 = -1072 kJ
CO2 (g) + 2SO2 (g) DHrxn
The final reaction for formation of CS2 is:
C(graphite) + 2S(rhombic)
CS2 (l)
V. Enthalpy of formation
Standard enthalpy of formation (DH0) is the heat change
that results when one
f
mole of a compound is formed from its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its most stable form is
zero.
DHf0 (O2) = 0
DHf0 (O3) = 142 kJ/mol
DH0 (C, graphite) = 0
f
DH f0 (C, diamond) = 1.90 kJ/mol
VII. Standard Heats of Formation
Example
Write an equation for the formation of C6H12O6 from
elements. The DHf0 for C6H12O6 is -1273.3 kJ/mole
B. Calculating DHf0 for a reaction
 To calculate DH rxn0 subtract the heats of formations
of the reactant multiplied by their stoichiometric
coefficient from heats of formation of the products
multiplied by their stoichiometric coefficents.
A. Hess’s Law
The standard enthalpy of reaction (DH0rxn ) is the enthalpy of a reaction carried
out at 1 atm.
aA + bB
DH0
=
rxn
[
cDH0 (C) +
f
cC + dD
dDH0 (D) ] - [
f
DH0
= S nDH0 (products)
rxn
f
aDH0 (A) +
f
bDH0 (B) ]
f
- S mDH0 (reactants)
f
Hess’s Law: When reactants are converted to products, the change in enthalpy
is the same whether the reaction takes place in one step or in a series of steps.
Example
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How
much heat is released per mole of benzene combusted? The standard enthalpy
of formation of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
Compound
DHf0 (kJ/mole)
C6H6 (l)
+49.04
O2 (g)
0
CO2 (g)
-393.5
H2O (l)
-187.6