Plan for Mon, 20 Oct 08
• Lecture
– Constant P and Constant V Calorimetry (6.2)
– Characteristics of enthalpy changes and
Hess’s Law (6.3)
• Q3, Ex1, Exp1 lab report returned
• Don’t forget there is a quiz on Wednesday!
Heat Flow
System:
1 L of water
T = 50oC
Surroundings:
T = 25oC
The liter of water
is allowed to cool,
until it reaches
room temperature.
1 L of water
T = 25oC
How much heat did the water lose?
Well, first we need to know
how much heat it had.
System:
1 L of water
T = 50oC
The liter of water is
allowed to cool, until it
reaches room
temperature, 25oC.
How much heat did
the water lose?
q  s  m  T
J
m  1 L = 1000 mL = 1000 g
s  4.184
g C
T  T f  Ti  50C+25C =  25C

q  4.184 J
1000 g  25C   104,600 J


g C
 104.6 kJ
Constant P Calorimetry
NaOH + HCl  H2O + NaCl; H = -58 kJ
The heat evolved in this reaction is trapped
in the water…this heat increases the
average Ek of the water molecules, leading
to an increase in the temperature of the
water.
qsystem = -qsurroundings
Formally, q is the amount of heat that must be
exchanged with the surroundings to return the
system to its original temperature.
In calorimetry we don’t let it escape…q goes
into raising the temperature of the water.
Example
• 5.00 g of ammonium nitrate is
dissolved in 500. mL of water at
25.00oC. What is the final
temperature of the water?
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Hsoln = +82.93 kJ/mol
Think of the reaction as the system and
the water as the surroundings…
Then the heat gained by the reaction is
exactly equal to that lost by the water:
 qrxn   qH 2O
Constant V Calorimetry
For reactions involving gases, such as
combustion.
The heat produced in the reaction is
transferred to the steel bomb, and
then to the surrounding water bath.
Constant V Calorimetry
-qrxn = qbomb + qH2O + qO2 + qproducts + …
-qrxn = qcalorimeter
qrxn = -CcalT
The “calorimeter constant” Ccal must be
obtained experimentally, using a
reference compound for which the E
of combustion is well-known.
Example: Combustion of Sucrose
• A 1.010-g sample of sucrose (C12H22O11)
is combusted in a bomb calorimeter with
Ccal = 4.90 kJ/oC. The initial temperature
was 24.92oC, and the final temperature
was 28.33oC. How much energy was
evolved as heat during this reaction?
• How much energy is evolved per mole of
sucrose combusted?
Enthalpy Changes
• Enthalpy is a measure of the potential energy stored in a chemical
system.
• A given molecule will always have the same kind of bonds in it, no
matter where or when or how it was made.
Methane, CH4,
always has four
C-H bonds.
H
H
C
H
H
Elemental oxygen, O2,
always consists of two O
doubly bonded O atoms.
O
• This means a given molecule at a given temperature and physical
state will always have the same enthalpy content, no matter where
or when or how it was made.
• Therefore, enthalpy changes associated with chemical or physical
processes are state functions…
• They depend only on the initial and final states of the system…not
on the steps taken to get from the initial state to the final state.
H
H
C
H
H
+ 2O
O
O
C
O
+ 2H
O
H
Consider the synthesis of nitrogen dioxide,
NO2, from its elements: N2 and O2.
N 2 (g) + 2O2 (g)  2NO 2 (g)
H1  68 kJ
N2 (g) + O2 (g)  2NO(g)
H2  180 kJ
+ 2NO(g) + O2 (g)  2NO2 (g)
H3  112 kJ
N2 (g) + 2O2 (g)  2NO2 (g)
H2 + H3  68 kJ
 H1
Hess’s Law
In going from a particular set of reactants to a
particular set of products, the change in
enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
This means we can use enthalpy changes for reactions
that we can easily measure to determine the enthalpy
changes for reactions that are not so easy to measure.
Two characteristics of enthalpy will be useful in this endeavor:
• Enthalpy is a state function. If a reaction is reversed, the
value of H is also reversed.
• Enthalpy is an extensive property. If the coefficients in a
balanced equation are multiplied by an integer, H must be
multiplied by the same integer.
NOTE: The value of H included with balanced chemical
equations is for the reaction as it is written.
The processes below represent an enthalpy change only
for the molar amounts present.
•
“Heat of Fusion” - enthalpy change associated with melting
H2O(s)  H2O(l) Hfus = +0.334 kJ/mol
•
“Heat of Vaporization” - enthalpy change associated with boiling
H2O(l)  H2O(g)Hvap = +2.26 kJ/mol
•
“Heat of Reaction” - enthalpy change associated with a chemical reaction
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Hrxn = -50.1 kJ/mol
•
“Heat of Solution” – enthalpy change associated with the dissolution of
ionic solids in water
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Hsoln = +82.93 kJ/mol
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
Hsoln = -26.2 kJ/mol
What is the
enthalpy
change
associated
with changing
graphite into
diamond?
Cgraphite  Cdiamond
H = ??
Using Enthalpy
• We can use the H for various reactions to
determine H for a composite reaction.
• Example:
C(s, diamond) + O2(g)
CO2(g); H = -396 kJ/mol
C(s, graphite) + O2(g)
CO2(g); H = -394 kJ/mol
Using H (cont.)
C(s, graphite) + O2(g)
+ CO2(g)
CO2(g)
H = -394 kJ/mol
C(s, diamond) + O2(g) H = +396 kJ/mol
C(s, graphite)
C(s, diamond)
H = +2 kJ
Hrxn > 0…..rxn is endothermic
C(s, diamond)
C(s, graphite)
H = -2 kJ
Hrxn < 0…..rxn is exothermic
Example
• Use the information below to determine H for the
reaction:
3 C(graphite) + 4 H2(g)  C3H8(g)
C3H8(g) + 5 O2(g)  3 CO2(g) + 4H2O(l) H = -2219.9 kJ
C (graphite) + O2(g)  CO2(g)
H = -393.5 kJ
H2(g) + 1/2 O2(g)  H2O(l)
H = -285.8 kJ
ANS: H = -104 kJ
Hess Hints
• Using Hess’s Law involves some degree of trial
and error when you are manipulating the given
chemical equations.
• Some tips for success:
– Work backward from the required reaction, using the
reactants and products to guide you in manipulating
the other given reactions.
– Reverse any reactions as needed to give the required
reactants and products in your final equation.
• Don’t forget to reverse the sign on H too also!!
– Multiply reactions to give the correct number of
reactants and products in your final equation.
• Don’t forget to multiply the H though, also!!