Download Theoretical Enthalpy

Theoretical Enthalpy:
Enthalpy Is a State Function
Applications of Hess’s
Law
Regardless of the pathway
the reaction takes, the enthalpy
of the reaction will always be
the same.
Consider:
N2(g) + 2O2(g)
N2(g) + 2O2(g) → 2NO2(g) ΔH = 68 KJ
It is very difficult to run this
reaction under normal laboratory
conditions. However, the following two
reactions are very easy to run in the
laboratory:
N2(g) + O2(g) → 2NO(g)
ΔH = 180 kJ
2NO(g) + O2(g) → 2NO2(g) ΔH = -112 kJ
Represented with an energy diagram, it is easy to
see how the two reactions can add to give the
enthalpy of the net reaction.
2NO(g)
2O2(g) + 2NO(g)
ΔH = -112 kJ
ΔH = 180 kJ
2NO2(g)
ΔH = 68 kJ
2NO2(g) ΔH = 68 KJ
Notice what happens when these
two reactions are added together:
N2(g) + O2(g)
→ 2NO(g)
Δ H = 180 kJ
2NO(g) + O2(g) → 2NO2(g) Δ H = -112 kJ
N2(g) + O2(g) + 2NO(g) + O2(g) →
2NO(g) +2NO2(g)
Δ H = 68 kJ
Simplified:
N2(g) + 2O2(g) → 2NO2(g)
ΔH = 68 kJ
Hess’s Law
If a reaction is carried out
in a series of steps, ΔH for the
reaction will be equal to the
sum of the enthalpy changes for
the individual steps.
N2(g) + 2O2(g)
1
Hess’s Law
Hess’s Law
 H is well known for many
reactions, and it is inconvenient to
measure H for every reaction in
which we are interested.
 However, we can estimate H using
H values that are published and
the properties of enthalpy.
Using Hess’s Law, we can
determine the change of
enthalpy of a reaction that can
not easily be measured, by
measuring simpler stepwise
reactions and adding their
enthalpies.
Example, the conversion of
graphite to diamond occurs
naturally in the earth’s crust
under extreme heat and pressure.
Using Hess’s Law find ΔH for:
C graphite (s) → C diamond (s)
Given:
C graphite (s) + O2(g) → CO2(g)
ΔH = -394 kJ
C diamond (s) + O2(g) → CO2(g)
ΔH = -396 kJ
Because H is a
state function, the
total enthalpy
change depends
only on the initial
state of the
reactants and the
final state of the
products.
Hess’s Law
1. Using the
diagram, write a
sequence of
thermochemical
equations that
would net the
enthalpy
change for the
combustion of
methane.
You must ensure that when you add
reactions together, you end up with the
correct reactants and the desired product
C graphite (s) + O2(g) → CO2(g)
ΔH = -394 kJ
CO2(g) → C diamond (s) + O2(g)
ΔH = -(-396 kJ)
C graphite (s) + O2(g) + CO2(g) →
C diamond (s) + O2(g) _+ CO2(g)
By convention:
C graphite (s) → C diamond (s)
ΔH = 2 kJ
2
Things to consider when
applying Hess’s Law:
• If a chemical reaction is reversed,
ΔH is also reversed.
• The net reaction must contain only
the reacting and produced species
• The magnitude of ΔH is directly
proportional to the quantities
described by the coefficients of a
balanced chemical reaction.
Reactions
ΔH (kJ)
Example:
Diborane(B2H6) is a highly reactive
hydride of boron. Calculate ΔH for the
synthesis of diborane from its elements,
according to the equation:
2B(s) + 3H2(g) → B2H6(g)
Given:
next slide
Practice:
Given:
(a) 2B(s) + 3/2O2(g) → B2O3(s)
-1273
(b) B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g)
-2035
NH3(g) → 1/2N2(g) + 3/2H2(g) ΔH = 46 kJ
(c) H2(g) + 1/2 O2(g) → H2O(l)
-286
(d) H2O(l) → H2O(g)
2H2(g) + O2(g) → 2H2O(g)
44
ΔH = -484kJ
Calculate ΔH for:
2N2(g) + 6H2O(g) → 3O2(g) + 4NH3(g)
Then graph using an energy diagram.
I find your lack of
understanding disturbing
A Theoretical
Approach
We have seen how to determine the enthalpy
of a reaction experimentally by calorimetry, and
for violent reactions by applying Hess’s Law to a
series of reactions carried out separately in a
calorimeter.
These methods have lead to extensive tables
listing the enthalpies of vaporization (∆Hvap),
enthalpies of fusion (∆Hfus), enthalpies of reduction
(∆Hred), enthalpies of combustion (∆Hcomb), etc…
However, a much more useful table can be
generated that lists the enthalpies of formation for
any pure substance that allows us to calculate the
expected enthalpy change when substances
undergo chemical reactions.
3
Standard Enthalpy
of Formation
The change in enthalpy that
accompanies the formation
of one mole of a compound
from its free elements in
their standard states.
Example:
1/2N2(g) + O2(g) → NO2(g) ΔHof =34 kJ/mol
Notice
• ΔHof is only for the production of
one mole of a substance
• ΔHof is zero for any pure element
in its standard state
1.Using information from
appendix C, write
thermochemical equations
for both Δ Hof and Δ H for
the following compounds:
A. Methanol (CH3OH)
B. CuO (s)
C. Solid rubidium chlorate
Standard Enthalpies of Formation
Standard enthalpies of formation, Hf,

are measured under standard conditions
(25°C and 1.00 atm pressure).
Determine the ΔH for the
previous reaction.
N2(g) + 2O2(g) → 2NO2(g)
ΔH =68 kJ
Remember that ΔHof is for only
one mole of a pure substance. The
enthalpy change associated with
the production of more than one
mole is the ΔH of the reaction.
2. Using information from
appendix C, write
thermochemical equations
for ΔHof of the following
compounds:
A.
B.
C.
D.
Methane (CH4)
Carbon dioxide
Water
Oxygen
4
Applying Hess’s Law
• The enthalpy change for any reaction
is the sum of the enthalpy changes for
the formation of each reactant and
product.
• Examine the following reaction for the
combustion of methane
C H 4 ( g )  2 O 2 ( g )  C O 2 ( g )  2 H 2 O (l )
Hrxn = ?
1. C ( s )  2 H 2 ( g )  CH 4 ( g )
ΔHof =-75 kJ/mol
2. C ( s )  O2 ( g )  CO2 ( g )
ΔHof =-394 kJ/mol
3. H 2 ( g )  12 O 2 ( g )  H 2O (l )
ΔHof =-286 kJ/mol
CH 4 ( g )  C ( s )  2 H 2 ( g ) - H1
C ( s )  O2 ( g )  CO2 ( g )
H 2
2 H 2 ( g )  O 2 ( g )  2 H 2 O (l )  H 3
CH 4 ( g )  2 O2 ( g ) 
CO2 ( g )  2 H 2O (l )
Hrxn  (  H 1 )  H 2  2H 3
By multiplying the heat of
formation by the number of
moles for the balanced reaction,
we can determine the enthalpy
change for the independent
steps, then use Hess’s Law to
determine the enthalpy for the
net reaction for the combustion
of methane.
Hrxn = (H2 + 2H3) - H1
Notice, the total enthalpy change is
the difference of the product enthalpies
minus the reactant enthalpies, each
multiplied by the number of moles in the
balanced reaction.
Hrxn (CH4) =
[Hf0(CO2) + 2Hf0(H2O)]
- [Hf0(CH4) + 2Hf0(O2) ]
5
Theoretical Calculation of H
We can use Hess’s law in this way:
3. 2.Calculate the enthalpy
change for the following
reactions:
H = nHf,products – mHf°,reactants
• Nitrogen dioxide gas bubbled
through water
where n and m are the stoichiometric
coefficients.
• Combustion of solid glucose
(C6H12O6)
Thermochemical
Stoichiometry
Using the change in enthalpy
for any chemical reaction, the
energy released by, or required by,
any reaction given any quantity of
reactant or product can be
calculated.
… you have now been sentenced to
die by chemical incineration for being
so sweet and chewy little Mr. Gummy
Bear. Do you have any last words?
3. Calculate the heats of the
reaction for the
decomposition of potassium
chlorate resulting in the
exothermic production of
O2 used to execute the
“gummy.”
ohh
No!!!
Example:
If the metabolism of glucose
is the same as the combustion of
C6H12O6(s), how much heat is
produced by the metabolism of
1.00 g of glucose?
4.How many grams of CH4
are needed to produce
enough heat to just boil
away 100. mL of water
under standard state
conditions?
6
Bond Enthalpy and Hess’s Law
• Yet another way to estimate H for a reaction
is to compare the bond enthalpies of bonds
broken to the bond enthalpies of the new
bonds formed.
Remember the basic premise behind a chemical
reaction:
Reactant Molecules (bonds broken)
atoms
Product Molecules (bonds formed)
Average Bond Enthalpies
• This table lists the
average bond
enthalpies for
many different
types of bonds.
• Average bond
enthalpies are
positive, because
bond breaking is
an endothermic
process.
Enthalpies of Reaction
•
•
CH4(g) + Cl2(g) 
CH3Cl(g) + HCl(g)
In this example, one
C—H bond and one
Cl—Cl bond are
broken; one C—Cl
and one H—Cl bond
are formed.
Enthalpies of Reaction
• Yet another way to
estimate H for a
reaction is to
compare the bond
enthalpies of
bonds broken to
the bond
enthalpies of the
new bonds formed.
• In other words,
Hrxn = (bond enthalpies of bonds broken) 
(bond enthalpies of bonds formed)
Average Bond Enthalpies
NOTE:
These are average
bond enthalpies,
not absolute bond
enthalpies; the
C—H bonds in
methane, CH4,
will be a bit
different than the
C—H bond in
chloroform,
CHCl3.
Enthalpies of Reaction
So,
Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)]
= (655 kJ)  (759 kJ)
= 104 kJ
7
5. Using the bond energies from the
“bond reference,” calculate the
change in enthalpy for reaction of
methane, chlorine and fluorine to
produce carbon dichloride
diflouride, hydrogen flouride and
hydrogen chloride. Also, predict
the relative bond lengths of all
reactants and products.
8