Section 3.0.
The 1st Law of Thermodynamics
(CHANG text – Chapter 4)
3.1. Revisiting Heat Capacities
3.2. Definitions and Concepts
3.3. The First Law of THERMODYNAMICS
3.4. Enthalpy
3.5. Adiabatic Expansion of an Ideal Gas
3.6. Thermochemistry
3.7. Bond Energies and Bond Enthalpies
3.6. Thermochemistry
Goals….
(1) To define enthalpy of reaction (ΔRH °)
(2) To understand how to calculate ΔRH from ΔfH °
for compounds
(1) To use Hess’s Law to calculate ΔfH °
3.6. Thermochemistry
……is the study of energy changes in chemical reactions….
• For a constant pressure process the heat of reaction (qp)
is equal to the enthalpy of reaction (ΔRH )
• Exothermic reaction (gives off heat) ΔRH
Endothermic (absorbs heat) ΔRH
= negative value
= positive value
• at a pressure of 1 bar and T = 298 K we say ΔRH = ΔRH °
the standard enthalpy of reaction
• ΔRH ° has units of kJ
• ΔRH ° is the enthalpy change when reactants in standard
state are converted to products in standard state
Exothermic
Endothermic
The standard enthalpy change for a chemical rxn. is defined as:
ΔRH ° =
v H ° ( products )
-
v H ° ( reactants )
H ° is the standard molar enthalpy
v is the stoichiometric coefficient (number of moles)
Example:
aA + bB
cC +dD
The standard enthalpy of reaction is ……………
ΔRH ° = + (c mol) H ° (C) + (d mol) H ° (D)
– (a mol) H ° (A) – (b mol) H ° (B)
Unfortunately we can’t measure the absolute values of H ° for
substances so we use standard molar enthalpy of formation
instead……… ΔfH °
so ΔRH ° =
v ΔfH °( products ) - v ΔfH °( reactants )
How can we calculate ΔRH ° ?
ΔfH °(products)
ΔfH °(reactants)
Enthalpy
Elements
Reactants
RH°
Products
We can calculate H° simply
by measuring the difference
between the heat of formation
(ΔfH °) of the products and
the reactants.
Example:
C (graphite) + O2(g)
ΔRH ° =
CO2(g)
RH° = – 393.5 kJ
v ΔfH °( products ) - v ΔfH °( reactants )
= (1 mol)ΔfH °(CO2) - (1 mol)ΔfH °( graphite) – (1 mol)ΔfH °(O2)
= – 393.5 kJ
Note:
we assign ΔfH ° = 0 for elements in their “most stable
allotropic forms” at a particular temperature
example at 298 K
ΔfH °(O2) = 0 kJ mol-1
ΔfH °( graphite) = 0 kJ mol-1
so ΔRH ° for combustion of graphite may be expressed as:
ΔRH ° = (1 mol) ΔfH °(CO2) = – 393.5 kJ
“Allotropy is the ability of a chemical to exhibit in a number
of different and physically distinct forms in its pure
elemental state.”
Carbon, for instance can exist as graphite, diamond
and fullerene.
“pure elemental substances
that can exist with different
crystalline structures”
So…..standard molar enthalpies of formation (ΔfH °) are
important values……..with them we can calculate the enthalpy
of a reaction (ΔRH ° )…………
How do we obtain the values of ΔfH ° for chemicals ?
There are two methods:
Direct Method
Indirect Method
The direct method only works for compounds that can be directly
synthesized from their elements…………this is not usually the case
so we have to use the indirect method.
Indirect Method for Determining ΔfH ° for Compounds
• based on Hess’s Law
• Hess’s Law is stated as follows: when reactants are converted
to products the change in enthalpy is the same
whether the reaction takes place in one step or
a series of steps
• remember enthalpy is a state function so it’s value is path
independent
let’s look at an example to understand this…………….
Let’s use Hess’s Law to calculate ΔfH ° for CO
C (graphite) + ½ O2(g)
CO(g)
We have the following two reactions to work with……………..
(1) C (graphite) + O2(g)
(2)
CO (g) + ½ O2(g)
CO2(g)
CO2(g)
ΔRH ° = -393.5 kJ
ΔRH ° = -283.0 kJ
• let’s take these two reactions and rearrange them so they
add up to give overall desired reaction shown above
• we need C(graphite) and ½ O2 in the reactants and CO in the
products
• let’s reverse equation (2); and then add up eqn (1) and reverse of (2)
ΔfH °(CO) = (-393.5 kJ + 283.0 kJ)/(1 mol) = -110.5 kJ mol-1
Hess’s Law continued……………
• so we add up the equations such that everything cancels out
to leave only the desired reactants and products
• if we reverse an equation we must change the sign of ΔRH °
• if we multiply the reaction by a coefficient we must also
multiply ΔRH ° by the same factor
Let’s work through one more example together…………
Calculate the standard molar enthalpy of formation of acetylene (C2H2)
from its elements.
2C(graphite) + H2 (g)  C2H2 (g)
The equations for combustion and the corresponding enthalpy
changes are:
(1) C(graphite) + O2(g)  CO2(g)
ΔRH ° = -393.5 kJ
(2) H2 (g) + ½ O2(g)  H2O (l)
ΔRH ° = -285.8 kJ
(3) 2C2H2 (g) + 5O2(g)  4CO2(g) + 2H2O (l)
ΔRH ° = -2598.8 kJ
Solution:
Need to rearrange and add equations 1, 2 and 3 so that:
1 - C, H2 are reactants and C2H2 is product
2 - O2, CO2 and H2O are eliminated
Solution continued…….
Note: C and H2 are already in reactants in eqns. 1 and 2 but
we need to reverse eq. 3 to get C2H2 into products.
1. Reverse equation 3 to get C2H2 on the product side
2. Now multiply eqn 1 by 4 and eqn 2 by 2 in order to
eliminate O2, CO2 and H2O
(1) 4C(graphite) + 4O2(g)  4CO2(g)
(2) 2H2 (g) + O2(g)
 2H2O (l)
ΔRH ° = - 1574 kJ
ΔRH ° = - 571.6 kJ
(3) 4CO2(g) + 2H2O (l)  2C2H2 (g) + 5O2(g) ΔRH ° = +2598.8 kJ
4 C(graphite) + 2 H2 (g)  2 C2H2 (g)
Thus ….2 C(graphite) + H2 (g)  C2H2 (g)
ΔRH ° = + 453.2 kJ
ΔRH ° = + 226.6 kJ
ΔfH °(C2H2(g)) = ΔRH °/(1 mol) = +226.6 kJ mol-1
most stable allotropic form
of carbon
• some compounds
have lower values of ΔfH °
than elements they are
formed from
(e.g. H2O (l))
• some compounds
have higher values
of ΔfH ° than
elements they are
formed from
(e.g. C2H2 (g))
*** compounds that have negative values for ΔfH ° are usually
more stable than those with positive values
WHY ?
Because neg. value for ΔfH ° means heat is released
during formation so must be supplied for decomposition.
Heat of Combustion: the heat involved in the oxidation
of a compound
…..basically a compound is burned in the presence of O2 to
produce CO2 and H2O
i.e.
CxHy + O2
CO2(g) + H2O
(g)
(l)
The heat of combustion for various foods gives us an indication
of their ability to store energy…………
Fuel
Hydrogen
Methane
Gasoline (octane)
Tristearin
(beef fat)
Glucose
Specific Enthalpy
(kJ/g)
142
55
48
cH°
(kJ/mol)
- 286
- 890
- 5471
38
16
Phys. Chem. With Applications in Biology – P.Atkins
A sample problem
Metabolism is the stepwise breakdown of food we eat to provide
energy for growth and function. A general overall equation for
this complex process represents the degradation of glucose
(C6H12O6) to CO2 and H2O:
C6H12O6 (s) + 6O2(g)  6CO2(g) + 6H2O(l)
Calculate the standard enthalpy of reaction at 298K.
Solution:
ΔRH °
=
v ΔfH °( products ) - v ΔfH °( reactants )
=
(6 mol)(ΔfH ° CO2) + (6 mol)(ΔfH ° H2O)
– (1 mol)(ΔfH ° C6H12O6) – (6 mol)(ΔfH ° O2)
= (6 mol)(–393.5 kJ/mol) + (6 mol)(–285.8 kJ/mol)
– (1 mol)(–1274.5 kJ/mol) – (6 mol)(0 kJ/mol)
= – 2801.3 kJ
3.6. Thermochemistry
Goals….
(1) To define enthalpy of reaction (ΔRH °)
(2) To understand how to calculate ΔRH ° from ΔfH °
for compounds
(3) To use Hess’s Law to calculate ΔfH °
Progress
(1) ΔRH ° is the enthalpy change when reactants in standard state
are converted to products in standard state
(2) ΔRH ° =
v ΔfH °( products ) - v ΔfH °( reactants )
(3) Break reaction down into several steps, rearrange individual
equations so can cancel out to leave required reactants and
products
Section 3.0.
The 1st Law of Thermodynamics
(CHANG text – Chapter 4)
3.1. Revisiting Heat Capacities
3.2. Definitions and Concepts
3.3. The First Law of THERMODYNAMICS
3.4. Enthalpy
3.5. Adiabatic Expansion of an Ideal Gas
3.6. Thermochemistry
3.7. Bond Energies and Bond Enthalpies
3.7. Bond Enthalpies (Section 4.7 in Chang)
For diatomic molecules the bond enthalpy…..has special meaning
since there is only one bond
e.g. N2 (g)
2N(g)
ΔRH ° = 941.4 kJ
so bond enthalpy may be assigned to that bond…..thus we call it
the bond dissociation enthalpy
For polyatomics……there is more than one bond so we refer to it
as the average bond enthalpy
e.g. for H2O the energy required to break first O-H bond
is different from energy required to break second……….
H2O (g)
OH(g)
H(g) + OH(g)
O(g) + H(g)
ΔRH ° = 502 kJ
ΔRH ° = 427 kJ
OH avg.
bond enthalpy
= 460 kJ/mol
Triple bond
stronger than
double which
are stronger
than single
ΔRH °, the enthalpy of reaction in the gas phase may be given by:
ΔRH ° =  BE (reactants) -  BE (products)
= total energy bonds broken – total energy bonds formed
BE is the average bond enthalpy
If ΔRH ° is positive than reaction is endothermic
and if negative than exothermic
Note: if all the reactants and products are diatomics then we will
be using bond dissociation enthalpies (which are known and
accurate) so we will get accurate value for ΔRH °
But if we are working with polyatomics….we will be using avg.
bond enthalpies so value obtained for ΔRH ° is only
reasonable approximation.
Endothermic
Exothermic
Sample Problem
Estimate the enthalpy of combustion for methane
CH4 (g) + 2O2(g)  CO2(g) + 2H2O (g)
At 298 K, 1 bar using bond enthalpies.
Solution
Calculate the no. of bonds broken and formed:
Bonds
broken
C-H
O=O
No. bonds
4
2
Bond enthalpy
(kJ/mol)
414
498.8
Enthalpy change
(kJ)
1656
997.6
Bonds formed
C=O
O-H
2
4
799
460
1598
1840
Solution Continued
ΔRH ° =  BE (reactants) -  BE (products)
= total energy bonds broken – total energy bonds formed
= [(1656 kJ + 997.6 kJ) – (1598 kJ + 1840 kJ)]
= – 784.4 kJ
We can also calculate ΔRH ° using ΔfH ° values:
ΔRH ° = (1 mol) ΔfH °CO2 + (2 mol) ΔRH °(H2O)
– (1 mol)ΔRH °(CH4) – (2 mol) ΔRH °(O2)
= (1 mol)(–393.5 kJ/mol) + (2 mol)(–241.8 kJ/mol)
– (1 mol)(–74.85 kJ/mol) – (2 mol)(0 kJ/mol)
= – 802.3 kJ