Download Standard Enthalpy of Formation

PHYSICAL CHEMISTRY
PRT 140
Semester I 2013/2013
PN. ROZAINI ABDULLAH
School of Bioprocess Engineering
UniMAP
ADOPTED SLIDE FROM: MR HUZAIRY & MS. ANIS ATIKAH
Standard Thermodynamic
Functions of Reaction
Introduction
This chapters explains how to use tables of
thermodynamic properties (for ex: G, H, and S)
for individual substances with respect to
reactions in equilibrium:
Standard States of Pure Substances
Standard States
- For a pure solid and pure liquid:
Standard state is defined as the state with
pressure P = 1 bar and temperature T, where T
is some temperature of interest.
thus, for each value of T, there is a single
standard state for a pure substance.
For example;
The molar volume of a pure solid or liquid at 1 bar
and 200 K is symbolized by
Superscript º (“naught”, “zero”, or “standard”)
 subscript 200 (temperature)
For a pure gas:
- The standard state at temperature T is chosen
as the state where P = 1 bar and the gas
behaves as an ideal gas.
In summary;
The standard-state pressure is denoted by Pº
Pº ≡ 1 bar
Standard Enthalpy of Reaction
Standard enthalpy (change) of reaction
- Is the enthalpy change for the process of transforming
stoichiometric numbers of moles of the pure, separated
reactants, each in its standard state at temperature T,
to stoichiometric numbers of moles of the pure,
separated products, each in its standard state at
temperature T.
- Often
-
(or
) is also called heat of reaction.
is defined in a similar manner.
For the reaction
The standard enthalpy change
is;
where
is the molar enthalpy of substance C in its standard
state at temperature T.
Eq. 4.94
For the general reaction;
Eq. (5.3)
Where the
are the stoichiometric coefficients
(+ve for products and –ve for reactants)
and
is the molar enthalpy of
standard state at T.
in its
For example;
for
=
- Stoichiometric coefficient
is dimensionless, so
the units of
are the same with
,
namely J/mol or cal/mol
- sometimes, the subscripts m or T are omitted 
-
depends on how the reaction is written, for ex:
the standard enthalpy of reaction
is twice that for;
For reactions 5.4 & 5.5;
- Although we cannot have half a molecule, we can have
half mole of O2, (5.5) is a valid way of writing a reaction
in chemical thermodynamics.
- The
at T = 298 K for;
(5.4) 
(5.5) 
- The factor mol-1 in
indicates 
that we are giving the standard enthalpy change per mole of
reaction as written, where the amount of reaction that has occurred
is measured by ξ, the extent of reaction.
- A
value is for ξ = 1 mol
- Since
=
, (Eq. 4.95)
 when ξ = 1 mol for (5.4), 2 mol of H2O is
produced;
 whereas when ξ = 1 mol for (5.5), 1 mol of H2O
is produced.
Aims to calculate
of a reaction from
tabulated thermodynamic data for reactants and
products.
However, the laws of thermodynamics only allow us to
measure changes in enthalpy, internal energy and
entropies (ΔH, ΔU and ΔS, not the absolute values of U,
H, and S, and we cannot tabulate absolute enthalpies of
substances.
 Instead, we can tabulate standard enthalpies of
formation
In summary
Phase abbreviations:
s
l
g
cr
am
cd
fl
solid
liquid
gas
crystalline (solids that have an ordered structure at
the molecular level)
amorphous (solids with a disordered structure)
condensed phase (either a solid or a liquid)
fluid phase (either a liquid or a gas)
Standard Enthalpy of Formation
Standard Enthalpy of Formation (or Standard
Heat of Formation)
of a pure substance at
T is for the process
 in which 1 mol of the substance in its
standard state at T is formed from the
corresponding separated elements at T, each
element being in its reference form.
- The reference form (or reference phase) of an element
at T is usually taken as the form of the element that is
most stable at T and 1-bar pressure.
For example;
The standard enthalpy of formation of gaseous
formaldehyde H2CO(g) at 307 K, symbolized by
is the standard enthalpy change
for the process;
- Gases on the left are in their standard states (unmixed, in pure state
at standard pressure Pº = 1 bar and 307 K).
- at 307 K and 1 bar, the stable forms of hydrogen and oxygen are
H2(g) and O2(g)  so, taken as reference forms of hydrogen and
oxygen.
- For an element in its reference form,
- For example;
of graphite is
is zero.
of the reaction
C(graphite, 307 K, Pº)  C(graphite, 307 K, Pº)
- nothing happens in this process, so, its
- For diamond,
is not zero, but is
is zero.
of
C(graphite, 307 K, Pº)  C(diamond, 307 K, Pº)
which experiment gives as 1.9 kJ/mol.
Standard enthalpy change
is;
Where
is the stoichiometric coefficient of substance i in the
reaction and
is the standard enthalpy of formation of
substance i at temperature T.
Prove it !!!
• Consider,
Direct conversion
Reactant
in their standard state
at T
Conversion of
reactants to
standard states
elements in
their reference
form
(1)
Products
in
their standard states at T
(2)
(3)
Elements in their standard
states at T
Conversion of
elements to products
The relation
H T 
v 
i
i
f
H T ,i
Example:
• Find
for the combustion of 1 mole
of the simplest amino acid, glycine,
NH2CH2COOH , according to
Answer:
• Substitution of Appendix
values into H T   vi  f H T ,i
i
[1/2(0) + 5/2(-285.830) +2 (-393.509) – (- 528.10) – 9/4 (0)]
kJ/mol
= -973.49 kJ/mol
Determination of Standard Enthalpies
of Formation and Reaction
Measurement of
The quantity
is
for isothermally
converting pure standard-state elements in their
reference forms to standard-state substance i.
To find
steps:
we must carry out the following
1.If any elements involved are gases at T and 1bar, we calculate
for the
hypothetical transformation of each gaseous element from an ideal gas
at T and 1 bar to a real gas at T and 1 bar
2.We measure
3.We use
and 1 bar
for mixing the pure elements at T and 1 bar
for bringing the mixture form T
4. We use calorimeter to measure
for the reaction in the state in which
the compound is formed from the mixed elements
5. We use
to find
for bringing the
compound from the state in which it is formed in step4 to T and 1 bar
6. If compound I is a gas, we calculate
for the hypothetical transformation
of I from a real gas to an ideal gas at T and 1 bar
• The net results of these 6 steps is the
conversion of standard-state elements at T to
standard-state i at T.
• The standard enthalpy of formation
is
the sum of these six
• Nearly all thermodynamics table list
at
298.15K (25°C)
• Some values of
are plotted in Fig1.
• A table of
is given in Appendix.
The standard enthalpy of formation
is the sum of these six ΔH’s.
Figure 1
values. The scales are logarithmic.
Nearly all thermodynamics tables list
at
298.15 K (25 ºC). A table of
is given in
the Appendix.
Example 5.1 (calculation of
from
data)
Find
for the combustion of 1 mole of the
simplest amino acid, glycine, NH2CH2COOH,
according to;
Solution;
We substitute Appendix
values into
(Eq 5.6) gives
as
= - 973.49 kJ/mol
⨳
 Now, do exercise at page 144
Calorimetry
- Step 4  find
of a compound;
- must measure ΔH for the chemical reaction that
forms the compound from its elements.
using calorimeter
For example: Combustion
- Reactions where some of the species are gases (ex:
combustion rxn) – studied in a constant-volume calorimeter
- Reactions not involving gases – studied in a constantpressure calorimeter.
The standard enthalpy of combustion
of a
substance is
for the reaction in which 1 mole of
the substance is burned in O2.
For example;
for solid glycine is
for reaction
in Example 5.1.
Some
values
are plotted in Figure 5.3.
(the products are CO2(g)
and H2O(l))
• An adiabatic bomb calorimeter is used to
measure heats of combustion.
This system is thermally insulated,
and does no work on its surrounding
therefore
q=0, w=0, ∆U=0
1) Start at 25oC
∆U =0
R+K at 25°C
(a)
(c)
∆rU298 = - Uel
P+K at 25°C + ∆T
2) Cooling
to
(b) Uel down
25oC
Ub=Uel=Vlt
R+K at 25°C
R= Mixtures of reactants
P= Product mixture
K= bomb wall + surrounding water bath
V= voltage
I= current
t= time
• Alternative procedure:
Imagine carrying out step (b) by supplying heat
qb to the system K+P (instead of using
electrical energy),then we would have
Heat capacity of the system K+P over
temperature range
Thus,
Example:
• Combustion of 2.016g of solid glucose at 25°C
in an adiabatic bomb calorimeter with heat
capacity 9550 J/K gives a temperature rise of
3.282°C. Find
of solid glucose.
Solution:
• ∆U= -(9550J/K)(3.282K)
= - 31.34k/J for combustion of 2.016g of glucose
The experimenter burned (2.016g)/(180.16g/mol)
=0.001119 mol
Hence ∆U per mole of glucose burned is:
(-31.34k/J)(0.001119mol)= -2801 kJ/mol
Relation between ΔHº and ΔUº
For a process at constant pressure,
ΔH = ΔU + P ΔV
ΔHº = ΔUº + Pº ΔVº
The changes in standard-state volume and internal energy:
The molar volumes of gases at 1 bar are much greater than
those of liquids and solids
 so only consider the gaseous reactants and products.
For example:
Neglect volumes of solid and liquid, A and E;
ΔVº =
So, by considering ideal gas 
gases C, D, and B
Hence,
c+d–b=
= RT / Pº for each
(change in number of moles of gas)
Thus, we have;
ng = number of moles of gas
So, from
Becomes
For example;
Has
So,
= 3 – 1 – 5 = -3
Example 5.3 - Calculation of
from
The reactions not involving gases,
is zero.
Hess’s Law
Study case:
Find the standard enthalpy of formation
gas at 25 ºC.

of ethane
*
Problem: Cannot react graphite with hydrogen and expect
to get ethane. So,
of ethane cannot be
measured directly. So, what now??
we can determine the heat of combustion of
ethane, hydrogen, and graphite.
+
Addition of these equations gives;
Therefore,
⨳
(The procedure of combining heats of several reactions to obtain
the heat of a desired reaction is Hess’s Law).
Example 5.4 – Calculation of
from
The standard enthalpy of combustion
of
C2H6(g) to CO2(g) and H2O(l) is -1559.8 kJ/mol.
Use this
and Appendix data on CO2(g) and
H2O(l) to find
Solution:
Thank you