PHYSICAL CHEMISTRY PRT 140 Semester I 2013/2013 PN. ROZAINI
ABDULLAH School of Bioprocess Engineering UniMAP ADOPTED SLIDE
FROM: MR HUZAIRY & MS. ANIS ATIKAH
Slide 2
Standard Thermodynamic Functions of Reaction
Slide 3
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:
Slide 4
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.
Slide 5
For example; The molar volume of a pure solid or liquid at 1
bar and 200 K is symbolized by Superscript (naught, zero, or
standard) subscript 200 (temperature)
Slide 6
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 P P 1
bar
Slide 7
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.
Slide 8
For the reaction The standard enthalpy change is; where is the
molar enthalpy of substance C in its standard state at temperature
T. For the general reaction; Eq. (5.3) Eq. 4.94
Slide 9
Where the are the stoichiometric coefficients (+ve for products
and ve for reactants) and is the molar enthalpy of in its standard
state at T.
Slide 10
For example; for =
Slide 11
- 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;
Slide 12
For reactions 5.4 & 5.5; -Although we cannot have half a
molecule, we can have half mole of O 2, (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 , the
extent of reaction.
Slide 13
-A value is for = 1 mol -Since =, (Eq. 4.95) when = 1 mol for
(5.4), 2 mol of H 2 O is produced; whereas when = 1 mol for (5.5),
1 mol of H 2 O is produced.
Slide 14
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 (H, U and S, 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
Slide 15
Phase abbreviations: s solid l liquid g gas cr crystalline
(solids that have an ordered structure at the molecular level) am
amorphous (solids with a disordered structure) cd condensed phase
(either a solid or a liquid) fl fluid phase (either a liquid or a
gas)
Slide 16
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.
Slide 17
For example; The standard enthalpy of formation of gaseous
formaldehyde H 2 CO(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 P = 1
bar and 307 K). - at 307 K and 1 bar, the stable forms of hydrogen
and oxygen are H 2 (g) and O 2 (g) so, taken as reference forms of
hydrogen and oxygen.
Slide 18
Slide 19
- For an element in its reference form, is zero. -For example;
of graphite is of the reaction C(graphite, 307 K, P) C(graphite,
307 K, P) - nothing happens in this process, so, its is zero. - For
diamond, is not zero, but is of which experiment gives as 1.9
kJ/mol. C(graphite, 307 K, P) C(diamond, 307 K, P)
Slide 20
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
!!!
Slide 21
Consider, Reactant in their standard state at T Products in
their standard states at T Elements in their standard states at T
(1) (2)(3) Direct conversion Conversion of reactants to standard
states elements in their reference form Conversion of elements to
products
Slide 22
The relation
Slide 23
Example: Find for the combustion of 1 mole of the simplest
amino acid, glycine, NH 2 CH 2 COOH, according to
Slide 24
Answer: Substitution of Appendix values into [1/2(0) +
5/2(-285.830) +2 (-393.509) (- 528.10) 9/4 (0)] kJ/mol = -973.49
kJ/mol
Slide 25
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 we must carry out the following steps:
Slide 26
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 for mixing the pure elements at T and 1 bar 3.We
use for bringing the mixture form T and 1 bar 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
Slide 27
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 (25C) Some values of are
plotted in Fig1. A table of is given in Appendix.
Slide 28
The standard enthalpy of formation is the sum of these six Hs.
Figure 1 values. The scales are logarithmic. Nearly all
thermodynamics tables list at 298.15 K (25 C). A table of is given
in the Appendix.
Slide 29
Example 5.1 (calculation of from data) Find for the combustion
of 1 mole of the simplest amino acid, glycine, NH 2 CH 2 COOH,
according to;
Slide 30
Solution; We substitute Appendix values into (Eq 5.6) gives as
= - 973.49 kJ/mol Now, do exercise at page 144
Slide 31
Calorimetry -Step 4 find of a compound; - must measure H 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 constant- pressure calorimeter.
Slide 32
The standard enthalpy of combustion of a substance is for the
reaction in which 1 mole of the substance is burned in O 2. For
example; for solid glycine is for reaction in Example 5.1. Some
values are plotted in Figure 5.3. (the products are CO 2 (g) and H
2 O(l))
Slide 33
An adiabatic bomb calorimeter is used to measure heats of
combustion. R+K at 25C P+K at 25C + T R+K at 25C U =0 r U 298 U el
q=0, w=0, U=0 R= Mixtures of reactants P= Product mixture K= bomb
wall + surrounding water bath This system is thermally insulated,
and does no work on its surrounding U b =U el =Vlt (a) (b) (c) r U
298 = - U el 1) Start at 25 o C therefore 2) Cooling down to 25 o C
V= voltage I= current t= time
Slide 34
Alternative procedure: Imagine carrying out step (b) by
supplying heat q b to the system K+P (instead of using electrical
energy),then we would have Thus, Heat capacity of the system K+P
over temperature range
Slide 35
Example: Combustion of 2.016g of solid glucose at 25C in an
adiabatic bomb calorimeter with heat capacity 9550 J/K gives a
temperature rise of 3.282C. Find of solid glucose.
Slide 36
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
Slide 37
Relation between H and U For a process at constant pressure, H
= U + P V H = U + P V 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.
Slide 38
For example: Neglect volumes of solid and liquid, A and E; V =
So, by considering ideal gas = RT / P for each gases C, D, and B
Hence, c + d b = (change in number of moles of gas)
Slide 39
Thus, we have; So, from Becomes n g = number of moles of
gas
Slide 40
For example; Has = 3 1 5 = -3 So,
Slide 41
Example 5.3 - Calculation of from The reactions not involving
gases, is zero.
Slide 42
Hesss Law Study case: Find the standard enthalpy of formation
of ethane gas at 25 C. * 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.
Slide 43
+
Slide 44
Addition of these equations gives; Therefore, ( The procedure
of combining heats of several reactions to obtain the heat of a
desired reaction is Hesss Law).
Slide 45
Example 5.4 Calculation of from The standard enthalpy of
combustion of C 2 H 6 (g) to CO 2 (g) and H 2 O(l) is -1559.8
kJ/mol. Use this and Appendix data on CO 2 (g) and H 2 O(l) to
find