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. For the general reaction; Eq. (5.3) Eq. 4.94

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.

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 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 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 H 2 O is produced;  whereas when ξ = 1 mol for (5.5), 1 mol of H 2 O 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 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)

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 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 Pº = 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.

- For an element in its reference form, is zero. -For example; of graphite is of the reaction C(graphite, 307 K, Pº)  C(graphite, 307 K, Pº) - 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, Pº)  C(diamond, 307 K, Pº)

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, 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

The relation

Example: Find for the combustion of 1 mole of the simplest amino acid, glycine, NH 2 CH 2 COOH, according to

Answer: Substitution of Appendix values into [1/2(0) + 5/2( ) +2 ( ) – ( ) – 9/4 (0)] kJ/mol = 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 we must carry out the following steps:

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

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 K (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 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, NH 2 CH 2 COOH, according to;

Solution; We substitute Appendix values into (Eq 5.6) gives as = 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 constant- pressure calorimeter.

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))

An adiabatic bomb calorimeter is used to measure heats of combustion. R+K at 25°C P+K at 25°C + ∆T R+K at 25°C ∆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

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

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) = k/J for combustion of 2.016g of glucose The experimenter burned (2.016g)/(180.16g/mol) = mol Hence ∆U per mole of glucose burned is: (-31.34k/J)( mol)= 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  = RT / Pº for each gases C, D, and B Hence, c + d – b = (change in number of moles of gas)

Thus, we have; So, from Becomes n g = number of moles of gas

For example; Has = 3 – 1 – 5 = -3 So,

Example Calculation of from The reactions not involving gases, is zero.

Hess’s Law Study case: Find the standard enthalpy of formation of ethane gas at 25 ºC.  * 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 C 2 H 6 (g) to CO 2 (g) and H 2 O(l) is kJ/mol. Use this and Appendix data on CO 2 (g) and H 2 O(l) to find

Solution:

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