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Miss Noorulnajwa Diyana Yaacob Email: [email protected]@unimap.edu.my 28 February 2011 Standard Thermodynamic Functions of Reaction

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Standard Thermodynamic Functions of Reaction

Miss Noorulnajwa Diyana YaacobEmail: [email protected] February 2011

Standard Thermodynamic Functions of Reaction1SubtopicsStandard states of pure substanceStandard Enthalpy of ReactionStandard Enthalpy of FormationDetermination of Standard Enthalpies of Formation and Reaction2IntroductionChemical reaction

To effectively apply this condition to reactions, we will need tables of thermodynamic properties (such as G, H, and S) for individual substances.In these tables, the properties are for substances in a certain state called STANDARD STATE.From table STANDARD STATE thermodynamic properties, we can calculate the changes in standard enthalpy, entropy and Gibbs Energy for chemical reaction3Standard state of Pure SubstancesFor a pure solid or a pure liquid, the standard state is defined as the state with pressure P= 1 bar and temperature T, where T is some temperature of interest.The symbol for standard state is a degree superscript with temperature written as subscript.Example: The molar volume of a pure solid or liquid at 1 bar and 200K

4For a pure gas, the standard state at temperature T is chosen as the state where P=1 bar and the gas behave as an ideal gas.Since real gases do not behave ideally at 1 bar, the standard state of a pure gas is a fictitious state.Summarizing, the standard state for pure substances are:Solid or liquid P= 1 bar, TGas P = 1 bar, T, gas idealThe standard-state pressure is denoted by P:

5Standard Enthalpy of ReactionStandard enthalpy (change ) of reaction ( )= enthalpy change for the process of transforming stoichiometric numbers of moles of the pure, separated reactant (each in its standard state at T) to stoichiometric numbers of moles of the pure, separated products (each in its standard state at same T) Is also called the heat of reaction

6The standard enthalpy change

Stoichiometric numbers(- for reactant and + for products)Molar enthalpy in its standard state7For Example:

The units = J/mol or cal/mol8Doubling the coefficients of a reaction doubles its

= -572 kJ/mol

= -286 kJ/mol9Standard Enthalpy of FormationThe standard enthalpy of formation (or standard heat of formation) of a pure substance at T = process in which one mole of a substances in its standard state at T is formed from the corresponding separated element at T( each element being in its reference form)Reference form (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

10 For example:The standard enthalpy of formation of gaseous formaldehyde at 307K symbolized by is the standard enthapy change for the process

The gases on the left are in their standard state(unmixed):each of their substance at standard pressure =1 bar and 307KThe most stable form of carbon is graphite11For an element in its reference form, = 0

Summarizing:Stoichiometric number12Consider,

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 formConversion of elements to products

13The relation

14Determination of Standard Enthalpies of Formation and Reaction1. Measurement ofThe quantity of ,is for isothermally converting pure standard-state elements in their reference form to 1 mole of standard-state substance i .To Find , we carried out 6 steps which are:

151.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 bar2.We measure for mixing the pure elements at T and1 bar3.We use for bringing the mixture form T and 1 bar4. We use calorimeter to measure for the reaction in the state in which the compound is formed from the mixed elements5. 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

16The 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 sumof 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

17Figure 1: values

18Example:Find for the combustion of 1 mole of the simplest amino acid, glycine, NH2CH2COOH , according to

19Answer: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

202. CalorimetryThe most common type of reaction studied calorimetrically is combustionHeat capacities also determined in a calorimeterReaction where some of the species are gases (combustion reaction)are studied in constant-volume calorimeterReaction not involving gases are studied in a constant-pressure calorimeter.

21The standard enthalpy of combustion of a substance is for the reaction in which 1 mole of the substance is burned in 02 .Some values are plotted in Figure 2

22Figure 2:Standard enthalpies of combustion at 25C. The products are CO2(g) and H20(l)

23An adiabatic bomb calorimeter is used to measure heats of combustion.

R+K at 25CP+K at 25C + TR+K at 25CU =0rU298 Uel q=0, w=0, U=0R= Mixtures of reactantsP= Product mixtureK= bomb wall+surrounding waterbath

This system is thermally insulated, and does no work on its surroundingUb=Uel=Vlt(a)(b)(c)rU298 = - Uel

24Alternative procedure:Imagine carrying out step (b) by supplying heat qb 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

25Example: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.

26Solution:U= -(9550J/K)(3.282K) = - 31.34k/J for combustion of 2.016g of glucoseThe experimenter burned (2.016g)/(180.16g/mol) =0.001119 molHence U per mole of glucose burned is:(-31.34k/J)(0.001119mol)= -2801 kJ/mol

273. Relation between H and UCalorimetric study of a reaction gives either H or U

For a process at constant pressure

Since the standard pressure P is the same for all substances, conversion of pure standard-state reactants to product is a constant-pressure

28

Products minus reactants29The molar volumes of gases at 1 bar are much greater than those of liquids or solid, so it is an excellent approximation to consider only gaseous reactant and products in applyingFor example:

Neglecting the volume of the solid and liquid

30The standard-state of a gas is an ideal gas, so

For each gases C, D and B. Therefore ,

Thus

ng/mol

31Example:For CO(NH2)2 (s), = -333.51 kJ/mol . Find of CO(NH2)2 (s).

32Solution:The formation reaction is:C (graphite) + O2 (g) + N2 (g) +2H2(g) = CO(NH2)2

= 0-2-2-1/2 = -7/2

= -333.51 kJ/mol (-7/2) (8.314 X 10-3 kJ/molK)(298.15K)

= -324.83 kJ/mol

33For reaction not involving gases, ng is zero and H is essentially the same as U .344. Hesss LawSuppose we want the standard enthalpy of formation of ethane gas at 25C.This is for 2C (graphite)+ 3H2 (g) C2H6(g) Unfortunately, we cannot react graphite with hydrogen and expect to get ethane, so the heat formation of ethane cannot be measure directly.HOW TO SOLVE THIS PROBLEM!!!

35We determine the heats of combustion of ethane, hydrogen, and graphite. The following values are found at 25CC2H6 (g) +7/2 O2 (g)2CO2 (g) +3H20(l) H298 = -1560 kJ/mol (1)C(graphite) +O2 (g) CO2(g) H298 = -393.5 kJ/mol (2)H2 (g) + O2(g) H2O(l) H298 = -286 kJ/mol (3)

Multiplying the definition by -1, 2, and 3 for reaction (1). (2), and(3), we get:

-(-1560kJ/mol) = -2Hm (CO2) - 3Hm (H2O) + Hm (C2H6) + 3.5Hm (O2) 2(-393.5kJ/mol) = 2Hm (CO2)-2 Hm (O2) - 2 Hm (C) 3 (-286kJ/mol) = 3Hm (H2O) - 3Hm (H2) 1.5Hm (O2)Addition of these equation gives:-85kJ/mol= Hm(C2H6)-2 Hm(C)-3Hm(O2)The quantity of the right side is the desired formation reaction2(C) (graphite) +3H2 C2H6

- 85kJ/molHesss Law(the procedure combining heats of several reactions to obtain the heat of desired reaction 36ExampleThe standard enthalpy of combustion of C2H6 (g) to CO2(g) and H2O(l) is -1559.8kJ/mol. Use this and Appendix data on H20 (l) and CO2(g) to find and of C2H6(g)

37SolutionCombustion means burning in oxygen. The combustion reaction for 1 mole of ethane isC2H6 (g) +7/2 O2 (g)2CO2 (g) +3H20(l)The relation gives for this combustion

Substitution of the values of of CO2 (g) and H20 (l) and gives at 298K -1559.8 kJ/mol = 2 (-393.51kJ/mol) +3 (-285.83kJ/mol)-(C2H6,g) -0 (C2H6, (g)= -84.7kJ/mol

38To find from we must write the formation reaction for C2H6 which is: 2(C) (graphite) +3H2 C2H6This reaction has = 1-3 =-2

= - 84.7 kj/mol (-2)90.008314 kj/molK)(298.1K) = -79.7 kj/mol

395. Calculation of Hid HreTo find of a gaseous compound or a compound formed from gaseous elements, we must calculated the difference between the standard-state ideal gas enthalpy and the enthalpy of the real gasLet Hre (T,P) be the enthalpy of the real gaseous substance at T and PLet Hid (T,P) is the enthalpy of hypothetical gas in which each molecule has the same structure as in the real gas.

40To find Hid Hre , we use the following hypothetical isothermal process at T:Real gas at P

real gas at 0 bar

ideal gas at 0 bar

ideal gas at P

(a)(b)(c)Wave a magic wand that eliminates intermolecular interactionReduce the pressureIsothermally increase the pressure

41The enthalpy change calculated from the integrated form of :

The quantity of Ure-Uid (both at the same T) is Uintermol ( intermolecular interaction to the internal energy)

Since, intermolecular interaction go to zero as P goes to zero in real gas Ure=Uid in the zero pressure limit. This situation also same for equation of state for the real gas approaches that for the ideal gas: (PV)re =(PV)id

42

Since H for an ideal gas is independent of pressure

This integral is evaluated using P-V-T data 43