Miss Noorulnajwa Diyana Yaacob 28 February 2011 Standard Thermodynamic Functions of Reaction.

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Miss Noorulnajwa Diyana Yaacob Email: noorulnajwa@unimap.edu.mynoorulnajwa@unimap.edu.my 28 February 2011 Standard Thermodynamic Functions of Reaction

Subtopics Standard states of pure substance Standard Enthalpy of Reaction Standard Enthalpy of Formation Determination of Standard Enthalpies of Formation and Reaction

Introduction Chemical 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 reaction

Standard state of Pure Substances For 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

For 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, T Gas P = 1 bar, T, gas ideal The standard-state pressure is denoted by P°:

Standard Enthalpy of Reaction Standard 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

The standard enthalpy change Stoichiometric numbers(- for reactant and + for products) Molar enthalpy in its standard state

For Example: The units = J/mol or cal/mol

Doubling the coefficients of a reaction doubles its = -572 kJ/mol = -286 kJ/mol

Standard Enthalpy of Formation The 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

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 307K The most stable form of carbon is graphite

For an element in its reference form, = 0 Summarizing: Stoichiometric number

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

Determination of Standard Enthalpies of Formation and Reaction 1. Measurement of  The 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:

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 2.We measure for mixing the pure elements at T and1 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 sumof 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

Figure 1: values

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(-285.830) +2 (-393.509) – (- 528.10) – 9/4 (0)] kJ/mol = -973.49 kJ/mol

2. Calorimetry  The most common type of reaction studied calorimetrically is combustion  Heat capacities also determined in a calorimeter  Reaction where some of the species are gases (combustion reaction)are studied in constant-volume calorimeter  Reaction not involving gases are 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 0 2.  Some values are plotted in Figure 2

Figure 2:Standard enthalpies of combustion at 25°C. The products are CO 2 (g) and H 2 0(l)

An adiabatic bomb calorimeter is used to measure heats of combustion. R+K at 25°CP+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 waterbath 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

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) = - 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

3. Relation between ∆H° and ∆U°  Calorimetric 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

Products minus reactants

The 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 applying For example: Neglecting the volume of the solid and liquid

The standard-state of a gas is an ideal gas, so For each gases C, D and B. Therefore, Thus ∆n g /mol

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

Solution: The formation reaction is: C (graphite) + ½ O 2 (g) + N 2 (g) +2H 2 (g) = CO(NH 2 ) 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

For reaction not involving gases, ∆n g is zero and ∆H° is essentially the same as ∆U°.

4. Hess’s Law  Suppose we want the standard enthalpy of formation of ethane gas at 25°C.  This is for 2C (graphite)+ 3H 2 (g) ⇨ C 2 H 6 (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!!!

We determine the heats of combustion of ethane, hydrogen, and graphite. The following values are found at 25°C C 2 H 6 (g) +7/2 O 2 (g) ⇨ 2CO 2 (g) +3H 2 0(l) ∆H° 298 = -1560 kJ/mol (1) C(graphite) +O 2 (g) ⇨ CO 2 (g) ∆H° 298 = -393.5 kJ/mol (2) H 2 (g) + ½ O 2 (g) ⇨ H 2 O(l) ∆H° 298 = -286 kJ/mol (3) Multiplying the definition by -1, 2, and 3 for reaction (1). (2), and(3), we get: -(-1560kJ/mol) = -2H° m (CO 2 ) - 3H° m (H 2 O) + H° m (C 2 H 6 ) + 3.5H° m (O 2 ) 2(-393.5kJ/mol) = 2H° m (CO 2 )-2 H° m (O 2 ) - 2 H° m (C) 3 (-286kJ/mol) = 3H° m (H 2 O) - 3H° m (H 2 ) – 1.5H° m (O 2 ) Addition of these equation gives: -85kJ/mol= H° m (C 2 H 6 )-2 H° m (C)-3H° m (O 2 ) The quantity of the right side is the desired formation reaction 2(C) (graphite) +3H 2 C 2 H 6 - 85kJ/mol Hess’s Law (the procedure combining heats of several reactions to obtain the heat of desired reaction

Example The standard enthalpy of combustion of C 2 H 6 (g) to CO 2 (g) and H 2 O(l) is -1559.8kJ/mol. Use this and Appendix data on H 2 0 (l) and CO 2 (g) to find and of C 2 H 6 (g)

Solution Combustion means burning in oxygen. The combustion reaction for 1 mole of ethane is C 2 H 6 (g) +7/2 O 2 (g) ⇨ 2CO 2 (g) +3H 2 0(l) The relation gives for this combustion Substitution of the values of of CO 2 (g) and H 2 0 (l) and gives at 298K -1559.8 kJ/mol = 2 (-393.51kJ/mol) +3 (-285.83kJ/mol)- (C 2 H 6,g) -0 (C2H6, (g)= -84.7kJ/mol

To find from we must write the formation reaction for C 2 H 6 which is: 2(C) (graphite) +3H 2 C 2 H 6 This reaction has = 1-3 =-2 = - 84.7 kj/mol – (-2)90.008314 kj/molK)(298.1K) = -79.7 kj/mol

5. Calculation of H id – H re To 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 gas Let H re (T,P°) be the enthalpy of the real gaseous substance at T and P° Let H id (T,P°) is the enthalpy of hypothetical gas in which each molecule has the same structure as in the real gas.

To find H id – H re, 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 interaction Reduce the pressure Isothermally increase the pressure

The enthalpy change calculated from the integrated form of : The quantity of U re -U id (both at the same T) is U intermol ( intermolecular interaction to the internal energy) Since, intermolecular interaction go to zero as P goes to zero in real gas U re= U id 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

Since H for an ideal gas is independent of pressure This integral is evaluated using P-V-T data

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