Table of contents The module calculates the thermochemical properties of a species, a mixture of species or a chemical reaction. accesses only compound databases. assumes all gases are ideal and ignores expansivities and compressibilities of solids and liquids. Section 1 Section 2 Section 3 Section 4 Section 5 Section 6 Section 7 Section 8 Section 9 Section 10 Section 11 Table of Contents Opening the module Pure substance property calculations (pure Cu) Isothermal standard state reactions (oxidation of copper) Using non-standard states in equilibrium reaction Non-isothermal non-equilibrium calculation (heating of pure (Al)) Two phase single component equilibrium (ideal binary system) Variable input amounts in non-isothermal reaction Pidgeon Process for the Production of Magnesium Aqueous applications Complex gases and condensed substances under high pressure 1
The module Click on in the main FactSage window. 2
Reactants window - entry of a species (pure Cu) Add a Product has two windows Reactants and Table Add a Reactant Open can only access compounds (not solutions) New Entry of reactant species Compound databases available All calculations shown here use the FACT compound databases and are stored in FactSage - click on: File > Directories > Slide Show Examples Go to the Table window 3.1
Table window thermodynamic properties of a species Open New Save Stop Calculation Summary of the Reactants window A multiple entry for T: min, max and step. The results also display the calculated transition temperatures. 3.2 Return to the Reactants window
Determination of most stable phase by Gibbs energy minimization Phase with lowest Gibbs energy is the most stable. 1358.00 K 2846.16 K -71236.5 J -216494.2 J Points on the solid lines for P = 1 atm are given in column «T» and «G» of the previous figure for copper. 3.3
Simple isothermal standard state reaction: oxidation of copper Entry of an isothermal standard state reaction: 4 Cu + O 2 = 2 Cu 2 O This example is stored in FactSage. Go to the menu bar and click on: File > Directories > Slide Show Examples and select file 2. Isothermal T throughout Non standard states checkbox is not selected Go to the Table window 4.1
Oxidation of copper at various temperatures Entry: T min =300K, T max =2000K and step=300k. Note the transition temperatures. The equilibrium constant column appears for an isothermal standard state reaction. DGº = -RT lnk. For the values of the gas constant R, click on the Units menu. 4.2
Simple chemical equilibrium: non standard state oxidation of copper a Cu(s) = X P O2 (g) = P Select non standard state Standard state reaction: P O2 (g) = 1.0 atm a Cu(s) = 1.0 5.1
Specifying an extensive property change to deal with chemical equilibrium For simple chemical equilibrium: Table provides DG using: 2 a 0 C u 2O D G D G R T ln 4 a P C u O 2 0 1 D G D G R T ln 4 X P and K eq and DG 0 = -RT ln K eq X 4 1 P when DG = 0. For the last entry: P O2 (g) = 10-12 a Cu(s) = 1 DG = 0, equilibrium 5.2
Heating Al from 300 K to the temperature T phase transition, Al (s) Al (l) (i.e. fusion) at 933.45 K DH fusion = T fusion DS fusion = 28649.1J - 17938.1J = 10711.0 J The equilibrium constant is not displayed because this is a non-isothermal nonequilibrium calculation. 6.1
Heating Al: creating the graphical display with Figure 1. Click on the menu Figure and select Axes... 2. A dialog box opens and provides you with a choice of axes for the figure. Click on Refresh for the default settings 6.2
Heating Al: graphical display of thermodynamic properties 933 T 0 D H = C (solid) dt + ΔH + C (liquid) dt T p fusion p 300 933 liquid 17.9 kj o ΔH fusion =10711.0 J Point the mouse to read the coordinates of the melting point. solid DH = T T 300 C (solid) dt p 933 K 6.3
Computation of Cu liquidus in an ideal binary system: data entry Cu(solid) = Cu(liquid) Equilibrium of the type: Me(pure solid,t) = Me (liquid,a(me) = x(me),t) a Cu(solid) = 1, pure solid copper a Cu(liquid) = X Selection of phases: phases from the FACT database; 2 compound databases are included in the Data Search but here only FACT data are selected. 7.1
Computation of Cu liquidus in an ideal solution: mixing databases 2. FACT and SGPS compound databases are selected 1. Click on the Data Search menu. 3. Follow the instructions if you want to add (or remove - this does not delete) a database to (or from) your list. 7.2
Computation of Cu liquidus in an ideal binary system: tabular and graphical output Calculated activity of Cu(liquid) in equilibrium (DG=0) with pure Cu(solid) at various temperatures T. For an ideal solution: a Cu(liquid) = X Cu(liquid) Liquid Liquidus line The 2 specified variables, T and DG, are highlighted. Liquid + Solid Note: When DG = 0, the reaction must be isothermal. 7.3
Variable input amounts in non-isothermal reaction and autobalance feature The following example shows how a variable amount of a reactant can be used to simulate an excess of this substance, i.e. its appearance among both the reactants and the products. A simple combustion reaction: CH 4 + <A> O 2 = CO 2 + 2 H 2 O + <A-2> O 2 The Alpha variable, <A>, is used to define the quantity of O 2. 8.0
Combustion of CH 4 in variable amount <Alpha> O 2 data entry The reactants are at 298 K but the products are at an unspecified T. Variable quantity <Alpha> The phase of each species is specified. The reaction is non-isothermal (except when T = 298 K). Hence: K eq will not appear as a column in the Table window. Setting DG = 0 is meaningless ( except when T = 298 K). 8.1
Combustion of CH 4 in variable amount <Alpha> O 2 adiabatic reactions Stoichiometric reaction (A = 2): CH 4 + 2 O 2 = CO 2 + 2 H 2 O with variable amount O 2 (A > 2): CH 4 + (A) O 2 = CO 2 + 2 H 2 O + (A + excess) O 2 Exothermic reaction Product flame temperature As <A> increases, the flame temperature decreases. Energy is required to heat the excess O 2. Adiabatic reaction: DH = 0 8.2
Heating the products of the methane combustion. «auto-balance» feature 8.3
Step wise heat balance in treating methane combustion Different thermodynamic paths, same variation of extensive properties (here DH). 2000 K Overall Process DH = - 210211.4 J Warming Products DH = + 592107.1 J 298 K Isothermal Heat DH = - 802318.5 J 298 K 8.4
Pidgeon Process for the Production of Magnesium Apparatus Schema: Equilibrium Mg partial pressure developed at the hot end of the retort 1423 K Water-cooled vacuum connection also condenses alkalis MgO-SiO 2 phase diagram: Note: MgO(s) and SiO 2 (s) can not coexist they react to form (MgO) 2. SiO 2. 9.1
Pidgeon Process for the Production of Magnesium: Data Entry In the reaction, the hydrostatic pressure above the condensed phases: MgO(s), Si(s) and SiO 2 (s2) is 1 atm i.e. has no effect. The reaction is isothermal (same T throughout), hence DG = 0 gives equilibrium. The activity of SiO 2 is: a SiO2 = X. The partial pressure of Mg (g) is: P Mg(g) = P atm. Allotrope s2(solid-2) has been selected for SiO 2 in order to fully specify the phase if the phase is not completely specified, equilibrium calculations (DG = 0) can not be performed. 9.2
Equilibrium Mg partial pressure developed at the hot end of the retort Note: There are an infinite number of values of (P Mg(g), a SiO2 (s2)) which satisfy K eq. Here we select 3 special cases. Standard state reaction at 1423 K: P Mg eq = 1 atm and a SiO2 (s2) = 1 DG º = 221.39 kj = -RT ln K eq, hence K eq = 7.4723 10-9 DG º > 0 but Mg can be produced by reducing P Mg(g) and/or a SiO2. At equilibrium(dg=0), when a SiO2 (s2)=1.0 and T=1423 K, P Mg eq =8.6443 10-5 atm At equilibrium(dg=0), when P Mg eq =1atm and T=1423 K, a SiO2 (s2) eq=7.4723 10-9 When DG=0, T=1423 K and a SiO2 (s2)=0.006317 then P Mg(g) eq =1.0876 10-3 atm. This value of a SiO2 (s2) is taken from the next page calculation. 9.3
Computation of SiO 2 activity when MgO coexists with (MgO) 2 SiO 2 Pure MgO SiO 2 (s2) at activity X Pure (MgO) 2 SiO 2 Gives the equilibrium value of the activity of SiO 2 (s2) at 1423 K: a SiO2 (s2)=0.006317 Isothermal and at equilibrium 9.4
Alternative way to calculate equilibrium Mg partial pressure Remember, DG = 0 (equilibrium) calculations are only meaningful for isothermal reactions («T» throughout). Magnesium production is enhanced by: reducing the total pressure (<0.0010876 atm); reducing a SiO2 this is done automatically due to (MgO) 2 SiO 2 formation, but the addition of say CaO (slag formation) reduces a SiO2 further. 9.5
Aqueous applications hydrogen reduction of aqueous copper ion The molality of Cu 2+ is given by X. H 2 (g) pressure is P atm. Click on Units to change Temperature to Celsius. 10.1
Aqueous applications hydrogen reduction of aqueous copper ion Click on Output to change display to E(volts) and define n, the number of electrons Standard state reaction at 25 C Equilibrium molality at various P H2 Standard state reaction at various temperatures Eº=-DG /nf, where F (= 96485 C/mol) is the Faraday constant. 10.2
Thermal balance for leaching of zinc oxide The reaction is exothermic: DH < 0. This entry calculates the product temperature for an adiabatic reaction: DH = 0. 10.3
Complex gases and condensed substances under high pressure The following two slides show how is used on a system with polymer formation in the gas phase (Na(l) Na 1 + Na 2 ) and on a pure substance system that is submitted to very high pressure (C). 11.0
Computation of Na and Na 2 partial pressure in equilibrium with liquid Na Both reactions are isothermal, hence DG=0 gives equilibrium. Na (l) Na (g) (monomer) 2 Na (l) Na 2(g) (dimer) 1. Calculate T when P Na = 1 atm 2. Calculate P Na when T = 1158 K 3. Calculate P Na2 when T=1158 K Na also forms a gaseous dimer Na 2 (g). The proportion of Na 2 /Na near the boiling point (1171.8 K) of Na is: 0.111/0.888 at 1158 K; and the total vapor pressure over Na(l) would be: P Na + P Na2 (0.888 + 0.111 @ 1). 11.1
Effect of high pressure on the graphite to diamond transition Where available, density (i.e. molar volume) data for solids and liquids are employed in REACTION (the VdP term) although their effect only becomes significant at high pressures. (However, unlike EQUILIB, compressibility and expansivity data are NOT employed.) At 1000 K and 30597 atm, graphite and diamond are at equilibrium (DG=0) Here, carbon density data are employed to calculate the high pressure required to convert graphite to diamond at 1000 K. The volume of diamond is smaller than graphite. Hence, at high pressures, the VdP term creates a favorable negative contribution to the enthalpy change associated with the graphite diamond transition. 11.2