FactSage Independent Study

FactSage Independent Study FactSage Independent Study 1 Question 1 The following amount of slag granules (298K) is melted in an induction furnace in a graphite crucible 112g SiO 2, 360g FeO, 100g CaO, 550g PbO, 100g Cu 2 O. What is the composition of the system at 1400K? 1.1 Define the Reactants Units - Make sure that Mass is in grams and Temperature is in Kelvin. The graphite (C) mass was not informed» Variable Amounts <A> Data Search Database will be provided on the exam FactPs, FTOxide, FTmisc 1.2 Compound Species Select All Gases, Pure Liquids and Pure Solids (given) Solution Species: select FTmisc CuLQ, FTOxide SlagA, FTOxid SpinA (given) After you select the solution species, FactSage will decide for you if you get a + or an I. Just trust her. Make sure that <A> = 0 in this first step (we are not checking this yet). 1.3 Calculate The composition of the system at 1400 K will be shown on the window. You will also get 0 moles of a bunch of things. Make sure to find all of them! KAREN PESSE 1

2 Question 2 FactSage Independent Study Is there an interaction possible with the graphite crucible and the molten slag (assume T crucible = 1400K)? Explain. To discover this information, we are going to activate <A> in the second window to analyse the influence of Carbon together with the molten slag, varying from 0g to 100g with steps of 10g. Taking this information out of the table would take too long so we are going to plot a graph J. ü Calculate -> Output -> Plot -> Plot Result: ü On the Results Processor (blue window), Select Species. ü On the Species Selection, select Pure Solids and Solutions (except Gas) which are higher than zero. This just means that they are there (obviously). There are no Pure Liquids in this selection (they are all zero). ü You can plot this graph in gram or moles (as long as you know how to interpret). ü Obs.: If you are asked for Elemental Composition, then you must select Elements instead of Pure Solids, Solutions and Pure Liquids. Press OK and let s define the axis: v On the Y axis you can put the amount of species in grams (quite useful). You will vary as from 0g until the amount you want. You will later see in the graph that none of them grow more than 600, and then you can refine your graph to this value. v On the X axis, put the Alpha (amount of Carbon that you give). If you give 0 100 10, nothing more obvious than use the same parameter (or not, f* the system). KAREN PESSE 2

v And here is the graph we have to analyse: FactSage Independent Study Amount that your crucible dissolves Amount of Carbon added v Copper is a noble material, and it will be the first to be removed from the slag (red line), as you move forward adding Carbon. v As the copper oxide reduces, the Slag decreases, because the slag is a mixture of oxides that also contain copper oxides, and when they reduce, they leave. v This is sort of an interpretation (the amount of crucible that dissolves), but you can also have reactions in which you add reductive agents, for example Carbon, and you would get similar results. v On the meantime, the Iron Oxide (FeO) also reduces, affecting the Spinel line (Black - SPINA) which is FeO combined with Fe 2 O 3. FeO will be converted into Fe 2 O 3 and then react again to form Fe 3 O 4, increasing the black line (10g). Eventually Fe 3 O 4 will stop being produced, decreasing the line of Spinel together with its own line. v When the FeO reacts with Fe2O3, both in the Slag, the amount of Slag decreases, and Spinel is formed (so spinel increases) at 10g. This is further reduced (20g-30g), and then iron oxide goes again to the slag, so the spinel reduces (20g) but the slag increases again. KAREN PESSE 3

FactSage Independent Study 3 Question 3 Ignore the interaction with the crucible: What is the influence of the partial pressure of oxygen on the system? v First step: lets ignore the interaction with the crucible: Edit -> Delete Reactant C Right click on the cross of Gas to get a window of selection. Right click on the cross of O 2 -> Activity -> log10 (activity). Define the Fixed Partial Pressure: Range of values from -10 to -1. 1 After defining the range of partial pressure that you want to analyze -> Calculate -> Output -> Plot -> Plot Results -> Species (Select). Select the relevant Species (Pure Solids and Solutions) by selecting the ones that have a relevant mass. A very important step in this part is to select the species number of which species you want to be in your X axis (the red sun) which in this case was #7, but can be another number. v To select the Y axis, you can choose mole or grams (although mole is recommendable) and by try and error, define the Y axis. If you put between 0g and 1500g you will see that as from 1200g there is nothing else interesting. If you put 0-15 mole, you will see that there is nothing relevant after 10 moles. The same works for the X axis. The decision of log 10 (activity) is intuitive, since we want to check the activity of O 2 in the system. So, let s analyze the graph J 1 The normal oxygen content of the air is 21% of 1 atm. Thus, at a normal partial pressure, oxygen cannot exceed 0.21 atm, which is the order of 10-1 atm (hence the upper limit) when you keep working with a total partial pressure of 1 atm (as we usually do in pyrometallurgy). The lower limit 10-10 is just a random very low number that means extremely reducing conditions. Lower than this is practically not reachable. So, in between 10-1 and 10-10 you have a complete range of partial pressure of oxygen that are practically feasible. KAREN PESSE 4

FactSage Independent Study v So, the influence of the partial pressure of the Oxygen on the system will be that if you work oxidative (from left to right adding O 2 to the system) then the copper metal is oxidized to form copper oxide (which is dissolved in the slag). Other elements will also be oxidized and dissolved in the slag hence the slag increases. KAREN PESSE 5

FactSage Independent Study FactSage Independent Study Part 2 1 Question 4 Copper Scrap Consider copper scrap with the following composition: Cu 76, Pb 2, Fe 20, Zn 1, Ni 1 (unit = wt%). The refinement happens in an ISA-Smelter, using cokes (C) and CH 4 as fuel, excess of air to work oxidizing, silica (SiO 2) as flux 1, volatile compounds are CH 4, and we work with pure silica (?). Initial Temperature 25 C. Let s first see what happens if we add an insufficient amount of fuel: 0,1g CH 4 and 0,1g C. Determine the mass of SiO 2 required to form 2FeO.SiO 2 (Fayalite), and the amount of the variable amount <A> of N 2. The units were specified to be grams, degrees Celsius and Joules. The specified amount of Fe is 20g. Thus, if we have 2Fe in an equation and 1 SiO 2, you can find the amount of mols of iron and compare to the mols of iron oxide and then find the amount of moles of silica. T he mass of SiO 2 will be 10.75 grams: 2Fe + SiO 2 + O 2 FeO. SiO 2 2Fe SiO 2 2 55,85 28,08 + (2 16) 20 grams x [28,08 + (2 16)] 20 = 10,75g 2 55,85 Take a periodic table with you to the exam just in case ; -) To calculate the variable amount of <?A> N2, you have to know that air is made of 79%N 2 and 21%O 2. You can find the third term with the equation 2 : [( 79 2 14) 1] 100 = 3,29 21 100 2 16 1 Silica as a flux you want to be able to gather the oxides that you go t from your impurities into one single phase: you will blow air through this liquid metal phase and impurities such as iron are oxidized (so now you are looking at the other way - before we were looking at reducing oxides and now we will oxidize metals). These iron oxides cannot exist on its own, that is why you need t o add a flux (seen in exercises of phase diagrams). It is necessary to have a certain liquidus temperature, and the flux will be lowering this temperature so you can work with the slag and not just have solid Iron Oxide (somewhere appearing in your reactor). 2 This number one comes from the statement of <1A> O2. You will vary the amount of Oxygen, and the amount of Nitrogen is linked to this amount because of the composition of air. Number 2 is related to 2 moles of N 2, and 2 moles of O 2. You are working with masses (in grams) thus you first need to convert the masses to moles and then use the composition of air ; -) KAREN PESSE 1

2 Question 5 FactSage Independent Study Calculating the value of <A>: To oxidize 20g of Fe, how much do we need of O2? 20g Fe? g O 2 2 Fe + O 2 2FeO 2 56 16 2 20g x [ 2 56 20 ] = [16 2 ] x = 5,7 6 x - We had to estimate how much of oxygen we could add so you have an idea of in which order of magnitude the oxygen is (and not start with 100 for example). 3 Question 6 - Calculate the Equilibrium (on the exam it will be clear what it is expected which phases are present, what are their compositions, ΔH of the system and the spinel activity) We want to purify the copper: you will see that there are two phases copper liquid and slag, and you want iron and silica to be in the slag. i. At Menu - Products compound species right click on gas to open the Selection window. Clear all the species and select only the ones pre-determined on the slide (H 2, N 2, O 2, H 2O, CO, CO 2, Zn, Pb). ii. iii. At Menu Products Compound Species right click on pure liquids and select only SiO2(s) and Fe3O4(s) (given in the slides). Do the same for pure solids. Calculate: 94,7% of copper is a lot more than we had to start from. The 1,5%w Fe is also a lot less than what you started from = succeeded in purifying the copper! Watch out: too much oxygen > copper will be oxidized as well > start to use copper as fuel and copper will end up in the slag! The slag is usually discarded afterwards so if it contains a lot of copper that is a waste. Even though the amount of Fe is much lower than in the beginning (20 wt% to 1,5 wt%), you still have Zn and Pb and Ni that are impurities. Even though the amount of spinel is 0 grams formed, the system is very close to forming it because the activity is as almost one (0,028); Viscosity needs to be kept minimal = spinel cannot be too large -> keep track of activity for spinel. KAREN PESSE 2

4 Question 7 FactSage Independent Study Required amount of fuel? What is the effect of an increase in amount of oxygen <A>? Vary <A>: 6 8 0.5 You have an impure copper scrap that you re -melt, and want to get a copper phase as pure as possible. i. Define <A> between 6 and 8, ticking every 0.5 and calculate. ii. Select Species: Select relevant Species with gram and wt% different than zero; We want to see how O 2 influences the reduction of copper, thus we select the composition of elements in the copper liquid to see on the graph if they will increase or decrease with increase of O 2. It is also interesting to select the pure Copper in the slag (ELEMENTS). iii. Define axis: analysing the activity x gram the program often gives you weird maximum and minimum value, and then you need to adjust by trial and error. RESULTS: At <A>=6 the composition of copper liquid is of 94 wt% of copper, but 1.5% of Iron which is good when you started with 20 wt%. It is already a pretty good refinement (not ideal but it is ok). At <A>=8 we have 98 wt% of Copper, which is good because it means we are purifying. Thus; the amounts of impurities decrease as you increase the amount o f O 2. KAREN PESSE 3

FactSage Independent Study The heat production (heat that is required) to your system even for <A> = 8, the h is positive: A = 6 A = 8 Reduction Oxidation So if you will not add heat to your reactor, it will cool down (not advisable everything solidifies) you need to add carbon, extra gas and oxygen. If A increases, the amount of Cu in the slag increases (unwanted). The ΔH is still positive, so the reaction takes place but absorbs energy from the surroundings Is the use of more air an option? Nop! If you use more air, you will reduce ΔH. However, you will also increase the usage of Copper as fuel (oxidize the copper), which we do not want. If A increases the amount of Cu in the slag increases (unwanted). For A=8 we had a high spinel activity dangerous. We have to keep that in mind. 3 We also have 10 wt% copper oxide in the slag Still not enough heat. You need to add heat keep the correct working temperature. (a=0.51), which is big loss. to your system to If you use more air you will only increase the copper a mount in the slag; Maybe in the end the system will become exothermic but that is not the purpose of this reactor you want to keep the copper in the metallic phase and not oxidize it and use it as a fuel. 3 When solid particles (and spinel particles are solid) start to form (and a phase starts to form whenever the activity becomes 1), the viscosity of the slag increases. When the viscosity of the slag in the reactor becomes too high, it might become very difficult (or almost impossible) to work wi th the slag (for example tapping the slag). KAREN PESSE 4

5 Question 8 FactSage Independent Study Burn Coal for Heating Generation Choose <A> = 7 (for O2) fixed for the oxidation of impurities But to burn C also O 2 required Per gram C, we need 2.66g O 2 needed (reacts to form CO 2) 4 Choose variable amount <A> for C and <2.66A + 7> for O 2. 5 Also adjust the amount of N 2: <8.774A+23.03>. Run this case for <A> = 2 5 1 Conclusion? You want to get an idea of how much oxygen or air is needed to oxidize the main impurity Fe to FeO, and this will bind with the Silica you also added. We have 20 grams of Iron, and for that to be oxidized to FeO you need 5.7 grams of oxygen (which we approximated and said it is 6g) and then calculate the equilibrium for this variable amount A = 6. - To burn C you need O 2. Per gram of C, you will need (a periodic table): C + O 2 CO 2 12 16 2 1 gram C x x = 2,66g O 2 This amount multiplied by 3.29 which is the ratio in the air, gives us N 2: <8.774A+23.03>. The point of adding this fuel was having enough heat generated by this burning so that the ΔH would go from positive to negative at a certain point. If we vary the amount of Carbon that is added as fuel to be burned, ΔH becomes zero between the amount of Carbon added 3 and 4 (A=3 and A=4) and that is what we want to know here (where we added the Carbon). 4 You cannot forget the number 7 First you have the pre-factor that you need to multiply with the variable amount of carbon for the stoichiometric number of Oxygen (2.66). The 7 does not depend on the variable amount of coal that is added, but is the amount of Oxygen that is required for the oxidation of the impurities <A> = 7 (for the refinement). 2.66 BURNING OF THE COAL; 7 REFINEMENT. 5 An then because we work with just air, we also need to calculate the correct numbers for the NITROGEN. KAREN PESSE 5

FactSage Independent Study We can also check on the results tab the two A values that we still have limited copper losses decent refinement and less tendency for spinel formation cause the activity will be lower. A = 4: 6.3% of copper loss in the slag; spinel activity a=0.39775; ΔH negative A = 5: 6.3% of copper loss in the slag; spinel activity a= 0.41; ΔH negative Now if you calculate for A = 4 and you are sure that ΔH is negative (because you do have some heat losses also so is best to go for a negative ΔH), and you have 100 ton of feed (industrial lance) you would need 60k cubic meters of air (a lot)! You can calculate this number by saying that you have started working with 100g (which was your initial reactants amount). From that you can calculate the number of moles of air from the grams that you have, from the grams of carbon; with a stoichiometric number that you calculated you can calculate the number of grams of oxygen, you can calculate the number of moles and assume ideal gas 1 mol = 22.4L and one cubic meter = 1000L; 100ton =100g*10 6 -> 60K. 6 Excess of oxygen Enriching the air with oxygen (more than 20% O 2), makes it possible to reduce the amount of gas and fuel needed for the refinement. <A> is again the amount of added O 2 Run the case for <A>=7, first with 0.1g C and CH 4 Investigate an oxygen enrichment of 50vol%O 2-50vol%N 2: <A> O 2 and <0.875A> N 2 It is sort of expensive to get pure air in an industry and you need a very large amount, so what can we do about it? Work with enriched air, so the amount of oxygen will be more than the 21% that you usually have. Let s consider no extra addition underestimation of the amount of fuel that is added and check how the influence of the amount of adding Oxygen here is. Note that for this 50-50 volume percent mixture the prefactor for Nitrogen will change because previously you had 21-79 volume ratio between the two and now you have 50-50. In this scenario we have air with an unusual composition: 50vol% O 2 and 50vol% N 2. Once again assuming these are ideal gasses, we equate vol% to mol%. MM(O 2) = 32g/mol and MM(N 2) = 28g/mol. 1 2 mol O 2 32 g mol = 16g O 2 1 2 mol N 2 28 g mol = 14g N 2 14g 16g = 0,875 So, mass-wise, we need 0,875 times as much N 2 as we do O 2. KAREN PESSE 6

RESULTS FactSage Independent Study Without fuel The amount of oxygen that was needed for the refinement is still the same because you still want to convert your iron in the copper metallic phase towards iron oxide, so the amount of oxygen required remains the same, but we see that you have a lower spinel activity and still a very decent refinement; however the reaction is not very exothermic. 7 Excess of oxygen + Heat Run now with <A> C and 7g of O 2 for the refinement We are going to vary the amount of C again, while keeping CH 4 constant. This means we have to take the production of CO 2 into account again. As we ve calculated before, <A> grams of C means we require <2,667A> grams of O 2. This gives us a very similar exercise as last time, all we have to change in the reagents is that for N 2 we ll multiply by 0,875 instead of 3,29. This gives us an amount of O 2 equal to <2,667A+7> grams and amount of N 2 equal to <2,34A+6,125> grams. Thus <2.667A+7> O 2 and <2.34A+6.125> N 2 Calculate for <A> = 0.4 1.2 0.2 CONCLUSION If you now add fuel, change the pre-factors according to this 50-50 ratio of mixture, and also vary the amount of Carbon, you will see that the heat is balanced much faster. Previously we would have to add 3 to 4 grams of coal and burn it, now you have a balance situation before 1 gram of coal is added. This is 4 time less coal than before. Also, less than half of the amount of oxygen that you needed previously will be required. Then 100 ton of feed requires only 13 thousand cubic meters (which is also 4 times less) in this scenario. Just by enriching the air with some oxygen you can easily reduce the amount of reactants that are required! Activity of spinel is low Only 5% Cu 2O in the slag Copper Liquid 97% pure KAREN PESSE 7

12/15/2016 Independent Study 2 - FactSage 1 Question 1 - Equilib Is the composition of the slag at high temperature (1200 C) different than that at 25 C? Explain your answer. The first part is all given, you just have to enter everything correctly in the Reactants Window. Make sure that you are selecting the correct Units (gram, Celsius and atm were the units given), and Database (FactPS, FToxid and FTmisc) at Data Search. Select gas species and all pure liquids. Do NOT select pure solids (total of 44 species). Select solution species: Base-Phase FToxid-SLAGA, FToxid-SPINA, FTmisc- PbLQ and FTmisc-CuLQ. Let Factsage decide if it will be + or I. Click Show and Selected for better display. Some information: Quenched slags look black and glassy. At high temperatures the slag is liquid; then a cold sampling bar is put in the slag to quench the slag that sticks onto it. When you quench something from a very high temperature very fast, you freeze the high temperature microstructures. In that way, you look at the microstructure as it was at a high temperature. can Also, the fact that there are solid particles in the slag (which is liquid at high temperature) means that your slag is being operated at a temperature beneath the liquidus t emperature (because liquidus temperature means that on the temperature above that liquidus temperature everything is liquid so there would not be spinel particles). The liquidus temperature is sometimes so high that they opt to work underneath it and thus work with spinel particles, otherwise the machines would deteriorate). The influence of the solids in the slag is clear on the increase viscosity though; Back to the exercise: The Temperature you want to work with is 25 C and 1200 C; Karen Louise de Sousa Pesse 1

12/15/2016 At higher temperatures we can find two phases so you have the slag phase with one composition and the liquid phase with another com position: You can see that there is a very limited amount of copper in the slag; we had 250 grams of copper oxide in the beginning (quite a lot the weight percentage initially was very high); We can see that most of the copper goes towards the copper liquid metallic phase; and so the amount of copper oxide in the slag is reduced at higher temperature. This happens because it is the most unstable oxide (most noble metal that is present in the system) and this can be seen in the Ellingham diagram. Karen Louise de Sousa Pesse 2

12/15/2016 2 Question 2 - Equilib What is the influence of the partial pressure of oxygen on the composition of the slag system and other potentially present phas es? Use plots of the ELEMENTAL composition for this. Vary the partial pressure of oxygen from 10-10 to 10-1. Explain this behavior. The first step (that was already explained in a previous pdf) is the following: Right click on the cross of gas (Menu Products Compound Species) A new window will appear (called Selection Equilib) with several gases. Right click on the cross of O2 Activity log10 (activity). Define the partial pressure ranging values between -10 to - 1. 1 Click OK Afterwards, let s analyze our results. Calculate Output Plot Plot Results Species (Select). ELEMENTS. It is very clear on the question that it is required the ELEMENTAL composition, thus let s separate the 1 The normal oxygen content of the air is 21% of 1 atm. Thus, at a normal partial pressure, oxygen cannot exceed 0.21 atm, which is the order of 10-1 atm (hence the upper limit) when you keep working with a total partial pressure of 1 atm (as we usually do in pyrometallurgy). The lower limit 10-10 is just a random very low number that means extremely reducing conditions. Lower than this is practically not reachable. So, in between 10-1 and 10-10 you have a complete range of partial pressure of oxygen that are practically feasible. Karen Louise de Sousa Pesse 3

12/15/2016 Select the relevant species (not Pure Solids and solutions but ELEMENTS because it is asked so) by selecting the ones that have a relevant mass (gram and wt%) different than zero: In this part there is a very important step, which I call the RED SUN just so you can remember. The red sun is when you have to define the number (#) that will be equivalent to the partial pressure of the GAS YOU WANT TO ANALYZE (which here is O 2). This number can be 1, 2, 7 or whatever other number FactSage decides to give you, you have to look up on this table. In some cases, the RED SUN will not appear to you this means that your Graph axis does not include the log10 of activity maybe you are plotting Grams x Temperature, I don t know). You can either make a plot of the different phases (one phase elemental composition and then another plot of the other phases that are present) because if you have multiple phases and multiple elements one plot can become very heavy and difficult to interpret the figures; Karen Louise de Sousa Pesse 4

12/15/2016 Everything is plotted; graph is heavy and difficult to interpret. REDUCTIVE OXIDATIVE Look from right to left (oxidative to reductive) when interpreting your graphs. For the slag phase you see a variation in the amount of iron and oxygen, etc., increase of lead and increase of copper (which is logical according to the Ellingham diagram because copper is more noble than lead, and lead is more noble than iron, and so on) All of a sudden it appears as if there is an increase in element (Pb for example), which is not necessarily possible but is NOT an increase in the total amount of lead but in the weigh percent of lead (it was supposed to appear but Factsage gave a horrible white color). This means that the amount of slag phase increased, so the ratio of Lead towards the full amount of slag will increase because the full amount of slag will decrease. If you get stupid colors, change them. Karen Louise de Sousa Pesse 5

12/15/2016 REDUCTIVE OXIDATIVE Graph for the other phases that are present, mostly the copper liquid phase. You see that the copper phase only starts forming from one point on (looking from oxidative to reductive, when the log of activity is -05) and you can link this to the point in which copper is starting to decrease from the slag phase on the other graph. The amount of lead increases on the copper liquid from that same point on. If you look at the absolute amount (grams) you can link it better to graph that you obtained. Karen Louise de Sousa Pesse 6

Independent Study B - FactSage 1 Question 3 We want to reduce this slag to obtain a liquid Cu -Pb alloy with a 50wt% - 50wt% composition. What are the industrially most used reductive compounds? (3 non-metallic species) Carbon, Carbon Monoxide gas and Hydrogen gas; A. Make a list of the reduction reactions of Cu 2O and PbO with these reductive agents at 1200 C. Give the enthalpy of these reactions and the Gibbs free energy change. Use only the FACT database and assume standard states. Present all these data in an organized table with all the reactions (also specify the state/phase under which the different reagents and products are encountered at this temperature) and the calculated thermodynamic data. a) Which reduction reaction has the highest driving force? So, the most negative ΔG; this will be the tricky question of the exam in which you have to reason about what she is asking; For example: Which is the most exothermic reaction? Most negative ΔH. If you need the Cp of any elements, add the Species on the Reactants window double click on it to see all the specifications. b) What is the difference / resemblance between the three reductive agents? (Also consider the practical applicability regarding the addition of the reagents for this. Also think about the influence on the retention of the temperature in the reactor.) You are looking at a certain reaction - and not just entering your reactants and giving the final circumstances to see what actually forms - you are interested in a specific reaction for which you want the ΔG and ΔH so you use the reaction module. Specify in the table which phase or state is encountered in that temperature (1200 C) whether the oxide or the metal is solid or liquid, and also the thermodynamical data (ΔG and ΔH); FactSage gives all information for you, even the balance of the equation. Karen Louise de Sousa Pesse 1

H (J) G (J) PbO (l) + ½ C (s) Pb (l) + ½ CO 2 (g) -13 227.6-117 114.4 PbO (l) + CO (g) Pb (l) + CO 2 (g) -95 988.5-73 977.9 PbO (l) + H 2 (g) Pb(l) + H 2O (g) -65 596.0-84 893.7 Cu 2O (s) + ½ C (s) 2Cu (l) + CO 2 (g) -7 369.1-138 314.2 Cu 2O (s) + CO (g) 2Cu (l) + CO 2 (g) -90 130.0-95 177.6 Cu 2O (s) + H 2 (g) 2Cu (l) + H 2O (g) -59 737.5-106 093.5 Making this table is simple: you know you have 3 reductive compounds and two elements you want to reduce. Combine them and let Factsage calculate the equilibrium in case you don t know. After filling in your reaction, you get a blank window, then press calculate to get the results. Results Depending on how many moles of a certain element you inserted, you might get different numbers and you have to be careful about that. For 1 C + 2 Cu 2O you get twice higher values like H, you just have to make sure to compare everything double, especially if you are using the automatic balance from FactSage. You can see on the blue line the most stable state of your element (solid; liquid or gas) at that specific temperature of 1200 C. In the ΔG and ΔH values you can see that: copper oxide reduction with carbon has the most negative ΔG so has the highest driving force. The fact that is negative does not mean that is the lowest, but that is spontaneous. ΔG Cu < ΔG Pb for each reducing agent. The ΔG for the reduction of copper oxide is always smaller than the one for ΔG for lead oxide reduction, this is also in accordance with the Ellingham diagram, that the copper oxide will more easily be reduced by any reducing agent because is more noble. For Co you have the most heat release. Is so much heat that is released that you might need cooling as well but you don t need extra heating as encountered in our previous exercise, you don t have to add carbon or extra oxygen for that. Carbon is solid and CO and H 2 are gases. Add something solid to a liquid is not ideal, because the reaction is very slow. If you have gasses, you can apply via lance in your molten bath, and Karen Louise de Sousa Pesse 2

the molten bath will be mixed so you have a lot of reactions taking place very fast; However, CO and H 2 are dangerous because CO is toxic and H 2 can be explosive. Remember being on Holidays? Oh, I member Karen Louise de Sousa Pesse 3

FactSage Independent Study B Question 3 Question 3 - Equilib 1. We want to obtain a liquid Cu-alloy with a composition of approximately 50wt% Cu and 50wt% Pb at 1200 C from the initial slag by addition of the reductive agents at 40 C. How much of each reactant is required for this? How much heat is released for each reactant? (leave p O2 unspecified) Don t forget that on the last exercise you were working only with FactPS database. Here we must go back to the initial settings (FactPS, FToxid and FTmisc). Add a second stream 1 at 40 C, and do this for the three reductive agents separately. I will do first for Carbon. Add <100+A> if <A> does not work 2. Work with 0 to 100 with steps of 10 0 100 10 <A>. Calculate, get your results and Plot a graph: Plot the Elemental Composition of your Copper liquid, because the question is how much reactant we need for this?. The graph should be (Alpha, wt%), so you can see how many grams of A you need for 50%A. To select the species, you need to have an insight of what we want to see. To know how much of Carbon is needed to make an alloy 50% Cu and 50%Pb, you know that Carbon is our Alpha variable, so this should be in your graph; you also know that you want percentages (wt%); you don t want to know about gas or slag, but the liquid copper alloy (CuLQ), so you can either select on the FTmisc section (which means miscellaneous by the way) or Elements section and Pb_CuLQ + Cu_CuLQ which is the exact same stuff. 1 All reactants in a given stream are grouped together. They have the same temperature and total pressure. If you change the temperature or pressure of any one member of the stream, then the temperature or pressure of all the other members of the stream will be changed to the same common value. 2 If you just add A you might get an error that states that A should be between 0 and 1. KAREN LOUISE DE SOUSA PESSE 1

Percentage of Copper and Lead For Carbon we need approximately 15 grams, as we can see in the graph, thus we sho uld calculate again for A = 15 The composition of alloy will then be: 48.8 wt% Lead and 51 wt% Copper, H = 8*10 5 J (don t forget that you also want how much heat is released ). This is an endothermic reaction, thus not spontaneous, and would need some external heat to happen. LESS Carbon MORE For the reductive agent Carbon Monoxide (CO): Around 75 grams of CO are needed. When A = 75g, we have 51,5 wt% Pb and 48% Cu. H = 8.3 * 10 5 Why is H positive: you added the reductive agents at 40 C, thus the amount of energy that you got from the reductive reaction (from reducing the oxide copper and oxide lead) is not enough to heat up your reductive agent from 40 C. For the reductive agent Hydrogen gas (H 2): H 2 = 5.25g H = 9.1 * 10 5 J Cu = 49.9 wt% Pb = 49.9 wt% KAREN LOUISE DE SOUSA PESSE 2

2. Which reductants produces the purest 50-50 Cu-Pb alloy? If you list these things up, you see that the carbon give s more pure alloy but the difference in between the 3 is very small and this is because the carbon has the biggest driving force as was shown in the previous question. To know the exact amount, you can pass your cursor on the intersection of the graph, and under the graph you can see your x and y axis. C CO H 2 49.919487wt% 49.913846wt% 49.917538wt% EXTRA INFORMATION Selecting Base-Phases notice that in the window Menu you have to select Solution Species. This should not be a random selection, and to be aware of what you are doing here is a small explanation. Let s say you selected FTmisc-PbLQ and FTmisc-CuLQ, which stands for FTmisc database (Miscellaneous) and the PHASE information of Pb or Cu liquid. This will then give you the results for Copper Liquid and Lead Liquid: KAREN LOUISE DE SOUSA PESSE 3