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Lecture 8: Thermo chemistry applications in metal extraction Contents Problem 1: Carbothermic reduction Problem 2: Heat of reaction in extraction of Zn Problem 3: Electric melting Problem 4: Heat of formation of compound Problem 5: Heat of formation of compound Problem 6: Production of SiC Problem7: Electric furnace to produce CaC Problem8: Hall Heroult cell Problem 9: Heat of formation Problem 1 : Carbothermic reduction Carbothermic reduction is commonly used to produce metals from oxides. Calculate the heat required (in Kcal/Kg of Zinc) to produce Zinc from reduction of ZnO with carbon.The reactants, ZnO and C, enter at 250°C , whereas Zinc vapor and CO gas leave at 1027 .use the following data: Zn S Zn 1 ; Tm 692.5K; ∆H 1740 cal /g. mole Zn 1 Zn g ; Tm 1180K; ∆HV 27565 cal /g. mole Components ZnO(S) C(S) Zn(S) CO(g) Zn(1) Zn(g) ∆H° (Cal/g.mol) ‐83800 0 0 ‐26420 ‐‐‐‐ ‐‐‐‐‐‐‐‐ Cp (Cal/g.mol. K) 11.71+(1.22 X 10 – 3 T) –(2.18 x 105 T ‐2) 0.026+(9.307X10‐3 T)‐ (0.354x105 T ‐2) 5.35+(2.40x10‐3T) 6.79+(0.98x10‐3T)‐(0.11x105 T‐2) 7.50 5.00 Solution: ZnO S C s Zn g CO g Reactants are at 298 K and products at 1300 K ∆HR K ∆H° ∑ H H p ∑ H H R 1 Reactants are at 298K; hence sensible heat in reactants is zero To calculate heat content in products, following scheme may be adopted Zn 298 Zn 692.5 ; sensible heat Zn 692.5K Zn 692.5 K ; Latent heat of fusion, Zn L , 692.5 Zn 1180 Zn v ; latent heat of evaporation Zn v 1180 K Zn l .1180 K sensible heat Zn v 1300K ; Sensible heat. cpdt Sensible heat in CO at 1300K = Heat of reaction at 298 K = 57380 Performing calculations on sensible heat of products by using Cp value and putting these values in equation 1 ∆HR K 101081 1555.1 Z Atomic weight of Zinc = 65 Problem 2: Heat of reaction in extraction of Calculate ∆H at 1250 K for the following reaction ZnS s CaO s Zn g CaS s CO g . Use the data of problem 1 for latent heats ∆H° (cal/g.mol) Cp (cal/g.mol K) CO(g) ‐26420 6.79+(0.98x10 ‐3T) – (0.11 x 10 5 T ‐2) Zn(1) Zn(g) ZnS(S) C(S) ‐‐‐‐‐‐ ‐‐‐‐‐‐ 0 ‐‐‐‐‐ 7.50 5.00 5.35 +(2.40 x 10 ‐3 T ) 4.10 + (1.02 x 10 ‐3 T) –(2.10 x 105T ‐2 ) ZnS(S) ‐48200 12.16+(1.24x10 ‐3T)‐(1.66 x 105 T ‐2) CaO ‐151600 11.86+(1.08 x 10 ‐3T)‐(1.66 x 105 T ‐2) CaS ‐110000 10.20+(3.8 x 10 ‐3 T) Components C s Solution : 63380 First calculate ∆H° . Now calculate sensible heat in products .The products are raised from a reference temperature 298 k to 1250 K .Use respective Cp values and integrate from 298 K to 1250 K. ∆H 55559 27733 Similarly ∆H ∆HR . 91206 . . Problem 3: Electric melting An electric melting furnace is used to melt copper scrap .The scrap is initially at 25°C .The overall power consumption is 300 kW ‐hr /ton of molten copper ,when heated to a temperature of 1523 K. Estimate the thermal efficiency of this furnace . melting point of copper 1356 K, Latent heat of melting =12970 J/g. mole Cp solid Cu 22.64 6.28 x 10 g 3T J Cp liquid Cu g. mole K 31.38 J g. mole K Solution: To estimate thermal efficiency,one has to calculate theoretical power consumption in heating scrap from 298 K to 1523 K. Power required = sensible heat required to raise scrap temperature from 298 K to 1356 K + Latent heat of melting + sensible heat to raise temperature of liquid copper from 1356 to 1523 K. Power required theoretically = 208.5 kw‐hr /ton of copper and thermal efficiency = 69.5% Problem 4: Heat of formation of compound Given the following heats formation (in cal /g.mole): Ca, Si, 2O 25200 Ca, Si, 3O 377900 Si, 2O 201000 Write each of these in the form of a thermo chemical equation .From these equations determine the heat of formation of CaO per kilogram of Ca. This type of problem is very useful in metal extraction .In cases when heat of formation of a compound is not known, then it can be determined from the heat of formation of different stepwise reactions leading to formation of compound. In the above problem one has to find heat of formation of CaO Let us form the Thermo chemical equations: SiO CaO Ca Si Si O 3O CaSiO ; ∆H° 25200 (1) CaSiO ; ∆H° 377900 (2) ∆H° 201000 (3) SiO ; Subtraction equation 1 and 3 from equation 2, w.ge 1 Ca O ∆H CaO 15700 cal⁄g. mol ° 3.8 x10 kcal⁄KgCa Problem of 5: Heat of formation of compound Given the following heats of formation Fe, O 1151cal/g of Fe. S, 3O 2930 cal/g of S . Fe, S, 4O 1456 cal/g of FeSO4. Required the molar heat of formation of FeSO4. from FeO and SO Now you can try this problem and get heat of formation of FeSO Problem 6: Production of silicon carbide Calculate the heat of the reaction for production of silicon carbide: SiO 3C SiC 2CO 63.1 The reactants and products are at 1973 K. Components ∆H° (J/g.mol) Cp (J/g.mol K) SiO2 841386 C Sic Co ‐‐‐‐‐‐‐‐ 117208 112352 46.945 + 34.309 x 10‐ 3 T 11.297 x 10 5 T – 2) 17.154 + 4.268 x10 109.266 + 14.67 x 10‐3T – 39.468 x 105T‐2) 28.409+4.1x10‐3 T 0.460 x 105T 2 ) Problem of this type are illustrated .Practice it to calculate heat of reaction 565.945 J Problem 7: Electric furnace to produce CaC An electric furnace is used to produce CaC as per the following reaction: CaO 3C CO CaC2 Power consumption is 4 kilowatt ‐hour /Kg of CaC .The CaC reacts with H O according to the reaction CaC 2H O Ca OH C H Given ∆H° Keal /Kg. mol CaC CaO 14600, and CO 26840 151600 Mean heat capacity of cold water =0.53cal/g° C Required: 1. The minimum Power to produce 1 ton of CaC per hour. 2. The electro – thermal energy efficiency of the furnace operation. 3. If 200 g. of CaC is treated with 20 kg. Of cold H O , calculate the rise in temperature? Calcium carbide is a very important regent and is used to desulphurize pig iron. You can try; the answers are: A) 2001.4 kwh/ton CaC B) 50% C) 4.3°C. Problem 8 Hall Heroult cell Calculate the heat of the reaction taking place in the Hall‐Heroult cell at a temperature of 1000 Al O 3C 2A l 3CO Although this temperature is below the melting point Al O , the alumina is actually dissolved in molten cryolite .Assume, therefore that heat content of the Al O includes its heat of fusion .Heat of fusion of Al O and Al is 1046.5 and 1004.6 J/g. mol S respectively .Melting point of Al is 659°C . Components Al2O3 ∆H° (J/g.mol) Cp (J/g.mol K) ‐380000 106.608 + 17.782 x10‐3T‐28.535x105T‐2 C Al(s) Al(1) ‐‐‐‐‐‐‐‐‐‐ ‐‐‐‐‐‐‐‐‐ 17.154+4.268x10‐3T‐8.786x105T‐2 20.669+12.385x10‐3T 31.798 CO ‐26840 28.409 + 4.1 x10‐3T – 0.460 x 105T ‐ 2 The above problem has a relevance in electrolytic reduction of Al O to produce Al. Hall‐ Heroult cell is widely used to produce aluminum. Heat of reaction 1241.911 kg/ mol. Problem 9: Heat of Formation Will heat of formation be different when a compound forms A) From two different compounds and (b) when if forms from pure elements. Answer; Heat of formation will be different .For example Heal of formation of CaCO when formed from elements Ca, C and O Heat of formation is 43450 kcal when CaCO forms from CaO and CO . 289500 kcal