Week 2-Lecture 5 : Ellingham Diagram — Transcript

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1. 0:17We are going to get into something high temperature chemical reduction ok.
2. 0:22This is where Ellingham Diagram is coming.
3. 0:27Now, of course every metal oxide and metal sulfide can be heated to some really high
4. 0:38temperature, extremely high temperature and you can break it down to its corresponding
5. 0:45metal and let say oxygen right, but that is not going to be feasible.
6. 0:51See, if you want to heat something at 3000 degree centigrade, the cost for that is going
7. 0:59to be a lot.
8. 1:00It is not just you know you have to have that facility in a industrial scale, it is going
9. 1:06to cost a lot you know at the end you may get 1 rupees of metal and you ended up spending
10. 1:13let us say 1000 rupees.
11. 1:15So, it is not going to be practically possible to just heat anything any metal compounds
12. 1:23at any given temperature although you know that if you can heat it at a suitable temperature,
13. 1:28you can break it.
14. 1:29It is possible theoretically, but practically you are never going to do this.
15. 1:34So, this is when we need to understand the thermodynamics.
16. 1:40What can you do what type of reaction can you do and Net result you are going to get
17. 1:47exactly the same metal in reduced form, but without heating at a too high temperature.
18. 1:52You want to heat it at a moderate temperature instead of 3000 degree centigrade; let us
19. 1:57say you want to heat it at 1000 degree centigrade which is much more achievable and you can
20. 2:05get exactly same pure metal.
21. 2:08What you need to use some sort of reducing agent, when or how you determine that, that
22. 2:15is what we are going to discuss.
23. 2:16So, most of the metal in ore is present in metal oxide form ok.
24. 2:23Some of them are present in metal sulfide or metal halide form.
25. 2:29Usually what happens is we try to convert metal sulfides to their corresponding metal
26. 2:36oxide because that way thermodynamic parameter works out better.
27. 2:41So, if you understand how metal oxides are converted to their corresponding metal, you
28. 2:48will understand how metal sulfides are converted to their corresponding metal because metal
29. 2:54sulfides are converted first to metal oxide and then metal oxide is going to converted
30. 3:01to corresponding metal it is a two step process for metal sulfide.
31. 3:05For metal oxide, it could be one step process; for metal sulfide, it is usually two steps
32. 3:10process ok.
33. 3:11So, that is why we will just mainly try to focus on metal oxides ok.
34. 3:18All the details are given, you can read.
35. 3:21Now, since we are talking that we need to heat it at a very high temperatures something
36. 3:27like 1000 degree centigrade at least or 500 to 700 all those temperature.
37. 3:33So, these reactions are going to be spontaneous.
38. 3:37These are going to be a very fast reaction.
39. 3:42Since we are talking about really very high temperature, the reactions are going to be
40. 3:47and we are looking for a spontaneous reaction, reactions are going to be fast.
41. 3:51So, kinetic parameter how fast it is, it is something we do not need to worry because
42. 3:58it is going to be very fast, it is not a slow process it is not going to take days ok; it
43. 4:03will be converted very fast.
44. 4:05Only thing we really need to worry about is the thermodynamics whether it is a favorable
45. 4:10reaction or not.
46. 4:12What is a favorable reaction?
47. 4:14So, we have I am sure you have learned before to some extent for any reaction we try to
48. 4:21calculate G, free energy change and this is equated with delta H fantastic minus T delta
49. 4:34S. Now usually this delta H term is a small one, it is not a huge term or the contribution
50. 4:46is not that huge compared to T delta S. So, we can kind of neglect this term because it
51. 4:52is you know it is kind of negligible contribution.
52. 4:56What all we need to understand or we need to deal with is this T delta S. Since temperature
53. 5:04is always a positive quantity like 25 degree centigrade, 100 degree centigrade of course,
54. 5:12you can you have to convert it to Kelvin and then 500, 1000 essentially what it boils down
55. 5:21to is you have to look at the delta S. What is delta S?
56. 5:27That is the entropy change.
57. 5:28How to easily relate the entropy change?
58. 5:32Let us say this reaction carbon plus oxygen giving you let say carbon monoxide ok.
59. 5:46Now this is a solid right; this is a solid, this is a gas.
60. 5:57You cannot see, this is a solid.
61. 6:01Carbon is a solid, oxygen is gas.
62. 6:05So, solid plus gas is giving you a gas ok.
63. 6:14In fact, one equivalent of gas is giving you two equivalent of gas.
64. 6:21So, entropy is increasing; that means, you know more degrees of freedom you are having.
65. 6:27So, delta S is going to be positive for this reaction of course, you most of you now that.
66. 6:33Now, for such reaction delta S is going to be positive plus and this minus, T is always
67. 6:41plus T delta S term this term is going to for this reaction is going to be positive
68. 6:47combined, it is going to be negative.
69. 6:52For a spontaneous reaction, we will look for negative delta G ok.
70. 6:59So, for a spontaneous reaction delta S should be positive and delta G should be negative
71. 7:13that is what exactly we are looking for.
72. 7:15Now, the problem is somewhere else, where is the trouble?
73. 7:20Metal oxide that reaction you will find or metal plus oxygen when you are reacting metal
74. 7:29with oxygen, it is a gas that is oxygen you are reacting metal oxide most often is a solid.
75. 7:39Metal plus oxygen going to metal oxide, delta S is not going to be positive that is what
76. 7:48the problem is otherwise I mean I guess we could have lot less of a problem anyway.
77. 7:56So, criterion for spontaneity is very simple as we say we need a delta G that is negative
78. 8:02for a favorable reaction.
79. 8:04So, equilibrium constant has to be greater than 1 for this cases, kinetics is not important
80. 8:11as reductions are done at high temperature and these reactions are very fast.
81. 8:18So, delta G as I wrote in here delta G equals delta H minus T delta S for the formation
82. 8:25of metal oxide metal is in solid form, oxygen is in gaseous form.
83. 8:33So, what we get is metal oxide solid.
84. 8:40Now for these sort of reaction, we have delta S negative.
85. 8:46Of course, oxygen gas is used up solid plus gas is going to solid.
86. 8:52So, entropy change is going to be you know negative.
87. 8:57So, it is not a spontaneous reaction that we are going to get right.
88. 9:02So, for these cases if temperature if you are increasing let us say you are heating
89. 9:08from 100 degree centigrade to 500 degree centigrade, your delta G is becoming more and more positive.
90. 9:15See delta G may be your starting from let us say negative something; negative let us
91. 9:21say 500, but as you are increasing temperature delta G is becoming more and more positive.
92. 9:27If the absolute values still could be negative, but it is becoming more positive -500 to -500,
93. 9:35-300, -200 as you are going to increase the temperature ok.
94. 9:41First the free energy change that is delta G increases with increase in the temperature.
95. 9:47Let me show you the show you the curve for one case.
96. 9:52So, let us say this is the calcium oxide curve.
97. 9:55Now delta G for calcium oxide formation was at 500 degree C, let say this is something
98. 10:04like -1200 or so, kilo Joule per mole delta G. As you are increasing temperature 500 to
99. 10:12let say 2500, you can see that the delta G is increasing.
100. 10:18Of course, you started with negative value that is why it is saying still negative because
101. 10:260 is here.
102. 10:27It is still negative value, but overall delta G is becoming more and more positive.
103. 10:36You started with somewhere -1200 or 1300, you are going to get up to let us say over
104. 10:42here at 2500 degree C you are going to -400.
105. 10:45So, delta G is increasing with respect to temperature.
106. 10:51If you are heating keep heating the reaction mixture, the reaction is becoming more and
107. 10:56more unfavorable, right.
108. 11:01Now so, delta G is becoming more positive that is understood, this is the 0.
109. 11:12This is delta G equals 0 anything below that is thermodynamically feasible means it is
110. 11:21a favorable process for any spontaneous reaction, for anything feasible whether it can happen
111. 11:28or not, you should have delta G negative ok.
112. 11:34Anything below 0, you can have these metal oxide formation
113. 11:39Let us say for example, silver oxide; we were saying that silver oxide can be heated to
114. 11:46get or to give you silver plus oxygen right so, that means, at this temperature where
115. 11:55delta G is becoming positive.
116. 11:57So, it is little negative right over here; if you heat it let us say 50 degree centigrade,
117. 12:06now this delta G is becoming more than 0, means positive.
118. 12:12The moment it is becoming positive, the silver oxide will not exist and silver oxide, it
119. 12:19will break down to silver.
120. 12:22If delta G is not negative that species is not going to or that oxide is not going to
121. 12:28stay over there, it will disintegrate into corresponding metal and oxygen, right.
122. 12:36So, this is the temperature you are looking for, but as you see for most of the metal
123. 12:42that temperature let say copper oxide; yes, it is something 200 degree C, but that temperature
124. 12:48is very high.
125. 12:50For most of the metal oxide, it is not crossing this delta G0 at a reasonable temperature
126. 12:57right.
127. 12:58So, they will stay even if you keep on heating your ore, you are not going to get anything
128. 13:05out of it.
129. 13:06You can melt it, you are having a solid mixture you can melt it, but still you cannot get
130. 13:13your pure metal, right.
131. 13:16For a only it is possible let say for sodium azide as I was saying, sodium azide is explosive,
132. 13:22but silver; you have to be careful with sodium azide, but silver oxide you can just heat
133. 13:28off little bit and you can get pure silver, but as you see almost nothing else is crossing
134. 13:35up to 2500 degree C except this copper oxide.
135. 13:38So, just simple heating is not going to be enough, you need a reducing equivalent; you
136. 13:47need something to reduce.
137. 13:49What can you reduce; what can you use to reduce?
138. 13:52All the problem here is all this metal oxide curves of course, technically speaking you
139. 14:00can take one from the top ok.
140. 14:06For example, technically speaking you can take calcium and you can heat calcium with
141. 14:12magnesium oxide and calcium oxide will form because calcium oxide formation is more favorable
142. 14:20than magnesium oxide.
143. 14:22But in practically speaking, are you going to do that.
144. 14:25I mean, are you going to sacrifice calcium to get magnesium?
145. 14:31No, it is not possible.
146. 14:34Although thermodynamically it is it should technically work out calcium oxide formation
147. 14:39at any point looks like going to be more favorable.
148. 14:41So, delta G as long as delta G is more negative for some metal oxide formation.
149. 14:49You can kind of trade off, you can just sell one to get another one.
150. 14:54You can sell calcium to get magnesium, but it is not a feasible process because it is
151. 15:00not going to be economically favorable ok.
152. 15:05You can argue that anything else you can take, but you know technically speaking even if
153. 15:11there is some price difference; if you are getting let say 50 rupees worth of metal by
154. 15:17using 40 rupees worth of another metal, you are not going to do that you need 50 rupees
155. 15:23worth of metal by using 50 paisa of worth of something some reducing agent.
156. 15:31What all you need to know is what is a suitable reducing agent.
157. 15:35This is where charcoal carbon all the industry you see, all the charcoal; all those you know
158. 15:43metallurgical extraction charcoal works out, why charcoal works out?
159. 15:48Now as I was saying charcoal works out because it is just working opposite way charcoal is
160. 15:56solid, oxygen is gas one equivalent of charcoal and half a equivalent of oxygen is giving
161. 16:05you one equivalent of carbon monoxide, right.
162. 16:08So, which is gas?
163. 16:10So, this is a favorable very very; delta S is going to be positive for this reaction.
164. 16:20Over here what you are saying metal plus oxygen going to metal oxide.
165. 16:25This metal oxide is going to be solid, but this carbon monoxide is going to be gas or
166. 16:32it is gas right.
167. 16:33So, technically speaking that is what the advantage of charcoal carbon monoxide.
168. 16:41CO is gas, but metal oxide any form M2O3 or MO2 or whatever it is going to be solid.
169. 16:49So, as you are increasing the temperature, this curve is becoming more and more positive
170. 16:57metal oxide curves are becoming more and more positive, but the green one over here which
171. 17:03is not written that is for your carbon plus oxygen going to carbon monoxide.
172. 17:09This is a having; this is having a downward slope means that reaction is going to help
173. 17:17almost the mankind to give the pure metal at a very cheap rate right; that is, what
174. 17:24all we are looking for.
175. 17:26We need an alternate solution or we need a solution which can work out for most of the
176. 17:33metal right.
177. 17:34So, what we are trying to say one more time, I think few of you are having queries how
178. 17:38delta G is going to be positive more positive with respect to temperature more and more
179. 17:46positive.
180. 17:47So, some reaction if you are looking at this reaction, metal solid is going and reacting
181. 17:51with gas to give you metal oxide right.
182. 17:54This is the reaction where you are going to have delta S negative because gas is consumed
183. 18:02and you are forming solid.
184. 18:04Delta S is going to be negative.
185. 18:07Now if delta S is going to be negative, the delta G is becoming more and more positive
186. 18:13if you are increasing the temperature.
187. 18:14Delta S is negative here, negative.
188. 18:17So, as you increase the temperature from let us say 25 to 100 to 300, you are going to
189. 18:23have more and more delta G positive.
190. 18:26So, if you are starting from let us say over here 1100 or 1200, it is -1100 or -1200, it
191. 18:34is becoming more and more positive.
192. 18:36So, -1200 you are getting let us say 800 or 600 and so on.
193. 18:41It is becoming more positive does not necessarily mean that value it itself will be becoming
194. 18:47absolutely positive from -1200, it is going to -1000, -600, -400 and so on.
195. 18:55Eventually it can reach to the 0 delta G is 0 which is kind of you know the point where
196. 19:02you will be able to break it, disintegrated.
197. 19:07Now since the other reaction which you are going to look at is the carbon reacting with
198. 19:15oxygen that is the charcoal we are talking about.
199. 19:18It is a good reducing agent and that is a good reducing agent basically because this
200. 19:24carbon monoxide gas is formed ok.
201. 19:28Carbon monoxide gas from half a equivalent of oxygen, one equivalent of carbon monoxide
202. 19:33is formed.
203. 19:34Therefore, delta S is going to be more positive.
204. 19:37So, delta S is going to be positive and thereby with respect to if delta S is going to be
205. 19:43positive for carbon plus oxygen this carbon monoxide formation.
206. 19:47So, delta G is becoming more and more negative as you can see, as you are increasing the
207. 19:55temperature.
208. 19:56So, you keep on increasing the temperature delta G is becoming more and more negative
209. 20:00this is just opposite to all most all the metal oxide curve that you see in here.
210. 20:08Technically speaking wherever this curve, this green curve is crossing the red curve.
211. 20:14Let us say, this is the temperature for zinc oxide, this is the point where carbon monoxide
212. 20:21curve carbon going to carbon monoxide curve is crossing.
213. 20:25So, right after that temperature right after that cross section, carbon monoxide formation
214. 20:30is more favorable.
215. 20:32Delta G for carbon monoxide formation let us say over here or here are more favorable.
216. 20:39So, if this is the 1000 degree centigrade where the green curve is crossing the red
217. 20:44curve at that point right after that point, you can react metal oxide ok; you can react
218. 20:54metal oxide with carbon, you can get carbon monoxide plus corresponding metal that is
219. 21:04what all I think Ellingham diagram has to tell you.
220. 21:10At any temperature or the temperature where it is crossing, any temperature above that
221. 21:16let us say this is crossing at 1000 degree C, 1001 degree centigrade if you are heating
222. 21:24carbon monoxide formation is more favorable compared to the metal oxide formation.
223. 21:30So, metal oxide will be reducing to corresponding metal, carbon will get oxidized to carbon
224. 21:38monoxide right ok.
225. 21:41So, that is the crucial temperature we are going to look at.
226. 21:45So, carbon plus let say metal oxide going to carbon monoxide plus metal.
227. 21:57Now you see that at some point this curve is breaking or seems like breaking, what is
228. 22:03happening there?
229. 22:04It is a change of state definitely.
230. 22:07So, metal this was solid metal it iself was solid, oxygen was gas.
231. 22:17Now if up to let us say for example, calcium oxide calcium metal was in the solid form
232. 22:23up to 1500 degree centigrade; after 1500 degree centigrade, what happen that calcium is becoming
233. 22:31calcium liquid; solid to calcium liquid right.
234. 22:35So, you are initially you are having metal solid plus oxygen gas giving you metal oxide
235. 22:44solid right.
236. 22:45Delta S for this reaction was negative right.
237. 22:53Now let us say, let say metal equals your calcium up to 1500 degree centigrade that
238. 23:00is what is happening; above 1500 degree centigrade, you have metal liquid right plus oxygen gas
239. 23:13giving you metal oxide solid; that solid remains as solid right.
240. 23:19Now, what is happening here?
241. 23:22It is a solid plus gas was giving you solid now it is a liquid which is having more often
242. 23:29entropy more free material liquid, solid to liquid.
243. 23:34So, liquid plus gas is giving you solid.
244. 23:37So, delta S is going to be so, delta S is going to be more negative.
245. 23:47Let us say just double negative let us say, what is?
246. 23:55So, if delta S is becoming more negative because it is a liquid plus gas is going to solid.
247. 24:02Previously solid plus gas was going to solid, now it is a liquid plus gas more freely you
248. 24:09know freely, free flow material is going into the solid material delta S becoming negative.
249. 24:17So, you just relate to delta G equals delta S minus T delta S.
250. 24:22If it is more negative, delta G is becoming more and more positive.
251. 24:26So, it was going positive, but the slope changes.
252. 24:33So, the it was going very slow positive, it was coming with respect to temperature it
253. 24:39was going very slow slope was less the point calcium melts the slope increases; that means,
254. 24:47it is becoming more and more unfavorable quickly this is happening or I mean this change is
255. 24:53better or slope is becoming more and thereby you are going to get more unfavorable or more
256. 25:03unfavorable delta G ok.
257. 25:06Same is true if it is becoming gas instead of liquid; if metal is metal instead of metal
258. 25:12being liquid if metal is becoming gas at certain temperature, you will get further increment
259. 25:18in the slope alright.
260. 25:22So, now other things to look at from the periodic; from that Ellingham diagram is very simply
261. 25:30these are the metal which are electro positive metal.
262. 25:36If you are looking at that electro chemical series these are the one which are going to
263. 25:42be in the on the top; that means, they tend to preferably stay in metal oxidized form
264. 25:47right.
265. 25:48So, the your seen the reduction potential is very less for these cases; that means,
266. 25:55they will tend to go to the reduced form; silver oxide will tend to go to silver, but
267. 26:02calcium oxide will prefer to stay in calcium 2 plus right.
268. 26:07It is there is of course, some I mean clear correlation with those electrochemical series
269. 26:14and that of the Ellingham diagram.
270. 26:19The more positive material which tends to stay in positive state electro positive material
271. 26:25is at the bottom those which are less electro positive there on the top ok.
272. 26:32As I said previously any metal will reduce the oxide of other metal which is above in
273. 26:37the Ellingham diagram ok.
274. 26:39Now of course, we need to also understand how carbon can be oxidized to carbon monoxide
275. 26:45or carbon dioxide and how they are plots are relating to each other right.
276. 26:53Now, this is usually as you are discussing usually this is the metal oxide fog plot metal
277. 26:59going to metal oxide and this is the usually carbon going to carbon monoxide plot.
278. 27:04So, at any temperature above these crossing point above this point, you will be having
279. 27:11carbon monoxide formation.
280. 27:13Let us say at this temperature you will have carbon monoxide formation and metal oxide,
281. 27:18you will be converted to corresponding metal right.
282. 27:22Now, if you are to relate how it is with carbon monoxide plus oxygen reacting with carbon
283. 27:29dioxide this is of course, another reaction which is possible in these cases.
284. 27:34If you are taking charcoal and reacting with metal oxide, you are going to get usually
285. 27:41a mixture of carbon monoxide and carbon dioxide right.
286. 27:44So, for carbon monoxide case, it is a gas plus gas giving you a gas, but it is a one
287. 27:51and half equivalent of gas is giving you a one equivalent of gas.
288. 27:55So, delta S is going to be negative.
289. 27:59Unlike the carbon case carbon solid plus oxygen giving you carbon monoxide here carbon monoxide
290. 28:05plus oxygen delta S is going to be negative ok.
291. 28:09Thereby the plot you see is over here, this purple color one over here.
292. 28:14So, with respect to temperature as you an increasing temperature it is becoming more
293. 28:21and more unfavorable ok.
294. 28:24It is just all depends on delta S whether delta S is negative or delta S is positive.
295. 28:30In this case one and half equivalent of gas giving you one equivalent of gas right so,
296. 28:35this is the purple line over here for carbon monoxide going to carbon dioxide.
297. 28:41Now, the one in green here is the carbon going to carbon monoxide, what is happening for
298. 28:49carbon going to directly carbon dioxide.
299. 28:51So, carbon is solid oxygen is gas it is going to gas right delta S remain constant; one
300. 29:00gas is going to another gas.
301. 29:03So, overall, it is going to stay constant.
302. 29:08Delta S should be I mean you know, it is delta of delta S right.
303. 29:15So, usually you can it 0, but it is better to put constant it is not changing it is a
304. 29:22change that we are looking for.
305. 29:23Now with respect to temperature, I think it is better to say 0 may be with.
306. 29:29So, all the delta H will matter ok.
307. 29:32So, delta S is not changing anything.
308. 29:35Now that’s the, that’s the curve for carbon going to carbon dioxide, this is carbon monoxide
309. 29:41going to carbon dioxide and this is the one going carbon to carbon monoxide.
310. 29:46Now, clearly the idea here is very simple which is most negative for all these area
311. 29:53when we say you in over here.
312. 29:58During these period up to the 710 degrees c or so, carbon monoxide to carbon dioxide
313. 30:05formation is more favorable because CO to CO2 graph carbon monoxide to carbon dioxide
314. 30:12graph as you can see, it is in the purple.
315. 30:15So, it is more negative.
316. 30:16So, all you need to consider is which is more negative which curve is more negative ok.
317. 30:24If you are looking for any reaction below 710 degree C, it is better to utilize carbon
318. 30:30monoxide to carbon dioxide curve or that is the reaction which is more favorable.
319. 30:36Above 710 degree C anything, you do it is going to be carbon to carbon monoxide formation
320. 30:44ok.
321. 30:45So, of course, you have the option to choose, but practically speaking usually we are heating
322. 30:52the ore at a you know kind of a high temperature.
323. 30:58So, we do not really worry about everything mostly carbon going to carbon monoxide is
324. 31:03the one you need to worry about any way.
325. 31:06Now I think same thing we have discussed already.