1 00:00:00,000 --> 00:00:00,016 The following content is provided under a Creative 2 00:00:00,016 --> 00:00:00,022 Commons license. 3 00:00:00,022 --> 00:00:00,038 Your support will help MIT OpenCourseWare continue to 4 00:00:00,038 --> 00:00:00,054 offer high quality educational resources for free. 5 00:00:00,054 --> 00:00:00,072 To make a donation or view additional materials from 6 00:00:00,072 --> 00:00:00,088 hundreds of MIT courses visit MIT OpenCourseWare at 7 00:00:00,088 --> 00:00:00,110 ocw.mit.edu. 8 00:00:00,110 --> 00:00:23,370 PROFESSOR: OK, let's get started. 9 00:00:23,370 --> 00:00:27,210 Why doesn't everyone go ahead and take 10 more seconds on 10 00:00:27,210 --> 00:00:29,130 this clicker question. 11 00:00:29,130 --> 00:00:29,310 It should look familiar. 12 00:00:29,310 --> 00:00:33,190 You were pretty split on this question on Friday, so we're 13 00:00:33,190 --> 00:00:36,630 hoping after learning a little bit more about delta g of 14 00:00:36,630 --> 00:00:39,520 formation, we have at least one direction 15 00:00:39,520 --> 00:00:41,560 that wins out here. 16 00:00:41,560 --> 00:00:42,450 OK, great. 17 00:00:42,450 --> 00:00:46,200 So now we're up to 85% of you, and hopefully in just a minute 18 00:00:46,200 --> 00:00:47,970 we'll be up to a 100%. 19 00:00:47,970 --> 00:00:51,920 But if delta g of formation here is less than zero, that 20 00:00:51,920 --> 00:00:55,000 means we're talking about a spontaneous reaction when we 21 00:00:55,000 --> 00:00:56,510 form the compound. 22 00:00:56,510 --> 00:00:59,770 So if it spontaneously forms the compound, that must mean 23 00:00:59,770 --> 00:01:02,060 that the compound is going to be stable 24 00:01:02,060 --> 00:01:03,390 relative to its elements. 25 00:01:03,390 --> 00:01:07,030 So that's kind of the last thing we went over in terms of 26 00:01:07,030 --> 00:01:14,820 topics on Friday, and we'll pick up right there today. 27 00:01:14,820 --> 00:01:15,730 All right. 28 00:01:15,730 --> 00:01:18,750 Here we have our notes for today, so reminder exam 2 is 29 00:01:18,750 --> 00:01:19,340 on Wednesday -- 30 00:01:19,340 --> 00:01:21,200 I don't think I need to remind anyone of that. 31 00:01:21,200 --> 00:01:23,930 But just please, don't come to this room, make sure that you 32 00:01:23,930 --> 00:01:27,120 go to Walker to take the exam. 33 00:01:27,120 --> 00:01:29,620 And also if you have any questions or you feel like 34 00:01:29,620 --> 00:01:32,010 there's anything you don't understand, I have office 35 00:01:32,010 --> 00:01:35,200 hours today from 2 to 4 in my office. 36 00:01:35,200 --> 00:01:38,320 Your TAs have also all moved their office hours so they 37 00:01:38,320 --> 00:01:39,940 fall before the exam. 38 00:01:39,940 --> 00:01:42,590 So make sure you get any of your questions addressed 39 00:01:42,590 --> 00:01:44,190 before you're trying to sit there and figure 40 00:01:44,190 --> 00:01:47,310 it out in exam time. 41 00:01:47,310 --> 00:01:49,260 All right, so today we're going to pick up where we left 42 00:01:49,260 --> 00:01:52,770 off on thermodynamics, so that was talking about free energy 43 00:01:52,770 --> 00:01:54,080 of formation. 44 00:01:54,080 --> 00:01:56,350 After that we're going to talk a little bit more about the 45 00:01:56,350 --> 00:01:58,810 effect of temperature on spontaneity. 46 00:01:58,810 --> 00:02:00,990 We touched upon this on Friday, but we're going to 47 00:02:00,990 --> 00:02:04,810 formalize exactly in which cases temperature can or can 48 00:02:04,810 --> 00:02:07,260 not affect whether a reaction is spontaneous or 49 00:02:07,260 --> 00:02:08,810 non-spontaneous. 50 00:02:08,810 --> 00:02:12,010 Then we're going to look a little bit into thermodynamics 51 00:02:12,010 --> 00:02:13,800 and biological systems. 52 00:02:13,800 --> 00:02:16,920 Two examples that I wanted to talk about were ATP coupled 53 00:02:16,920 --> 00:02:19,980 reactions -- those are very important in biology. 54 00:02:19,980 --> 00:02:22,430 And also, thinking about the idea of hydrogen bonding, 55 00:02:22,430 --> 00:02:25,090 we're going to combine our thoughts on bond enthalpies, 56 00:02:25,090 --> 00:02:28,570 and also our ideas of bonding that we had from thinking 57 00:02:28,570 --> 00:02:30,160 about covalent and ionic bonds. 58 00:02:30,160 --> 00:02:34,040 All right, so let's finish up first with 59 00:02:34,040 --> 00:02:35,830 free energy of formation. 60 00:02:35,830 --> 00:02:39,670 So as 85% of you just told me, when we have a case where 61 00:02:39,670 --> 00:02:43,220 delta g of formation is less than zero, what we're talking 62 00:02:43,220 --> 00:02:47,100 about here is a compound that is thermodynamically stable 63 00:02:47,100 --> 00:02:49,470 relative to its elements. 64 00:02:49,470 --> 00:02:52,370 So that means we know the inverse as well, which is when 65 00:02:52,370 --> 00:02:55,920 we're talking about a case where delta g is now greater 66 00:02:55,920 --> 00:02:59,040 than zero, the compound is going to be thermodynamically 67 00:02:59,040 --> 00:03:01,840 unstable relative to its elements. 68 00:03:01,840 --> 00:03:05,090 So any time you're looking at delta g of formation, just 69 00:03:05,090 --> 00:03:08,330 remember to think about this is just the delta g of a 70 00:03:08,330 --> 00:03:10,380 reaction where you're forming a compound. 71 00:03:10,380 --> 00:03:12,710 So it should make sense that if it's negative, it's going 72 00:03:12,710 --> 00:03:14,950 to be spontaneous and you're going to have a stable 73 00:03:14,950 --> 00:03:17,970 compound here. 74 00:03:17,970 --> 00:03:21,050 So we could look at any number of examples, really we could 75 00:03:21,050 --> 00:03:24,290 pick any compound to look at, but up on the screen here I'm 76 00:03:24,290 --> 00:03:26,990 just putting the compound, the formation of benzene, we've 77 00:03:26,990 --> 00:03:28,710 looked at benzene a lot. 78 00:03:28,710 --> 00:03:31,500 So benzene has a delta g of formation of 124 79 00:03:31,500 --> 00:03:32,830 kilojoules per mole. 80 00:03:32,830 --> 00:03:39,900 Is benzene stable or unstable compared to its elements? 81 00:03:39,900 --> 00:03:42,000 I can't really tell the difference between stable and 82 00:03:42,000 --> 00:03:43,540 unstable when you say it, so everyone start at 83 00:03:43,540 --> 00:03:44,740 the same time, go. 84 00:03:44,740 --> 00:03:46,420 STUDENT: Unstable. 85 00:03:46,420 --> 00:03:47,780 PROFESSOR: OK, excellent. 86 00:03:47,780 --> 00:03:50,260 Benzene is unstable compared to its elements. 87 00:03:50,260 --> 00:03:53,090 So that means the reverse reaction here is going to be 88 00:03:53,090 --> 00:03:55,320 stable -- or the reverse reaction is going to be 89 00:03:55,320 --> 00:03:57,460 spontaneous where we actually have the 90 00:03:57,460 --> 00:03:59,770 decomposition of benzene. 91 00:03:59,770 --> 00:04:01,450 So when we're seeing this, when we're seeing that the 92 00:04:01,450 --> 00:04:04,470 reverse reaction is spontaneous, a question that 93 00:04:04,470 --> 00:04:07,100 might immediately come to us is well, why did we just spend 94 00:04:07,100 --> 00:04:09,350 all the time talking about benzene because clearly 95 00:04:09,350 --> 00:04:11,040 benzene's just going to break down, right. 96 00:04:11,040 --> 00:04:14,850 Why do we form benzene and not have it immediately decompose 97 00:04:14,850 --> 00:04:15,790 into its elements. 98 00:04:15,790 --> 00:04:19,160 Thermodynamically that is what should happen and it is what 99 00:04:19,160 --> 00:04:22,630 does happen, but the reality is that this reaction, this 100 00:04:22,630 --> 00:04:26,320 decomposition of benzene is actually very, very slow. 101 00:04:26,320 --> 00:04:29,550 It's so slow that, you know we use benzene all the time in 102 00:04:29,550 --> 00:04:33,200 organic reactions, we don't see it break down even when we 103 00:04:33,200 --> 00:04:35,770 heat it up, is because even though this is 104 00:04:35,770 --> 00:04:39,630 thermodynamically a non-spontaneous reaction to 105 00:04:39,630 --> 00:04:43,040 form benzene, it's very slow for the actual decomposition 106 00:04:43,040 --> 00:04:45,960 of benzene to occur. 107 00:04:45,960 --> 00:04:48,730 So this is just another case, and I'll keep saying this, and 108 00:04:48,730 --> 00:04:51,360 when Professor Drennan starts talking about kinetics, she'll 109 00:04:51,360 --> 00:04:52,920 keep repeating this as well. 110 00:04:52,920 --> 00:04:55,820 What we want to keep in mind is that delta g tells us 111 00:04:55,820 --> 00:04:58,900 whether a reaction will happen or whether it won't happen. 112 00:04:58,900 --> 00:05:02,710 It tells us absolutely nothing about how long it takes for 113 00:05:02,710 --> 00:05:03,790 that reaction to happen. 114 00:05:03,790 --> 00:05:07,100 It tells us nothing about the rate of the reaction. 115 00:05:07,100 --> 00:05:10,370 We'll keep seeing examples of this, so hopefully no one will 116 00:05:10,370 --> 00:05:13,840 be confused by the time we do get to kinetics. 117 00:05:13,840 --> 00:05:16,310 So if we're talking about calculating delta g for any 118 00:05:16,310 --> 00:05:18,690 reaction, now we actually have several ways to do it. 119 00:05:18,690 --> 00:05:21,240 The first way is very analogous to thinking about 120 00:05:21,240 --> 00:05:22,910 delta h for the reaction. 121 00:05:22,910 --> 00:05:24,950 We can just look up a table where we have 122 00:05:24,950 --> 00:05:26,480 delta g as a formation. 123 00:05:26,480 --> 00:05:29,390 So we can take the sum of the delta g of formation of the 124 00:05:29,390 --> 00:05:32,880 products, and subtract from it the delta g of formation of 125 00:05:32,880 --> 00:05:34,220 the reactants. 126 00:05:34,220 --> 00:05:36,960 We also have another way if maybe we don't have that 127 00:05:36,960 --> 00:05:38,740 information available to us. 128 00:05:38,740 --> 00:05:43,220 We can also take a look at using this reaction or this 129 00:05:43,220 --> 00:05:46,490 equation right here, which is telling us that delta g of a 130 00:05:46,490 --> 00:05:49,110 reaction is equal to the change in enthalpy 131 00:05:49,110 --> 00:05:51,050 minus t delta s. 132 00:05:51,050 --> 00:05:53,050 So that tends to be very helpful, especially when we 133 00:05:53,050 --> 00:05:57,040 want to take into consideration temperature. 134 00:05:57,040 --> 00:05:58,900 So let's take into consideration temperature. 135 00:05:58,900 --> 00:06:01,880 We've done this a little bit so far, but let's really take 136 00:06:01,880 --> 00:06:04,260 a look at some reactions where temperature's going to make a 137 00:06:04,260 --> 00:06:05,250 big difference. 138 00:06:05,250 --> 00:06:07,290 So the reaction we're going to look at here is the 139 00:06:07,290 --> 00:06:10,660 decomposition of sodium bicarbonate, or sodium bicarb 140 00:06:10,660 --> 00:06:14,890 here, and it decomposes it into sodium carbonate, plus c 141 00:06:14,890 --> 00:06:17,820 o 2, carbon dioxide, and water. 142 00:06:17,820 --> 00:06:20,500 So we can think about calculating that delta g of 143 00:06:20,500 --> 00:06:21,770 this reaction. 144 00:06:21,770 --> 00:06:24,280 I'll tell you that the change in enthalpy, this is actually 145 00:06:24,280 --> 00:06:25,730 an endothermic reaction. 146 00:06:25,730 --> 00:06:26,630 It requires heat. 147 00:06:26,630 --> 00:06:28,420 It's plus 135 . 148 00:06:28,420 --> 00:06:30,110 6 kilojoules per mole. 149 00:06:30,110 --> 00:06:32,300 So let's go to a clicker question and I want you to 150 00:06:32,300 --> 00:06:36,010 pick out from several choices which of the changes in 151 00:06:36,010 --> 00:06:41,440 entropy seem reasonable to you here. 152 00:06:41,440 --> 00:06:43,640 So what would you predict the delta s for 153 00:06:43,640 --> 00:06:51,250 this reaction to be? 154 00:06:51,250 --> 00:07:06,900 All right, let's do 10 more seconds on that. 155 00:07:06,900 --> 00:07:10,030 OK, so we've got the majority, but not everyone. 156 00:07:10,030 --> 00:07:13,800 So let's take a look at why this is the correct answer, 157 00:07:13,800 --> 00:07:16,920 that it should be plus 0.334 . 158 00:07:16,920 --> 00:07:20,710 If we're going from 2 moles of solid, to 1 mole of solid, 159 00:07:20,710 --> 00:07:25,020 plus 2 moles of gas, are we increasing or 160 00:07:25,020 --> 00:07:26,600 decreasing the disorder? 161 00:07:26,600 --> 00:07:28,340 STUDENT: [INAUDIBLE] 162 00:07:28,340 --> 00:07:29,770 PROFESSOR: We're increasing the disorder. 163 00:07:29,770 --> 00:07:32,400 If we're getting to a more disordered state, then we're 164 00:07:32,400 --> 00:07:35,500 going to have a positive change in entropy. 165 00:07:35,500 --> 00:07:37,190 We're going to increase the disorder. 166 00:07:37,190 --> 00:07:40,910 The only one with a positive delta s is this choice here. 167 00:07:40,910 --> 00:07:45,365 So if we switch back to our notes, we can see that, in 168 00:07:45,365 --> 00:07:54,060 fact, so what we see is that it's 0.334 kilojoules per k 169 00:07:54,060 --> 00:07:59,240 per mole is our delta s, so we can go ahead and calculate our 170 00:07:59,240 --> 00:08:01,860 delta g for the reaction, so we're just plugging 171 00:08:01,860 --> 00:08:03,520 in our delta h. 172 00:08:03,520 --> 00:08:06,620 So our delta h is 135 . 173 00:08:06,620 --> 00:08:07,880 6. 174 00:08:07,880 --> 00:08:10,280 And we're talking about to start with let's talk about 175 00:08:10,280 --> 00:08:11,370 room temperature. 176 00:08:11,370 --> 00:08:18,050 So 298 k times 0.334 . 177 00:08:18,050 --> 00:08:21,430 So what we end up having for the delta g of our reaction is 178 00:08:21,430 --> 00:08:22,860 that its 36 . 179 00:08:22,860 --> 00:08:24,520 1 kilojoules per mole. 180 00:08:24,520 --> 00:08:26,740 So this is at room temperature. 181 00:08:26,740 --> 00:08:30,030 So is our reaction spontaneous or non-spontaneous at room 182 00:08:30,030 --> 00:08:30,700 temperature? 183 00:08:30,700 --> 00:08:31,970 STUDENT: Non-spontaneous. 184 00:08:31,970 --> 00:08:32,740 PROFESSOR: Non-spontaneous. 185 00:08:32,740 --> 00:08:34,800 All right, but let's take a look at a different 186 00:08:34,800 --> 00:08:35,230 temperature. 187 00:08:35,230 --> 00:08:37,560 For example, let's look at baking temperature. 188 00:08:37,560 --> 00:08:40,700 So if we think about baking cookies, we maybe bake them at 189 00:08:40,700 --> 00:08:44,910 350 degrees fahrenheit, so that would be 450 kelvin -- 190 00:08:44,910 --> 00:08:48,830 our ovens are usually set to fahrenheit and not kelvin. 191 00:08:48,830 --> 00:08:50,700 So if we think about this reaction that this 192 00:08:50,700 --> 00:08:53,100 temperature, first of all let me point out why we would be 193 00:08:53,100 --> 00:08:55,860 talking about baking cookies for this 194 00:08:55,860 --> 00:08:56,990 particular reaction here. 195 00:08:56,990 --> 00:08:59,310 Does anyone know what another name for sodium bicarb is? 196 00:08:59,310 --> 00:09:00,840 STUDENT: [INAUDIBLE] 197 00:09:00,840 --> 00:09:02,410 PROFESSOR: Yeah, it's just baking soda. 198 00:09:02,410 --> 00:09:04,910 So this is the reaction that causes your cookies or your 199 00:09:04,910 --> 00:09:06,700 cakes to actually rise. 200 00:09:06,700 --> 00:09:10,010 So we're producing gas here, and when we're producing that 201 00:09:10,010 --> 00:09:12,850 gas when this sodium bicarb decomposes, we're forming 202 00:09:12,850 --> 00:09:15,750 these pockets of gas in our baked goods. 203 00:09:15,750 --> 00:09:19,630 So if we bake our cookies at room temperature, obviously, 204 00:09:19,630 --> 00:09:24,020 they don't bake and they also don't rise because the sodium 205 00:09:24,020 --> 00:09:26,550 bicarb, clearly it was non-spontaneous at room 206 00:09:26,550 --> 00:09:28,850 temperature, this reaction just doesn't happen. 207 00:09:28,850 --> 00:09:31,730 But if we take a look at baking temperature now, we're 208 00:09:31,730 --> 00:09:34,140 going to plug in this new temperature, which is 450 k, 209 00:09:34,140 --> 00:09:38,130 and plug this into our delta g reaction. 210 00:09:38,130 --> 00:09:41,410 What we find is now our delta g for the reaction 211 00:09:41,410 --> 00:09:43,250 is negative 14 . 212 00:09:43,250 --> 00:09:46,100 7 kilojoules per mole. 213 00:09:46,100 --> 00:09:48,550 So in this case, we are dealing with a spontaneous 214 00:09:48,550 --> 00:09:51,200 reaction, which is good, because this means that when 215 00:09:51,200 --> 00:09:54,480 we put our cookies in and we turn it to 350 fahrenheit or 216 00:09:54,480 --> 00:09:58,760 450 k, the baking soda will decompose and we'll got our 217 00:09:58,760 --> 00:10:01,520 cookies to rise a little bit. 218 00:10:01,520 --> 00:10:03,920 All right, so one thing that I want you to notice when we 219 00:10:03,920 --> 00:10:06,790 were talking about the case with the decomposition of 220 00:10:06,790 --> 00:10:10,660 sodium bicarb is that the delta h for that reaction and 221 00:10:10,660 --> 00:10:13,940 the delta s, they both had the same sign. 222 00:10:13,940 --> 00:10:16,830 And something that you can keep in mind in general is any 223 00:10:16,830 --> 00:10:20,130 time that both delta h and delta s have the same sign, 224 00:10:20,130 --> 00:10:23,430 it's actually possible to switch from spontaneous to 225 00:10:23,430 --> 00:10:26,560 non-spontaneous or vice versa just by changing the 226 00:10:26,560 --> 00:10:29,330 temperature of the reaction. 227 00:10:29,330 --> 00:10:31,300 And we can think about this graphically. 228 00:10:31,300 --> 00:10:34,370 If we assume that delta h and delta s are independent of 229 00:10:34,370 --> 00:10:36,300 temperature, which is a good assumption to a first 230 00:10:36,300 --> 00:10:39,810 approximation, in this case we find that the delta g of the 231 00:10:39,810 --> 00:10:43,650 reaction is the linear function of the temperature. 232 00:10:43,650 --> 00:10:46,430 So that means we can go ahead and graph what we saw for the 233 00:10:46,430 --> 00:10:49,090 case of baking soda. 234 00:10:49,090 --> 00:10:51,520 So our first point here was that we saw at room 235 00:10:51,520 --> 00:10:55,380 temperature, about 298 k, the delta g was positive, it was 236 00:10:55,380 --> 00:10:58,190 36 kilojoules per mole. 237 00:10:58,190 --> 00:11:01,450 We also saw that once we heated it up to baking 238 00:11:01,450 --> 00:11:05,690 temperature, we actually had the delta g now at negative 15 239 00:11:05,690 --> 00:11:07,010 kilojoules per mole. 240 00:11:07,010 --> 00:11:10,650 So since this is linear, we can actually draw a straight 241 00:11:10,650 --> 00:11:13,950 line right through here, and we can think about the fact 242 00:11:13,950 --> 00:11:17,270 that we have this temperature here where anything below this 243 00:11:17,270 --> 00:11:20,260 temperature has a positive delta g, and anything above 244 00:11:20,260 --> 00:11:23,700 this temperature is going to have a negative delta g. 245 00:11:23,700 --> 00:11:26,760 And let's actually think about the fact that this is a line 246 00:11:26,760 --> 00:11:29,610 and see what our slope and our y-intercept is going to mean. 247 00:11:29,610 --> 00:11:33,390 We can say that delta g is equal to negative delta s t 248 00:11:33,390 --> 00:11:36,290 plus delta h, if we want to put it in the formation of the 249 00:11:36,290 --> 00:11:37,810 equation for line. 250 00:11:37,810 --> 00:11:40,290 So what this tells us is that our slope is going to be 251 00:11:40,290 --> 00:11:44,330 negative delta s here, and if we think about our 252 00:11:44,330 --> 00:11:47,880 y-intercept, that's going to be the change in enthalpy for 253 00:11:47,880 --> 00:11:51,190 the reaction. 254 00:11:51,190 --> 00:11:54,040 So again, what I want to point out is that any time we're at 255 00:11:54,040 --> 00:11:56,230 a temperature that's lower than this change in 256 00:11:56,230 --> 00:11:59,230 temperature here, we're going to find that delta g is 257 00:11:59,230 --> 00:12:01,700 greater than zero, and we're going to be dealing with a 258 00:12:01,700 --> 00:12:04,870 non-spontaneous reaction. 259 00:12:04,870 --> 00:12:08,220 However, if we move our temperature up and up and up 260 00:12:08,220 --> 00:12:10,750 so we go across the graph this way, eventually we'll hit a 261 00:12:10,750 --> 00:12:13,820 point where if we're above that temperature, we'll find 262 00:12:13,820 --> 00:12:16,670 that delta g is less than zero, and now we have a 263 00:12:16,670 --> 00:12:21,340 spontaneous reaction. 264 00:12:21,340 --> 00:12:24,280 So we can actually think about what this temperature is. 265 00:12:24,280 --> 00:12:26,410 We can call this T star. 266 00:12:26,410 --> 00:12:29,040 This is the temperature at which our reaction switches, 267 00:12:29,040 --> 00:12:31,620 whether it's spontaneous or non-spontaneous. 268 00:12:31,620 --> 00:12:33,970 And if you're thinking about trying to get a reaction to 269 00:12:33,970 --> 00:12:36,380 go, it's very important to be able to calculate what this 270 00:12:36,380 --> 00:12:37,500 temperature is. 271 00:12:37,500 --> 00:12:40,320 A lot of times in organic chemistry laboratories, they 272 00:12:40,320 --> 00:12:43,100 need to heat up reactions -- part of why they do that is 273 00:12:43,100 --> 00:12:45,590 kinetics, but the other part is sometimes they need to make 274 00:12:45,590 --> 00:12:48,380 a reaction go from being non-spontaneous to 275 00:12:48,380 --> 00:12:49,280 spontaneous. 276 00:12:49,280 --> 00:12:53,160 So let's think about how to calculate T star or this 277 00:12:53,160 --> 00:12:54,410 change in temperature. 278 00:12:54,410 --> 00:12:56,880 So we're talking about this threshold temperature, so 279 00:12:56,880 --> 00:12:59,210 we're talking about where delta g is going to be equal 280 00:12:59,210 --> 00:13:02,480 to zero, because if we set delta g equal to zero, we know 281 00:13:02,480 --> 00:13:04,650 that anything on one side of that temperature is going to 282 00:13:04,650 --> 00:13:06,910 be spontaneous, and the other side is going to be 283 00:13:06,910 --> 00:13:09,600 non-spontaneous. 284 00:13:09,600 --> 00:13:12,850 So if we do this, we can just rewrite our reaction, delta g 285 00:13:12,850 --> 00:13:17,270 equals delta h minus t delta s, and let's plug in our zero 286 00:13:17,270 --> 00:13:20,490 for delta g there, and now rearrange our reaction so that 287 00:13:20,490 --> 00:13:23,260 we're talking about this threshold temperature. 288 00:13:23,260 --> 00:13:26,670 So that T star is just going to be equal to the change in 289 00:13:26,670 --> 00:13:29,940 enthalpy divided by the change in entropy. 290 00:13:29,940 --> 00:13:32,640 So it's very easy for us to calculate, so let's go ahead 291 00:13:32,640 --> 00:13:34,950 and do this for the case with baking soda. 292 00:13:34,950 --> 00:13:39,340 And for baking soda what we saw was that delta h was 136 293 00:13:39,340 --> 00:13:43,890 kilojoules per mole, and the change in entropy was 0.334 294 00:13:43,890 --> 00:13:46,920 kilojoules per k mole. 295 00:13:46,920 --> 00:13:49,710 And that means if we do that simple division, then what we 296 00:13:49,710 --> 00:13:55,090 end up as our temperature star is 406 kelvin. 297 00:13:55,090 --> 00:13:58,090 So basically, what this tells us is if we tried to bake our 298 00:13:58,090 --> 00:14:02,420 cookies below 406 kelvin they would not rise, if we bake 299 00:14:02,420 --> 00:14:05,800 them above 406 kelvin they will rise because this 300 00:14:05,800 --> 00:14:07,550 reaction is now spontaneous. 301 00:14:07,550 --> 00:14:12,290 All right, so this was a case where we had seen that the 302 00:14:12,290 --> 00:14:16,310 delta h and the delta s both had a positive value. 303 00:14:16,310 --> 00:14:19,560 But let's take a look at what happens when now delta h and 304 00:14:19,560 --> 00:14:22,070 delta s are both negative. 305 00:14:22,070 --> 00:14:24,860 So we can think about this just by plotting it on our 306 00:14:24,860 --> 00:14:27,560 graph again -- we don't have actual values, but we can 307 00:14:27,560 --> 00:14:29,430 think about what the sign should be. 308 00:14:29,430 --> 00:14:33,170 So if we talk about our zero point where temperature is 309 00:14:33,170 --> 00:14:37,980 absolute zero, if we have a positive, or excuse me, if we 310 00:14:37,980 --> 00:14:40,870 have a negative delta h now and we're at temperature 311 00:14:40,870 --> 00:14:44,480 equals zero, what is delta g going to be? 312 00:14:44,480 --> 00:14:47,070 Negative or positive? 313 00:14:47,070 --> 00:14:48,290 Yeah, it's going to be negative. 314 00:14:48,290 --> 00:14:51,690 So if we have negative delta h and we're at temperature of 315 00:14:51,690 --> 00:14:54,960 zero, our s term completely falls out, so we're definitely 316 00:14:54,960 --> 00:14:57,310 going to start with a negative delta g. 317 00:14:57,310 --> 00:15:00,480 As we increase the temperature higher and higher, at some 318 00:15:00,480 --> 00:15:03,050 point that delta s term is going to become greater than 319 00:15:03,050 --> 00:15:06,090 our delta h term, so at some point we're going to flip to 320 00:15:06,090 --> 00:15:09,730 where the reaction has now a positive delta g. 321 00:15:09,730 --> 00:15:12,510 So you can draw this into your graphs in your notes where 322 00:15:12,510 --> 00:15:15,240 we're going to start, in this case if delta g is negative 323 00:15:15,240 --> 00:15:18,270 where delta g starts negative, and then when it hits that 324 00:15:18,270 --> 00:15:20,180 threshold temperature, anything above that 325 00:15:20,180 --> 00:15:22,420 temperature is going to be positive. 326 00:15:22,420 --> 00:15:24,970 So if we had actual numbers we could plug those into our 327 00:15:24,970 --> 00:15:27,290 graph, but we should be able to understand what the general 328 00:15:27,290 --> 00:15:30,120 trend is going to be even in a hypothetical case where we're 329 00:15:30,120 --> 00:15:37,100 just dealing with is it positive or is it negative. 330 00:15:37,100 --> 00:15:40,420 So what we see is that below this temperature, we have a 331 00:15:40,420 --> 00:15:43,460 spontaneous reaction, and above this temperature we have 332 00:15:43,460 --> 00:15:45,330 a non-spontaneous reaction. 333 00:15:45,330 --> 00:15:48,460 That was flipped from the case where we had delta h and delta 334 00:15:48,460 --> 00:15:49,910 s both be positive. 335 00:15:49,910 --> 00:15:54,190 All right, so let's actually summarize what all of our four 336 00:15:54,190 --> 00:15:57,860 different scenarios could be if we're dealing with delta 337 00:15:57,860 --> 00:15:59,560 h's and delta s's. 338 00:15:59,560 --> 00:16:02,370 So to start with, why don't you tell me what you think if 339 00:16:02,370 --> 00:16:06,750 we have a reaction where we have a negative delta h and we 340 00:16:06,750 --> 00:16:10,510 have a positive delta s, do you think that this will be a 341 00:16:10,510 --> 00:16:13,360 reaction that's never spontaneous, always 342 00:16:13,360 --> 00:16:15,950 spontaneous, or will this be one of these cases where the 343 00:16:15,950 --> 00:16:18,090 spontaneity depends on temperature? 344 00:16:18,090 --> 00:16:40,430 All right, let's take 10 more seconds on this. 345 00:16:40,430 --> 00:16:41,090 OK, great. 346 00:16:41,090 --> 00:16:42,310 So most of you got it. 347 00:16:42,310 --> 00:16:45,730 So let's go back to the slides and think a little bit about 348 00:16:45,730 --> 00:16:46,760 why this is. 349 00:16:46,760 --> 00:16:50,820 So, this case is always spontaneous, because in this 350 00:16:50,820 --> 00:16:54,450 case we have a negative delta h, which contributes to a 351 00:16:54,450 --> 00:16:58,550 negative delta g, and a positive delta s, but since 352 00:16:58,550 --> 00:17:02,910 the equation says minus t delta s, that means a positive 353 00:17:02,910 --> 00:17:04,850 delta s is also going to contribute to a 354 00:17:04,850 --> 00:17:06,190 negative delta g. 355 00:17:06,190 --> 00:17:08,320 So regardless of what the temperature is, we're going to 356 00:17:08,320 --> 00:17:10,360 have a spontaneous reaction. 357 00:17:10,360 --> 00:17:13,600 So we say this is always spontaneous -- delta g is less 358 00:17:13,600 --> 00:17:15,580 than zero at all temperatures. 359 00:17:15,580 --> 00:17:18,840 All right. 360 00:17:18,840 --> 00:17:22,110 So let's look at the reverse case here where delta h is now 361 00:17:22,110 --> 00:17:25,830 greater than zero, and delta s is less than zero. 362 00:17:25,830 --> 00:17:29,120 Is this always, never, or sometimes spontaneous? 363 00:17:29,120 --> 00:17:32,470 STUDENT: [INAUDIBLE] 364 00:17:32,470 --> 00:17:34,230 PROFESSOR: Nope. 365 00:17:34,230 --> 00:17:38,770 So, this is going to be never spontaneous in this case. 366 00:17:38,770 --> 00:17:41,840 So for the same reason, because delta h is greater 367 00:17:41,840 --> 00:17:45,270 than zero, that contributes to a positive delta g, and delta 368 00:17:45,270 --> 00:17:47,790 s being negative also contributes to a 369 00:17:47,790 --> 00:17:48,900 positive delta g. 370 00:17:48,900 --> 00:17:52,550 So we'll say again, that this delta g is greater than zero 371 00:17:52,550 --> 00:17:53,850 at all temperatures. 372 00:17:53,850 --> 00:17:57,930 All right, so now we have a case where delta h is greater 373 00:17:57,930 --> 00:18:01,240 than zero and delta s is greater than zero as well. 374 00:18:01,240 --> 00:18:03,770 Is this going to be always, never, or sometimes 375 00:18:03,770 --> 00:18:04,550 spontaneous? 376 00:18:04,550 --> 00:18:06,360 STUDENT: Sometimes. 377 00:18:06,360 --> 00:18:07,170 PROFESSOR: Sometimes, good. 378 00:18:07,170 --> 00:18:09,720 So, even before thinking, you can just remember if the signs 379 00:18:09,720 --> 00:18:11,420 are the same, we can have a dependence 380 00:18:11,420 --> 00:18:13,050 on temperature here. 381 00:18:13,050 --> 00:18:16,710 So what we find is that this is sometimes spontaneous. 382 00:18:16,710 --> 00:18:18,900 So we can think about when this happens. 383 00:18:18,900 --> 00:18:22,810 Would it be when the actual temperature is greater or less 384 00:18:22,810 --> 00:18:24,560 than that threshold temperature? 385 00:18:24,560 --> 00:18:27,890 STUDENT: [INAUDIBLE] 386 00:18:27,890 --> 00:18:29,060 PROFESSOR: All right, so I heard a lot of people say 387 00:18:29,060 --> 00:18:29,710 greater than. 388 00:18:29,710 --> 00:18:32,640 It's greater than, because when it's greater than that 389 00:18:32,640 --> 00:18:35,970 threshold temperature, that means that the delta s term is 390 00:18:35,970 --> 00:18:38,200 the one that's going to actually sort of take over, 391 00:18:38,200 --> 00:18:41,520 it's going to be at some point greater than the delta h term. 392 00:18:41,520 --> 00:18:45,210 The delta h is going to be making the delta g positive. 393 00:18:45,210 --> 00:18:48,290 That means it would make it non-spontaneous, but once 394 00:18:48,290 --> 00:18:51,380 delta s gets large enough, that's going to override, and 395 00:18:51,380 --> 00:18:53,890 now we're going to have a negative delta g only when the 396 00:18:53,890 --> 00:18:57,010 temperature is above that threshold temperature. 397 00:18:57,010 --> 00:19:01,000 So similarly, when we see that delta h is negative and delta 398 00:19:01,000 --> 00:19:04,570 s is negative, this is also going to be sometimes 399 00:19:04,570 --> 00:19:08,230 spontaneous, and specifically, it will be spontaneous or 400 00:19:08,230 --> 00:19:12,030 delta g will be less than 0 when the temperature is less 401 00:19:12,030 --> 00:19:13,580 than that threshold temperature. 402 00:19:13,580 --> 00:19:16,980 All right, so you should be able to look at any situation 403 00:19:16,980 --> 00:19:20,770 where you have or you figure out or you calculate the 404 00:19:20,770 --> 00:19:23,540 enthalpy and the enthropy change in the reaction, and 405 00:19:23,540 --> 00:19:26,055 you should right away, before doing any calculations, be 406 00:19:26,055 --> 00:19:29,040 able to know is this always spontaneous, is this never 407 00:19:29,040 --> 00:19:31,530 spontaneous, or is this something where I'm going to 408 00:19:31,530 --> 00:19:33,680 need to really take temperature into 409 00:19:33,680 --> 00:19:34,710 consideration. 410 00:19:34,710 --> 00:19:37,580 And if I do, you can actually calculate what that 411 00:19:37,580 --> 00:19:39,840 temperature is going to be where you flip from 412 00:19:39,840 --> 00:19:42,880 spontaneous to non-spontaneous or vice versa. 413 00:19:42,880 --> 00:19:46,920 All right, so shifting gears a little bit, let's take a look 414 00:19:46,920 --> 00:19:49,580 at a few examples where it's important to think about 415 00:19:49,580 --> 00:19:52,050 thermodynamics and biological systems. 416 00:19:52,050 --> 00:19:55,020 So there's two things I particularly want to focus on. 417 00:19:55,020 --> 00:19:59,700 So the first is the idea of ATP coupled reactions. 418 00:19:59,700 --> 00:20:01,950 So this is really important, because when we're talking 419 00:20:01,950 --> 00:20:05,040 about biological reactions, a lot of the reactions that take 420 00:20:05,040 --> 00:20:08,170 place in our body are actually non-spontaneous, so 421 00:20:08,170 --> 00:20:11,710 energetically we figure that the delta g for this reaction 422 00:20:11,710 --> 00:20:13,100 is actually positive. 423 00:20:13,100 --> 00:20:16,990 So I just show a schematic biological reaction here where 424 00:20:16,990 --> 00:20:20,020 we have some molecule that's broken up into 425 00:20:20,020 --> 00:20:21,030 two building blocks. 426 00:20:21,030 --> 00:20:23,820 So let's say, for example, this has a delta g that's 427 00:20:23,820 --> 00:20:25,030 greater than zero. 428 00:20:25,030 --> 00:20:27,580 We need to think about how it is that our body can actually 429 00:20:27,580 --> 00:20:29,740 make this reaction happen. 430 00:20:29,740 --> 00:20:31,940 And the way that it does this is that it takes the 431 00:20:31,940 --> 00:20:35,910 hydrolysis of ATP, and the hydrolysis of ATP is what's 432 00:20:35,910 --> 00:20:40,740 called that it's coupled to this non-spontaneous reaction. 433 00:20:40,740 --> 00:20:44,150 And when you have a spontaneous reaction coupled 434 00:20:44,150 --> 00:20:47,830 to a non-spontaneous reaction, you can potentially drive the 435 00:20:47,830 --> 00:20:49,860 reaction in the forward direction. 436 00:20:49,860 --> 00:20:53,690 Remember the hydrolysis of ATP is spontaneous, so what we're 437 00:20:53,690 --> 00:20:57,100 doing in this case is we're taking a spontaneous reaction 438 00:20:57,100 --> 00:21:00,300 where we give off energy, and we're coupling it to a 439 00:21:00,300 --> 00:21:03,700 non-spontaneous reaction that requires energy. 440 00:21:03,700 --> 00:21:06,820 So what we will hope is when we add up the total energy 441 00:21:06,820 --> 00:21:10,200 changes between these two reactions, if the sum of those 442 00:21:10,200 --> 00:21:13,550 two is a negative delta g total, now we can have this 443 00:21:13,550 --> 00:21:17,990 whole process move in the forward direction. 444 00:21:17,990 --> 00:21:21,140 So let's take a look at thinking about what we can do 445 00:21:21,140 --> 00:21:22,600 in terms of a coupled reaction. 446 00:21:22,600 --> 00:21:24,870 The first thing we need to do if we're thinking about using 447 00:21:24,870 --> 00:21:29,060 the hydrolysis of ATP is actually calculate what the 448 00:21:29,060 --> 00:21:33,080 delta g for the reaction of ATP hydrolysis is, and I 449 00:21:33,080 --> 00:21:35,250 picked 310 kelvin because that's the 450 00:21:35,250 --> 00:21:36,830 temperature of our bodies. 451 00:21:36,830 --> 00:21:41,220 So again, this reaction is taking adenosine triphosphate 452 00:21:41,220 --> 00:21:44,020 that has three phosphate groups and hydrolyzing it, so 453 00:21:44,020 --> 00:21:47,680 reacting it with water to form ADP plus 454 00:21:47,680 --> 00:21:51,810 phosphate plus acid here. 455 00:21:51,810 --> 00:21:54,490 So I'll tell you that the delta h of this reaction is 456 00:21:54,490 --> 00:21:56,460 negative 24 kilojoules per mole. 457 00:21:56,460 --> 00:21:59,770 This is what we actually calculated in class on Friday. 458 00:21:59,770 --> 00:22:05,370 And that the delta s is plus 22 joules per k, per mole, so 459 00:22:05,370 --> 00:22:09,270 we can go ahead and calculate what this delta g is here. 460 00:22:09,270 --> 00:22:14,000 So the delta g of this overall reaction is just going to be, 461 00:22:14,000 --> 00:22:19,260 of course, delta h minus t delta s, and our delta h is 462 00:22:19,260 --> 00:22:27,340 minus 24 kilojoules per mole, and we'll subtract 310 463 00:22:27,340 --> 00:22:30,590 kelvin times 0 . 464 00:22:30,590 --> 00:22:38,720 022 kilojoules over k mole. 465 00:22:38,720 --> 00:22:43,020 So if we do this, what we end up getting is a delta g of 466 00:22:43,020 --> 00:22:47,280 negative 31 kilojoules per mole of ATP. 467 00:22:47,280 --> 00:22:52,130 All right, so again, this is good, we got a negative 468 00:22:52,130 --> 00:22:53,890 number, that's what we were expecting. 469 00:22:53,890 --> 00:22:56,990 So that means that we will have energy available to use 470 00:22:56,990 --> 00:22:59,670 if we hydrolyize ATP and couple it to 471 00:22:59,670 --> 00:23:01,990 another reaction here. 472 00:23:01,990 --> 00:23:04,450 So what we saw for our delta g, negative 31 473 00:23:04,450 --> 00:23:05,910 kilojoules per mole. 474 00:23:05,910 --> 00:23:09,300 All right, so let's talk about a reaction that is a ATP 475 00:23:09,300 --> 00:23:12,400 coupled reaction, and there's just tons of examples of this 476 00:23:12,400 --> 00:23:13,060 in our body. 477 00:23:13,060 --> 00:23:16,100 One I picked because it has to do with glucose, which we have 478 00:23:16,100 --> 00:23:20,350 talked so much about, is the conversion of glucose to 479 00:23:20,350 --> 00:23:22,520 glucose 6 p. 480 00:23:22,520 --> 00:23:25,950 So 6 p -- p just stands for a phosphate group here, so what 481 00:23:25,950 --> 00:23:29,190 we're doing is taking, if we numbered the glucose carbons, 482 00:23:29,190 --> 00:23:32,330 this would be carbon number six, and we're putting a 483 00:23:32,330 --> 00:23:34,910 phosphate group on carbon number six. 484 00:23:34,910 --> 00:23:38,430 So first of all, why would our body need to do this reaction? 485 00:23:38,430 --> 00:23:40,770 It turns out that glucose, we know that we 486 00:23:40,770 --> 00:23:42,220 use glucose for energy. 487 00:23:42,220 --> 00:23:47,610 Glucose is somewhat apolar, so it can actually move in and 488 00:23:47,610 --> 00:23:49,890 out of our cells, because our cell walls, 489 00:23:49,890 --> 00:23:51,220 those are very greasy. 490 00:23:51,220 --> 00:23:53,820 So what we're actually doing here is we're putting a 491 00:23:53,820 --> 00:23:56,490 charged molecule onto the glucose. 492 00:23:56,490 --> 00:23:59,280 There's two negative charges in a phosphate group, this is 493 00:23:59,280 --> 00:24:01,010 going to make sure that our glucose is now 494 00:24:01,010 --> 00:24:02,950 it's very polar molecule. 495 00:24:02,950 --> 00:24:05,400 And now that we have a very polar molecule, we're not 496 00:24:05,400 --> 00:24:07,240 going to be able to have the glucose move in 497 00:24:07,240 --> 00:24:08,460 and out of the cell. 498 00:24:08,460 --> 00:24:10,610 So we keep it in our cell by putting this 499 00:24:10,610 --> 00:24:12,000 phosphate group on it. 500 00:24:12,000 --> 00:24:14,380 That's why we want to do this reaction, but let's think 501 00:24:14,380 --> 00:24:17,620 about the energy of actually doing it. 502 00:24:17,620 --> 00:24:21,510 And it turns out that this requires energy. 503 00:24:21,510 --> 00:24:25,630 The delta g for this reaction is 17 kilojoules per mole. 504 00:24:25,630 --> 00:24:28,700 All right, so that could be a problem if our body had not 505 00:24:28,700 --> 00:24:32,180 come up with a way to solve it, which it has, and what it 506 00:24:32,180 --> 00:24:34,940 actually does is it couples this reaction here with the 507 00:24:34,940 --> 00:24:38,040 conversion of ATP into ADP. 508 00:24:38,040 --> 00:24:41,560 And we just calculated that that has a delta g of negative 509 00:24:41,560 --> 00:24:44,270 31 kilojoules per mole. 510 00:24:44,270 --> 00:24:47,190 So this means we can think about delta g total, and by 511 00:24:47,190 --> 00:24:50,780 total we mean of this overall process here. 512 00:24:50,780 --> 00:24:54,650 So that's just equal to 17, and we're adding it to the 513 00:24:54,650 --> 00:24:58,070 other delta g, which is negative 31. 514 00:24:58,070 --> 00:25:01,550 So we get an overall delta g for this process of negative 515 00:25:01,550 --> 00:25:03,270 14 kilojoules per mole. 516 00:25:03,270 --> 00:25:06,370 All right, so that's one really simple example just to 517 00:25:06,370 --> 00:25:09,850 illustrate to you how these ATP coupled reactions work. 518 00:25:09,850 --> 00:25:14,060 Some ATP coupled reactions require one molar equivalent 519 00:25:14,060 --> 00:25:16,200 of ATP, some require a lot more. 520 00:25:16,200 --> 00:25:17,860 But essentially, the idea is the same. 521 00:25:17,860 --> 00:25:21,170 We have this reaction that's energetically unfavorable that 522 00:25:21,170 --> 00:25:21,940 we couple with an 523 00:25:21,940 --> 00:25:24,000 energetically favorable reaction. 524 00:25:24,000 --> 00:25:26,190 So, quite literally, this is what we mean when we talk 525 00:25:26,190 --> 00:25:29,840 about ATP as being the energy currency for the cell. 526 00:25:29,840 --> 00:25:33,090 We spend some ATP in order to get these non-spontaneous 527 00:25:33,090 --> 00:25:34,210 reactions to go. 528 00:25:34,210 --> 00:25:37,940 All right, so I also want to talk to a little bit about 529 00:25:37,940 --> 00:25:39,260 hydrogen bonding. 530 00:25:39,260 --> 00:25:42,510 This also is a topic that deals with the thermodynamics, 531 00:25:42,510 --> 00:25:45,480 and it also is related to the ideas that we talked about 532 00:25:45,480 --> 00:25:46,980 before in terms of thinking about 533 00:25:46,980 --> 00:25:50,880 different types of bonds. 534 00:25:50,880 --> 00:25:54,260 So if we talk about a hydrogen bond, a hydrogen bond, first I 535 00:25:54,260 --> 00:25:57,430 just want to be really clear, is not a covalent bond. 536 00:25:57,430 --> 00:26:01,030 So this h x bond here is a covalent bond, this is not a 537 00:26:01,030 --> 00:26:02,100 hydrogen bond. 538 00:26:02,100 --> 00:26:05,290 But a hydrogen bond can form when you have a partial 539 00:26:05,290 --> 00:26:08,480 positive on a hydrogen, and you have what's called a 540 00:26:08,480 --> 00:26:12,010 hydrogen bond donor atom, which has a partial negative 541 00:26:12,010 --> 00:26:14,330 on it, and also has a lone pair. 542 00:26:14,330 --> 00:26:17,820 Typically this y atom will be on a separate molecule. 543 00:26:17,820 --> 00:26:21,450 And what happens is you have that Coulombic attraction 544 00:26:21,450 --> 00:26:24,150 between the partial positive on the hydrogen and the 545 00:26:24,150 --> 00:26:27,780 partial negative on this hydrogen bond donor atom. 546 00:26:27,780 --> 00:26:30,720 So what you form between them is a hydrogen bond, and you'll 547 00:26:30,720 --> 00:26:33,020 notice that I drew it as a dashed line. 548 00:26:33,020 --> 00:26:36,360 H bonds or hydrogen bonds are drawn either as dashed lines 549 00:26:36,360 --> 00:26:38,530 or as dotted lines, and they're done so to 550 00:26:38,530 --> 00:26:41,930 differentiate them from covalent bonds, which we draw 551 00:26:41,930 --> 00:26:44,530 with this straight line here. 552 00:26:44,530 --> 00:26:48,100 So let's think about what can form hydrogen bonds. 553 00:26:48,100 --> 00:26:51,130 First of all, there's a requirement for what this h x 554 00:26:51,130 --> 00:26:52,510 bond can be. 555 00:26:52,510 --> 00:26:56,060 It has to be a really electronegative atom in this x 556 00:26:56,060 --> 00:26:58,600 here, because we need to we need to form that partial 557 00:26:58,600 --> 00:27:01,950 positive on the hydrogen, which means that the atom that 558 00:27:01,950 --> 00:27:04,770 it's covalentally bonded to needs to be pulling away some 559 00:27:04,770 --> 00:27:07,290 of that electron density from the hydrogen, such that we 560 00:27:07,290 --> 00:27:10,470 have a delta positive on a hydrogen atom. 561 00:27:10,470 --> 00:27:14,260 Similarly, the same atoms are what can be the y here, so it 562 00:27:14,260 --> 00:27:17,420 can either be a nitrogen or an oxygen or a fluorine. 563 00:27:17,420 --> 00:27:20,320 So these are the only hydrogen bond donors 564 00:27:20,320 --> 00:27:23,420 that can form h bonds. 565 00:27:23,420 --> 00:27:26,190 And we can think about the reason for this and the reason 566 00:27:26,190 --> 00:27:27,250 is pretty straightforward. 567 00:27:27,250 --> 00:27:30,320 We need to have a small atom, but it also needs to be really 568 00:27:30,320 --> 00:27:31,620 electronegative. 569 00:27:31,620 --> 00:27:33,910 This should make sense because we need to have this partial 570 00:27:33,910 --> 00:27:37,190 negative charge here to have that Coulomb attraction. 571 00:27:37,190 --> 00:27:39,730 And the other requirement is that we definitely need to 572 00:27:39,730 --> 00:27:43,180 have that lone pair of electrons so it can actually 573 00:27:43,180 --> 00:27:47,990 interact in this hydrogen bond here. 574 00:27:47,990 --> 00:27:50,770 So let's take a look at an example of a hydrogen bond, 575 00:27:50,770 --> 00:27:52,850 and that's between two water molecules. 576 00:27:52,850 --> 00:27:56,220 So let's go ahead and re-draw our water molecules -- they're 577 00:27:56,220 --> 00:27:58,640 just kind of randomly oriented there. 578 00:27:58,640 --> 00:28:01,440 But let's re-draw them as if they were going to have a 579 00:28:01,440 --> 00:28:03,090 hydrogen bond between them. 580 00:28:03,090 --> 00:28:05,800 And one thing I want to point out about hydrogen bonds is 581 00:28:05,800 --> 00:28:09,230 that they're strongest when you actually have all three 582 00:28:09,230 --> 00:28:12,470 atoms in a straight line like this, because it keeps the 583 00:28:12,470 --> 00:28:15,270 dipoles the strongest, so the partial positive here and the 584 00:28:15,270 --> 00:28:17,050 partial negative here can interact. 585 00:28:17,050 --> 00:28:24,480 So let's re-draw our water molecules like that. 586 00:28:24,480 --> 00:28:27,820 So we have our first water molecule here, and let's say 587 00:28:27,820 --> 00:28:30,060 we're going to be talking about this hydrogen in terms 588 00:28:30,060 --> 00:28:30,940 of hydrogen bonding. 589 00:28:30,940 --> 00:28:33,400 Obviously, we have four hydrogens up there that can 590 00:28:33,400 --> 00:28:37,555 take place or take part in h bonding, but we're just going 591 00:28:37,555 --> 00:28:39,050 to focus on this one here. 592 00:28:39,050 --> 00:28:41,570 So that means we want our straight line to be something 593 00:28:41,570 --> 00:28:44,370 like this to our second oxygen. 594 00:28:44,370 --> 00:28:48,540 So let's draw the hydrogens on this water molecule as well. 595 00:28:48,540 --> 00:28:52,820 All right, so thinking about what the Lewis structure of 596 00:28:52,820 --> 00:28:55,295 water is, how many lone pairs do we have 597 00:28:55,295 --> 00:28:56,850 on this oxygen here? 598 00:28:56,850 --> 00:28:57,820 STUDENT: [INAUDIBLE] 599 00:28:57,820 --> 00:28:58,680 PROFESSOR: Two lone pairs, great. 600 00:28:58,680 --> 00:29:02,485 So let's draw those in, and let's draw these 601 00:29:02,485 --> 00:29:04,020 lone pairs in as well. 602 00:29:04,020 --> 00:29:07,250 All right, and we're going to be talking about these two 603 00:29:07,250 --> 00:29:10,540 atoms in terms of our hydrogen bond, so does this have a 604 00:29:10,540 --> 00:29:13,860 delta plus or a delta negative on it? 605 00:29:13,860 --> 00:29:14,570 Delta plus. 606 00:29:14,570 --> 00:29:18,515 So this is a polar bond here, this o h bond, so we have a 607 00:29:18,515 --> 00:29:23,280 delta plus on our hydrogen, and we have a delta minus on 608 00:29:23,280 --> 00:29:24,010 our oxygen. 609 00:29:24,010 --> 00:29:26,410 So these also have delta plusses, of course, and this 610 00:29:26,410 --> 00:29:28,360 also has a delta minus. 611 00:29:28,360 --> 00:29:31,070 So because these two can interact and they're lined up 612 00:29:31,070 --> 00:29:35,210 to do so, what we can do is we can draw a hydrogen bond right 613 00:29:35,210 --> 00:29:39,630 in between this hydrogen and this oxygen atom here. 614 00:29:39,630 --> 00:29:43,420 So again, I pointed out that, in fact, hydrogen bonds are 615 00:29:43,420 --> 00:29:46,300 not as strong as covalent bonds. 616 00:29:46,300 --> 00:29:49,210 Since they're not as strong, if you remember thinking about 617 00:29:49,210 --> 00:29:52,750 the relationship between bond strength and bond length, 618 00:29:52,750 --> 00:29:55,960 would you expect a hydrogen bond to be longer or shorter 619 00:29:55,960 --> 00:29:57,080 than a covalent bond? 620 00:29:57,080 --> 00:29:58,560 STUDENT: Longer. 621 00:29:58,560 --> 00:30:01,410 PROFESSOR: Good, OK, it should be longer here. 622 00:30:01,410 --> 00:30:04,920 So this is our longer bond because it's our weaker bond. 623 00:30:04,920 --> 00:30:07,150 They're not held together as tightly. 624 00:30:07,150 --> 00:30:12,080 All right, so this is an example of an intermolecular 625 00:30:12,080 --> 00:30:16,010 hydrogen bond, it's between two different molecules. 626 00:30:16,010 --> 00:30:18,650 Hopefully, you can also see that if we're focusing on any 627 00:30:18,650 --> 00:30:23,340 single one water molecule, we could also form hydrogen bonds 628 00:30:23,340 --> 00:30:26,160 between other water molecules with either of these two 629 00:30:26,160 --> 00:30:30,120 hydrogens as well, and also with our other lone pair here. 630 00:30:30,120 --> 00:30:33,230 So any water molecule can actually form four different 631 00:30:33,230 --> 00:30:34,300 hydrogen bonds. 632 00:30:34,300 --> 00:30:38,460 All right, so let's talk a little bit about these 633 00:30:38,460 --> 00:30:41,730 hydrogen bonds in terms of their bond enthalpies here. 634 00:30:41,730 --> 00:30:44,200 I said that they were weaker than covalent bonds, so let's 635 00:30:44,200 --> 00:30:45,720 look at a few comparisons here. 636 00:30:45,720 --> 00:30:50,380 So we can look at an h o hydrogen bond versus an h o 637 00:30:50,380 --> 00:30:52,510 covalent bond. 638 00:30:52,510 --> 00:30:55,170 And what we find is that the bond enthalpies in kilojoules 639 00:30:55,170 --> 00:31:00,450 per mole, it's 20 kilojoules per mole for a hydrogen bond 640 00:31:00,450 --> 00:31:05,430 versus 463 kilojoules per mole for a covalent bond. 641 00:31:05,430 --> 00:31:07,730 So we're not talking about just a little bit weaker, 642 00:31:07,730 --> 00:31:09,620 we're talking about much, much weaker for the 643 00:31:09,620 --> 00:31:11,940 hydrogen bonds here. 644 00:31:11,940 --> 00:31:15,120 Similarly, if we look at nitrogen, for nitrogen 645 00:31:15,120 --> 00:31:19,860 hydrogen bonds, if we have an o h bound to a nitrogen, 646 00:31:19,860 --> 00:31:22,390 that's 29 kilojoules per mole. 647 00:31:22,390 --> 00:31:26,080 If it's an n h hydrogen bound to a nitrogen, that's 14 648 00:31:26,080 --> 00:31:29,670 kilojoules per mole, and if we're talking about an h n 649 00:31:29,670 --> 00:31:32,550 covalent bond, now we're talking about 388 650 00:31:32,550 --> 00:31:34,780 kilojoules per mole. 651 00:31:34,780 --> 00:31:39,260 All right, so again what we see in these cases is that the 652 00:31:39,260 --> 00:31:41,960 hydrogen bonds are much, much weaker. 653 00:31:41,960 --> 00:31:46,440 They tend to be as low as 5% of what the covalent bond is, 654 00:31:46,440 --> 00:31:49,620 so this is actually much weaker than any kinds of bonds 655 00:31:49,620 --> 00:31:53,460 we have within molecules, but it's also the strongest type 656 00:31:53,460 --> 00:31:55,070 of bond that we can have between 657 00:31:55,070 --> 00:31:56,420 two different molecules. 658 00:31:56,420 --> 00:31:59,800 And one thing that you find as in the case of water, is when 659 00:31:59,800 --> 00:32:02,650 you have lots of hydrogen bonds between molecules it 660 00:32:02,650 --> 00:32:06,510 changes the property of, for example, water here. 661 00:32:06,510 --> 00:32:08,940 It makes the boiling point much, much higher than you 662 00:32:08,940 --> 00:32:12,120 might expect if you looked at the other properties, because 663 00:32:12,120 --> 00:32:15,380 you have to actually break apart all of these individual 664 00:32:15,380 --> 00:32:17,510 hydrogen bonds, and even though it's not that much 665 00:32:17,510 --> 00:32:20,560 energy for one individual hydrogen bond, once you get a 666 00:32:20,560 --> 00:32:23,070 huge number of hydrogen bonds, you're talking about huge 667 00:32:23,070 --> 00:32:25,440 energies here. 668 00:32:25,440 --> 00:32:28,280 So let's look at some examples of where we see hydrogen bonds 669 00:32:28,280 --> 00:32:30,180 in biological processes. 670 00:32:30,180 --> 00:32:33,060 First thinking about proteins, they're absolutely hugely 671 00:32:33,060 --> 00:32:34,730 important in proteins. 672 00:32:34,730 --> 00:32:37,420 And in proteins, we're talking about a molecule that's so 673 00:32:37,420 --> 00:32:40,980 large that you don't just see h bonds in between two 674 00:32:40,980 --> 00:32:44,960 different molecules, what you actually see is what's called 675 00:32:44,960 --> 00:32:47,970 intramolecular hydrogen bonding, so hydrogen bonding 676 00:32:47,970 --> 00:32:50,190 within the protein molecule itself. 677 00:32:50,190 --> 00:32:52,875 And hydrogen bonding is incredibly important, it, in 678 00:32:52,875 --> 00:32:56,670 fact, is the shape of a protein, which I'm just 679 00:32:56,670 --> 00:32:59,390 showing an example here, histone deacetylase, and the 680 00:32:59,390 --> 00:33:06,350 histone deacetylase molecule actually has just countless 681 00:33:06,350 --> 00:33:09,970 hydrogen bonds in here, and the hydrogen bonds largely 682 00:33:09,970 --> 00:33:11,930 govern the shape of the molecule. 683 00:33:11,930 --> 00:33:14,790 So, if you're thinking about any protein that has a shape, 684 00:33:14,790 --> 00:33:17,780 so for example, we see these ribbon structures here, 685 00:33:17,780 --> 00:33:21,080 they're called alpha helices, you can see some arrows, which 686 00:33:21,080 --> 00:33:24,150 are just like a sheet structure or beta sheets, that 687 00:33:24,150 --> 00:33:28,320 actual shape is stabilized and formed in the first place by 688 00:33:28,320 --> 00:33:32,000 just many, many hydrogen bonds within the protein being in a 689 00:33:32,000 --> 00:33:34,890 correct orientation that allows for this particular 690 00:33:34,890 --> 00:33:36,410 shape to be stabilized. 691 00:33:36,410 --> 00:33:39,040 And we know that shape is so important when we're talking 692 00:33:39,040 --> 00:33:42,090 about proteins, and that's because the actual shape of 693 00:33:42,090 --> 00:33:44,610 the protein governs the interaction with other 694 00:33:44,610 --> 00:33:46,810 proteins or other small molecules. 695 00:33:46,810 --> 00:33:49,480 The shape of the protein is very important in terms of 696 00:33:49,480 --> 00:33:53,660 positioning the specific atoms that are involved in the 697 00:33:53,660 --> 00:33:57,150 chemistry that the enzyme carries out being in the 698 00:33:57,150 --> 00:33:58,560 correct position. 699 00:33:58,560 --> 00:34:00,230 So hydrogen bonds are very 700 00:34:00,230 --> 00:34:02,170 important in terms of proteins. 701 00:34:02,170 --> 00:34:04,570 We can also think of them in terms of sugars or 702 00:34:04,570 --> 00:34:05,060 polysaccharides. 703 00:34:05,060 --> 00:34:08,890 This is a nice dramatic example, thinking about the 704 00:34:08,890 --> 00:34:11,740 structure of a protein is a very microscopic way of 705 00:34:11,740 --> 00:34:13,180 thinking about hydrogen bonding. 706 00:34:13,180 --> 00:34:16,130 If we want to take it to the macroscopic level, we can talk 707 00:34:16,130 --> 00:34:18,470 about h bonding in trees. 708 00:34:18,470 --> 00:34:21,710 So if we're talking about trees, trees are made up -- a 709 00:34:21,710 --> 00:34:27,060 major part of a tree is the polysaccharide cellulose. 710 00:34:27,060 --> 00:34:29,150 So that's shown right here. 711 00:34:29,150 --> 00:34:31,230 Cellulose is a chain of glucose 712 00:34:31,230 --> 00:34:33,050 molecules linked together. 713 00:34:33,050 --> 00:34:36,180 It can be as many as 1,000 or more glucose molecules linked 714 00:34:36,180 --> 00:34:37,730 together covalentally. 715 00:34:37,730 --> 00:34:41,095 But what actually happens to those individual chains is 716 00:34:41,095 --> 00:34:43,520 that if you look at this picture here, all those dotted 717 00:34:43,520 --> 00:34:45,710 lines are actually hydrogen bonds. 718 00:34:45,710 --> 00:34:49,450 So the individual molecules are in a much larger chain of 719 00:34:49,450 --> 00:34:52,640 hydrogen bonded molecules and they're incredibly rigid. 720 00:34:52,640 --> 00:34:55,060 You have so many hydrogen bonds that now it takes a huge 721 00:34:55,060 --> 00:34:56,910 amount of energy to break all of these. 722 00:34:56,910 --> 00:34:59,010 So that accounts for why wood is such 723 00:34:59,010 --> 00:35:00,630 a hard, solid material. 724 00:35:00,630 --> 00:35:03,870 And also, I mean for any plant, and tree is the most 725 00:35:03,870 --> 00:35:07,000 dramatic, the structure of a tree, the macroscopic 726 00:35:07,000 --> 00:35:09,150 structure, can be explained by hydrogen bonding. 727 00:35:09,150 --> 00:35:11,940 It's hydrogen bonds that keep all of those molecules 728 00:35:11,940 --> 00:35:14,870 together in the forest and in trees. 729 00:35:14,870 --> 00:35:15,190 All right. 730 00:35:15,190 --> 00:35:18,220 So, that's proteins, that thinking about sugars. 731 00:35:18,220 --> 00:35:20,180 We can also talk about the importance of 732 00:35:20,180 --> 00:35:23,180 hydrogen bonding in DNA. 733 00:35:23,180 --> 00:35:25,730 So if you think about the characteristic DNA double 734 00:35:25,730 --> 00:35:30,600 helix, hydrogen bonds, we have in a double helix we have two 735 00:35:30,600 --> 00:35:34,140 strands of DNA that you can see are intertwined here. 736 00:35:34,140 --> 00:35:37,120 And those two strands are actually held together by 737 00:35:37,120 --> 00:35:38,860 hydrogen bonding. 738 00:35:38,860 --> 00:35:41,420 So specifically, it's hydrogen bonding between what are 739 00:35:41,420 --> 00:35:44,530 called complimentary bases within the DNA. 740 00:35:44,530 --> 00:35:48,810 So, for example, if we look at guanine and cytosine here, you 741 00:35:48,810 --> 00:35:52,430 can see that there's several places, when they're lined up 742 00:35:52,430 --> 00:35:54,730 like this, where we can have hydrogen bonds. 743 00:35:54,730 --> 00:35:58,080 So in terms of thinking about how these two molecules are 744 00:35:58,080 --> 00:36:00,890 lined up, how many h bonds would you expect 745 00:36:00,890 --> 00:36:04,020 between these two bases? 746 00:36:04,020 --> 00:36:06,810 Yeah, so it's lined up that there are three 747 00:36:06,810 --> 00:36:08,460 that can form here. 748 00:36:08,460 --> 00:36:13,840 So we have an o h hydrogen bond, an h n, and then an h o 749 00:36:13,840 --> 00:36:16,260 hydrogen bond that can form. 750 00:36:16,260 --> 00:36:18,470 So the other side of complimentary based pairs are 751 00:36:18,470 --> 00:36:22,100 AT base pairs, and if you look at the way these are lined up 752 00:36:22,100 --> 00:36:24,610 here, how many hydrogen bonds can form between 753 00:36:24,610 --> 00:36:27,050 A and T base pairs? 754 00:36:27,050 --> 00:36:29,620 Yup, so it it's two that we can form here. 755 00:36:29,620 --> 00:36:34,750 We have one between the h o, one between the n h, and you 756 00:36:34,750 --> 00:36:38,970 see that we've lost this nitrogen, this n h group here, 757 00:36:38,970 --> 00:36:41,410 and first of all this would be too far apart anyway for 758 00:36:41,410 --> 00:36:45,150 hydrogen bond, but we can't have a hydrogen bond form when 759 00:36:45,150 --> 00:36:46,920 we have a carbon h. 760 00:36:46,920 --> 00:36:50,955 Remember, it has to either be a nitrogen or an oxygen or a 761 00:36:50,955 --> 00:36:53,870 fluorine because those are the atoms that are going to pull 762 00:36:53,870 --> 00:36:57,190 away enough electron density from the hydrogen to give it a 763 00:36:57,190 --> 00:36:59,430 partial positive charge that it needs. 764 00:36:59,430 --> 00:37:02,810 So in terms of thinking about DNA, this is a really neat 765 00:37:02,810 --> 00:37:05,780 case to consider the actual thermodynamics or the bond 766 00:37:05,780 --> 00:37:09,150 enthalpies of the hydrogen bonds, because, of course, 767 00:37:09,150 --> 00:37:11,710 does not always stay in it's double helix -- when we're 768 00:37:11,710 --> 00:37:15,150 talking about transcription, we actually need to unzip or 769 00:37:15,150 --> 00:37:18,490 separate this helix into its two strands, so each 770 00:37:18,490 --> 00:37:20,720 individual strand can be copied. 771 00:37:20,720 --> 00:37:23,380 So it's really important that hydrogen bonds are strong 772 00:37:23,380 --> 00:37:26,750 enough to hold the DNA double strand together, but that 773 00:37:26,750 --> 00:37:29,860 they're not so strong that when you actually pull the 774 00:37:29,860 --> 00:37:33,200 hydrogen bonds apart to open up the double strand, that you 775 00:37:33,200 --> 00:37:35,500 actually, you don't want to break all of a covalent bonds 776 00:37:35,500 --> 00:37:37,460 in DNA as well. 777 00:37:37,460 --> 00:37:40,740 All right, so that's all we're going to say in terms of 778 00:37:40,740 --> 00:37:43,390 thinking about thermodynamics and biological systems. 779 00:37:43,390 --> 00:37:45,690 Hold on a sec, we have plenty of time left. 780 00:37:45,690 --> 00:37:47,730 I want to go over a few clicker questions. 781 00:37:47,730 --> 00:37:50,430 I didn't want to give you too long set of notes, because 782 00:37:50,430 --> 00:37:52,490 we're switching over to Professor Drennan's going to 783 00:37:52,490 --> 00:37:55,340 start lecturing on Friday, so I thought maybe I shouldn't 784 00:37:55,340 --> 00:37:58,150 have to have her finish up with my notes. 785 00:37:58,150 --> 00:38:01,150 So, Professor Drennan will continue talking about 786 00:38:01,150 --> 00:38:04,260 thermodynamics and thinking about equilibrium, and then 787 00:38:04,260 --> 00:38:06,450 she'll transition that into talking about kinetics. 788 00:38:06,450 --> 00:38:09,410 So that's actually what's going to start happening after 789 00:38:09,410 --> 00:38:11,610 the exam on Wednesday. 790 00:38:11,610 --> 00:38:14,320 But let's take a look, since we do you have a little bit of 791 00:38:14,320 --> 00:38:18,460 time left, at a few clicker questions just to make sure 792 00:38:18,460 --> 00:38:21,450 everyone has caught what we've gone over so far. 793 00:38:21,450 --> 00:38:25,210 And a few reviews for the exam, and I will have this 794 00:38:25,210 --> 00:38:29,480 clicker question or one of the ones following, be a clicker 795 00:38:29,480 --> 00:38:32,410 question quiz, so make sure you are still answering these 796 00:38:32,410 --> 00:38:34,620 questions here. 797 00:38:34,620 --> 00:38:37,740 So the first one covers what we went over today, so I want 798 00:38:37,740 --> 00:38:40,230 you to tell me and try to not look back in your notes for 799 00:38:40,230 --> 00:38:44,710 this, for a reaction where you have a positive enthalpy and a 800 00:38:44,710 --> 00:38:48,720 negative entropy change, would you expect this reaction to be 801 00:38:48,720 --> 00:38:51,070 never, always, or sometimes spontaneous? 802 00:38:51,070 --> 00:38:53,780 So let's take 10 seconds on that, that 803 00:38:53,780 --> 00:39:05,700 should be pretty fast. 804 00:39:05,700 --> 00:39:06,650 OK, excellent. 805 00:39:06,650 --> 00:39:08,220 90% is very good. 806 00:39:08,220 --> 00:39:10,990 So this should never be spontaneous, because we both 807 00:39:10,990 --> 00:39:14,650 have the entropy and the enthalpy term contributing to 808 00:39:14,650 --> 00:39:16,010 a positive delta g. 809 00:39:16,010 --> 00:39:19,710 All right, let's shift gears to a couple of questions that 810 00:39:19,710 --> 00:39:21,060 are review for the exam. 811 00:39:21,060 --> 00:39:23,110 So hopefully, these will all be very 812 00:39:23,110 --> 00:39:25,230 straightforward for you now. 813 00:39:25,230 --> 00:39:28,460 So I want you to think about bond lengths here, and tell me 814 00:39:28,460 --> 00:39:32,760 which molecule contains the shorter nitrogen oxygen bond. 815 00:39:32,760 --> 00:39:38,120 So, we're comparing n o minus 1, and n o 2 minus 1. 816 00:39:38,120 --> 00:39:40,180 So if you wanted to get started on this problem, 817 00:39:40,180 --> 00:39:43,200 what's the first thing you should do? 818 00:39:43,200 --> 00:39:44,860 Yeah, draw some Lewis structures. 819 00:39:44,860 --> 00:39:47,850 So that might be a good place to start in 820 00:39:47,850 --> 00:39:56,110 thinking about this. 821 00:39:56,110 --> 00:39:58,850 So I'll do that up here as well, but don't look if you 822 00:39:58,850 --> 00:40:01,310 want to try it on your own. 823 00:40:01,310 --> 00:40:04,130 Remember, the quiz points come for answering, not 824 00:40:04,130 --> 00:40:05,260 for getting it right. 825 00:40:05,260 --> 00:40:26,590 So try doing it on your own here. 826 00:40:26,590 --> 00:40:34,340 All right, let's go ahead and take 10 more 827 00:40:34,340 --> 00:40:49,270 seconds on this one. 828 00:40:49,270 --> 00:40:49,890 OK, hold on. 829 00:40:49,890 --> 00:40:51,440 Don't show the correct answer. 830 00:40:51,440 --> 00:40:53,020 You know we're going to re-poll and give you a little 831 00:40:53,020 --> 00:40:55,320 bit more time to see if we come to a consensus here. 832 00:40:55,320 --> 00:40:58,270 So let's take another 30 seconds or so on this while I 833 00:40:58,270 --> 00:41:32,430 finish drawing these up here. 834 00:41:32,430 --> 00:41:33,060 OK, good. 835 00:41:33,060 --> 00:41:38,410 So we've got 81% have it correct that in the n o minus 836 00:41:38,410 --> 00:41:41,500 1 bond here, this is going to be a double bond. 837 00:41:41,500 --> 00:41:46,200 STUDENT: What if [INAUDIBLE]. 838 00:41:46,200 --> 00:41:47,090 PROFESSOR: No. 839 00:41:47,090 --> 00:41:49,440 It would be a double bond here. 840 00:41:49,440 --> 00:41:51,880 So, if you follow just the Lewis structure rules, and you 841 00:41:51,880 --> 00:41:54,970 go ahead, you'll find that we end up having four bonding 842 00:41:54,970 --> 00:41:56,770 electrons available. 843 00:41:56,770 --> 00:42:00,370 So you can just go ahead and plug those in, and you end up 844 00:42:00,370 --> 00:42:01,820 with this many left. 845 00:42:01,820 --> 00:42:08,120 If we had a triple bond, then we would end up having more 846 00:42:08,120 --> 00:42:10,590 bonding -- or we would end up using more electrons than we 847 00:42:10,590 --> 00:42:12,190 have available for bonding here. 848 00:42:12,190 --> 00:42:18,060 All right, so we have a double bond in the case of n o. 849 00:42:18,060 --> 00:42:22,180 What is this bond that we have here? 850 00:42:22,180 --> 00:42:24,170 Yup, so it's actually a 1 . 851 00:42:24,170 --> 00:42:24,860 5 bond. 852 00:42:24,860 --> 00:42:26,600 What is this bond that we have here? 853 00:42:26,600 --> 00:42:28,360 STUDENT: [INAUDIBLE] 854 00:42:28,360 --> 00:42:31,260 PROFESSOR: All right, so we have two 1 . 855 00:42:31,260 --> 00:42:33,780 5 bonds in this case, how come these are 1. 856 00:42:33,780 --> 00:42:35,630 5 and not, for example, a double bond? 857 00:42:35,630 --> 00:42:36,840 STUDENT: [INAUDIBLE] 858 00:42:36,840 --> 00:42:38,580 PROFESSOR: Great, so that's the key. 859 00:42:38,580 --> 00:42:40,790 Even though we have double bonds in each of these 860 00:42:40,790 --> 00:42:44,350 structures, in this case here, we have resonance, so it's 861 00:42:44,350 --> 00:42:46,460 turns out to be, in fact, two 1 . 862 00:42:46,460 --> 00:42:47,620 5 bonds. 863 00:42:47,620 --> 00:42:50,410 So since this is the double bond, it's going to be the 864 00:42:50,410 --> 00:42:51,500 stronger bond. 865 00:42:51,500 --> 00:42:53,620 Since it's the stronger bond, it's also going to be the 866 00:42:53,620 --> 00:42:54,670 shorter bond. 867 00:42:54,670 --> 00:42:57,060 So make sure you can make those relationships between 868 00:42:57,060 --> 00:43:01,160 bond strength and bond length and thinking about if you have 869 00:43:01,160 --> 00:43:02,940 resonance or you don't have resonance in 870 00:43:02,940 --> 00:43:04,260 a particular situation. 871 00:43:04,260 --> 00:43:08,510 All right, let's try one more clicker question here. 872 00:43:08,510 --> 00:43:10,950 We'll have this be the last one for you. 873 00:43:10,950 --> 00:43:13,140 So this is one that hopefully you'll all get right, because 874 00:43:13,140 --> 00:43:15,670 we've gone over this again and again both in class and 875 00:43:15,670 --> 00:43:16,670 recitation. 876 00:43:16,670 --> 00:43:19,390 So let's talk about the hybridization of the specific 877 00:43:19,390 --> 00:43:23,230 carbon and oxygen atoms in ATP, so I want you to go ahead 878 00:43:23,230 --> 00:43:25,970 and tell me what these hyrbridizations are 879 00:43:25,970 --> 00:43:49,550 for these two atoms. 880 00:43:49,550 --> 00:43:51,580 OK, let's take 10 more seconds here, get those 881 00:43:51,580 --> 00:44:03,800 final answers in. 882 00:44:03,800 --> 00:44:06,790 OK, good, so 83% of you got this. 883 00:44:06,790 --> 00:44:08,100 Let's take a look at why. 884 00:44:08,100 --> 00:44:13,250 So carbon a is bonded to three things, but it is bonded to no 885 00:44:13,250 --> 00:44:17,580 lone pairs, so we need to have it be three hybrid orbitals, 886 00:44:17,580 --> 00:44:19,440 so it's s p 2. 887 00:44:19,440 --> 00:44:23,260 And oxygen is bonded to two atoms plus two lone paris. 888 00:44:23,260 --> 00:44:26,220 So we need four hybrid orbitals or s p 3. 889 00:44:26,220 --> 00:44:28,490 All right, so this was your quiz question, so as long as 890 00:44:28,490 --> 00:44:31,600 you answered it, you got your quiz points for today. 891 00:44:31,600 --> 00:44:35,780 And you can get going a little bit early today and finish 892 00:44:35,780 --> 00:44:37,850 studying for this exam.