00:00:05.190 --> 00:00:09.850 In this problem we need to do an energy balance on our reacting system in order to determine 00:00:09.850 --> 00:00:14.520 the amount of heat that needs to be removed from that reacting system, and in order to 00:00:14.520 --> 00:00:20.630 do this we are going to have to use information about the heats of formation of the compounds 00:00:20.630 --> 00:00:26.160 at 25 degrees C, which is standard, to tell us something about what the energy balance 00:00:26.160 --> 00:00:31.780 should be at an elevated temperature of 350 degrees C. To do this one we will just have 00:00:31.780 --> 00:00:38.629 to utilize the fact that the enthalpy is a state function. Now the problem statement 00:00:38.629 --> 00:00:45.629 tells us we are feeding 100 moles of nitrogen, and 300 moles of hydrogen, that is being 00:00:46.519 --> 00:00:53.519 fed to a reactor that operates at 350 degrees C and a conversion of 75%. So that fractional 00:00:53.569 --> 00:01:00.179 conversion means for the stoichiometric feed of 1 to 3. So we say that it is stoichiometric 00:01:00.179 --> 00:01:06.210 because the stoichiometry of the reaction equation is also 1 to 3. So that 75% conversion 00:01:06.210 --> 00:01:11.890 means that we will consume 75 moles of nitrogen, therefore we will have 25 moles of nitrogen 00:01:11.890 --> 00:01:17.610 left over. We will still have the stoichiometric amount of hydrogen, which is 75 moles. Then 00:01:17.610 --> 00:01:23.130 if we've consumed 75 moles of nitrogen, we should have formed twice the amount that amount of ammonia, 00:01:23.130 --> 00:01:30.130 so twice 75 is 150 moles of ammonia that are formed. Now we have specified our reactant 00:01:30.150 --> 00:01:36.490 and product streams, and again this reaction is occurring at 350 degrees C. We need to do an energy 00:01:36.490 --> 00:01:41.770 balance to determine how much heat needs to be added to our system, and in this case it will 00:01:41.770 --> 00:01:47.080 be removed, so we are going to expect a negative sign, following the convention that positive 00:01:47.080 --> 00:01:52.630 heat is added, to actually have this reaction operate isothermally at 350 00:01:52.630 --> 00:01:59.550 degrees C, where we are both feeding our reactants at 350 degrees C, and getting out products 00:01:59.550 --> 00:02:06.550 at 350 degrees C. So an energy balance for this open system is just that the amount of heat 00:02:07.520 --> 00:02:14.520 that is required, per mole of nitrogen for example that's fed, is going to be equal to the 00:02:15.360 --> 00:02:21.550 change in enthalpy for this stream. So we just need to be able to evaluate this enthalpy 00:02:21.550 --> 00:02:27.080 change, in order to evaluate the amount of heat that needs to be removed. So the concept 00:02:27.080 --> 00:02:32.750 behind how we are going to do this is shown schematically on this plot. So we have information that 00:02:32.750 --> 00:02:37.690 tells us we have enthalpies of formation of reactants and products, as we will talk about 00:02:37.690 --> 00:02:41.520 it in a minute, that tell us what the enthalpy change should be going from reactants 00:02:41.520 --> 00:02:47.520 to products at a temperature of 25 degrees C. And so when we are ultimately interested 00:02:47.520 --> 00:02:54.520 in is that same enthalpy change going from reactants to products at 350 degrees C. This 00:02:54.650 --> 00:03:00.180 is ultimately the information that we want. This is the heat of reaction that we are interested 00:03:00.180 --> 00:03:04.760 in. We can take advantage of the fact that enthalpy is a state function, and so we can 00:03:04.760 --> 00:03:11.280 choose any path we want to get from this point on the left at 350 degrees C to this open circle 00:03:11.280 --> 00:03:16.380 on the right. So the path we will choose. We have enough information to use, is that 00:03:16.380 --> 00:03:23.380 we can choose to cool down the reactants from 350 degrees C to 25 degrees C, have the reaction 00:03:24.630 --> 00:03:31.630 take place at 25 degrees C, and then heat up the products at 75% conversion from 25 00:03:31.849 --> 00:03:37.770 degrees C to the 350 degrees C. So we we just need to sequentially count these different 00:03:37.770 --> 00:03:43.380 enthalpic contributions. So here we will sort of color code these contributions. So this 00:03:43.380 --> 00:03:49.170 last leg which we will start with first of heating up the products to 25 degrees C to 00:03:49.170 --> 00:03:55.290 350 degrees C, we will have in blue ink. So this will be equal to the number moles of 00:03:55.290 --> 00:04:02.160 nitrogen in the product stream. So I'm going to use the final here, subscript f, of nitrogen, times 00:04:02.160 --> 00:04:08.550 the heat capacity for that nitrogen times the final temperature minus the reference 00:04:08.550 --> 00:04:15.550 temperature of 25 degrees C, and we have similar terms for hydrogen and ammonia that are in the 00:04:16.470 --> 00:04:21.570 product streams. Well we're basically heating up 25 moles of nitrogen, 75 moles 00:04:21.570 --> 00:04:28.570 of hydrogen, and 150 moles of ammonia from a reference temperature of 25 degrees C, so this 00:04:30.220 --> 00:04:37.220 is T ref, to a final temperature of 350 degrees C, and we use a constant heat capacity value 00:04:37.470 --> 00:04:43.290 for these ideal gases according to the heat capacities that are given in the problem statement. 00:04:43.290 --> 00:04:49.780 Alright so that is that last leg of heating up the product gasses. The leg before that we 00:04:49.780 --> 00:04:56.430 need to actually conduct the reaction, so we expect that that component will correspond to 00:04:56.430 --> 00:05:02.930 the enthalpy of the products minus the enthalpy of the reactants. So this leg is equal to 00:05:02.930 --> 00:05:09.930 the enthalpy of reaction per mole of ammonia that forms, so the enthalpy that is required or 00:05:09.990 --> 00:05:16.460 liberated when one mole of ammonia is formed, times 150 moles of ammonia that are actually formed 00:05:16.460 --> 00:05:23.460 in this reaction. Now the enthalpy of reaction is equal to the products minus the reactants. 00:05:23.919 --> 00:05:30.919 Ammonia has a heat of formation as given in the problem statement of -46 kJ/mol, and then molecular 00:05:32.530 --> 00:05:37.919 nitrogen and hydrogen have heats of formation of 0, because those are the states found in nature 00:05:37.919 --> 00:05:41.410 of hydrogen and nitrogen. If you look it up in the back of a textbook you 00:05:41.410 --> 00:05:47.229 will find that they have heats of formation of 0. So then our heat of reaction for producing ammonia from 00:05:47.229 --> 00:05:53.510 those elements is equal to -46 kJ/mol, the heat of formation. So we can just plug that 00:05:53.510 --> 00:05:59.139 in for this term. For the final term we will show in black here, and here we just need 00:05:59.139 --> 00:06:05.580 to cool down the product stream. So the initial number of moles of nitrogen, which is 100, 00:06:05.580 --> 00:06:10.990 and the initial number of moles of hydrogen, which is 300, times their heat capacities times the 00:06:10.990 --> 00:06:16.350 temperature difference here going form 25 degrees C at the end state minus the initial state 00:06:16.350 --> 00:06:23.350 of 350 degrees C. Our heat capacity information is given above, that it is 29 J/mol*K for hydrogen 00:06:25.080 --> 00:06:32.080 and nitrogen, 36 j/mol*K for ammonia. So now we have been able to specify all of our variables 00:06:32.979 --> 00:06:37.840 in order to calculate the enthalpy change according to this path, because we also have 00:06:37.840 --> 00:06:41.509 the heat of reaction. We also have the have all the number of moles. We have all the temperatures, 00:06:41.509 --> 00:06:48.509 and we have all heat capacities. So when we evaluate this, we get that delta H is equal to -8.0x10^3 00:06:50.199 --> 00:06:55.659 kJ to do this conversion of 100 moles of nitrogen and 300 moles of hydrogen and get 75% 00:06:55.659 --> 00:07:00.100 conversion. That is how much heat, according to the negative sign we need to remove for 00:07:00.100 --> 00:07:05.500 this exothermic reaction, and now I should have mentioned this in my initial set up to 00:07:05.500 --> 00:07:10.550 the problem statement that these 100 moles and 300 moles were actually per hour, and so we 00:07:10.550 --> 00:07:15.570 are going to need to remove heat over time. For the heat removed we just need to 00:07:15.570 --> 00:07:21.650 remove this amount of heat per hour, which we can do a simple unit conversion of hours 00:07:21.650 --> 00:07:28.650 to seconds to convert to kJ/s, or in another words kW. We will need to remove 2.2 kW to 00:07:29.460 --> 00:07:35.100 processes this amount of material. So to summarize this problem what we saw is that we want to calculate 00:07:35.100 --> 00:07:39.270 the amount of heat required that is just related to the enthalpy change associated 00:07:39.270 --> 00:07:44.479 with the reaction processes from the heats of formation at 25 degree C of the reactants 00:07:44.479 --> 00:07:50.710 and products. We know that delta H at 25 degrees C, and then we can just use heat capacity 00:07:50.710 --> 00:07:56.350 information, and here things are simple, because we had heat capacities that can be considered 00:07:56.350 --> 00:08:00.979 constant, in order to basically cool down the reactants to the reaction temperature and 00:08:00.979 --> 00:08:05.830 heat up the products to the reactant temperature. This allows us to calculate the actually enthalpy 00:08:05.830 --> 00:08:11.110 required at a temperature other then 25 degrees C, and therefore determine the heat load.
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