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Enthalpy Change of Reaction & Formation - Thermochemistry & Calorimetry Practice Problems
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00:00:00.000 in this video we're going to focus on 00:00:02.62900:00:02.639 how to calculate the enthalpy change of 00:00:05.59900:00:05.609 a reaction 00:00:07.00000:00:07.010 so consider the combustion reaction of 00:00:10.74900:00:10.759 ethanol c2h5oh in a combustion reaction 00:00:17.26900:00:17.279 you have a hydrocarbon which usually 00:00:22.37000:00:22.380 reacts with o2 the products of a 00:00:25.46000:00:25.470 combustion reaction is always going to 00:00:26.90000:00:26.910 be co2 and water now go ahead and pause 00:00:32.81000:00:32.820 the video and balance the reaction so 00:00:36.83000:00:36.840 notice that we have two carbon atoms on 00:00:39.29000:00:39.300 the left side whenever you're balanced 00:00:41.06000:00:41.070 in the combustion reaction and balance 00:00:43.10000:00:43.110 the carbon atoms first after that move 00:00:45.97900:00:45.989 on to the hydrogen atoms and save the 00:00:48.68000:00:48.690 oxygen atoms for last since we have two 00:00:51.65000:00:51.660 carbon atoms on the left side we need to 00:00:54.61900:00:54.629 put a two in front of co2 now notice 00:00:58.72900:00:58.739 that we have six hydrogen atoms on the 00:01:01.67000:01:01.680 left side 6 divided by 2 is 3 therefore 00:01:05.24000:01:05.250 we need to put a 3 in front of h2o now 00:01:09.53000:01:09.540 notice that we have a total of 7 oxygen 00:01:12.98000:01:12.990 atoms on the right side we have 4 from 00:01:15.53000:01:15.540 the two co2 molecules and 3 from the 3 00:01:19.03900:01:19.049 h2o molecules we also have 1 in ethanol 00:01:24.46000:01:24.470 now 4 plus 3 is 7 and 7 minus 1 is 6 00:01:28.88000:01:28.890 so we need 6 oxygen atoms from o2 6 00:01:34.37000:01:34.380 divided by 2 is 3 so we need to put a 3 00:01:36.35000:01:36.360 in front of o2 so now the reaction is 00:01:39.67900:01:39.689 balanced now let's say if you're given D 00:01:47.92000:01:47.930 the enthalpy change of formation for 00:01:50.81000:01:50.820 ethanol let's say it's 278 kilojoules 00:01:55.39900:01:55.409 per mole and for co2 it's negative 393 00:01:59.12000:01:59.130 and for h2o it's negative 286 so this is 00:02:04.63900:02:04.64900:02:07.63900:02:07.649 each substance how can you use this 00:02:10.49000:02:10.500 information to calculate the 00:02:12.80000:02:12.810 have to be changed for the entire 00:02:14.75000:02:14.760 reaction in order to do that and the 00:02:21.26000:02:21.270 enthalpy change of the entire reaction 00:02:22.94000:02:22.950 is the sum of the enthalpy change of 00:02:28.01000:02:28.020 formation for the products minus the sum 00:02:33.08000:02:33.090 of the enthalpy change of formation for 00:02:36.64000:02:36.650 the reactants the products is everything 00:02:47.44900:02:47.459 on the right side that includes the two 00:02:50.99000:02:51.000 co2 molecules and the three h2o 00:02:55.69900:02:55.709 molecules the reactants is everything on 00:03:00.97900:03:00.989 the left side that includes ethanol 00:03:06.46000:03:06.470 c2h5oh and the three oxygen molecules so 00:03:14.24000:03:14.250 now at this point we need to plug in the 00:03:17.72000:03:17.730 heats of formation for each substance so 00:03:22.81900:03:22.829 for CL 2 its negative 393 times 2 00:03:26.59900:03:26.609 because we have a 2 in front plus 3 00:03:32.96000:03:32.970 times negative 2 86 for water and 00:03:37.66000:03:37.670 ethanol is negative 278 now o 2 is a 00:03:42.97900:03:42.989 pure element and for any pure element in 00:03:46.34000:03:46.350 our natural state the enthalpy of 00:03:48.92000:03:48.930 formation is zero so this is going to be 00:03:52.30900:03:52.319 3 times zero so now we'll just have to 00:03:57.89000:03:57.900 do the map at this point two times a 00:04:00.89000:04:00.900 negative 393 that's about negative 786 00:04:06.31900:04:06.329 and three times negative 286 that's a 00:04:12.17000:04:12.180 negative 858 and then negative negative 00:04:17.59900:04:17.609 278 that's plus 278 and then the other 00:04:21.50000:04:21.510 one is just zero so now let's add these 00:04:24.64900:04:24.659 three numbers negative 00:04:25.67000:04:25.680 seven eight six minus eight five eight 00:04:27.85000:04:27.860 plus two seven eight 00:04:29.81000:04:29.820 you should get negative 1366 kilojoules 00:04:37.15900:04:37.169 per mole so that's the enthalpy of the 00:04:40.55000:04:40.560 reaction and so as you can see you can 00:04:43.58000:04:43.590 calculate it by taking the difference 00:04:45.92000:04:45.930 between the sum of the enthalpies of 00:04:48.77000:04:48.780 formation of the products minus the 00:04:50.27000:04:50.280 reactants so what exactly is the 00:04:57.26000:04:57.270 enthalpy of formation of reaction the 00:05:02.18000:05:02.190 enthalpy of formation is the same as the 00:05:07.15900:05:07.169 enthalpy of a certain type of reaction a 00:05:09.74000:05:09.750 reaction where a compound one mole of a 00:05:13.07000:05:13.080 compound is produced from its elements 00:05:17.57000:05:17.580 in our natural state so let's write the 00:05:21.71000:05:21.720 reaction the enthalpy information for 00:05:24.95000:05:24.960 the reaction of co2 so we're going to 00:05:28.99900:05:29.009 produce one mole of co2 on the right 00:05:31.64000:05:31.650 side and it has to be one you can't 00:05:34.27900:05:34.289 change it to two now the elements that 00:05:38.71900:05:38.729 make up CL two are carbon and oxygen and 00:05:42.05000:05:42.060 oxygen exist as Oh two in its natural 00:05:46.12900:05:46.139 state it's diatomic so the enthalpy of 00:05:53.54000:05:53.550 this reaction is equal to the enthalpy 00:05:57.64900:05:57.659 of formation for CL 2 now let's prove it 00:06:02.81000:06:02.820 will know that the enthalpy of reaction 00:06:04.39900:06:04.409 is always equal to the products minus 00:06:09.26000:06:09.270 the reactants that is the sum of the 00:06:11.54000:06:11.550 products minus the sum of the reactants 00:06:14.52900:06:14.539 so it's going to be CL 2 which is 00:06:18.56000:06:18.570 product that is two reactants carbon 00:06:21.56000:06:21.570 plus oxygen by the way this is a gas 00:06:26.80000:06:26.810 this is a gas and this is a solid so the 00:06:32.89900:06:32.909 enthalpy information for CL 2 and the 00:06:34.73000:06:34.740 last example we said it was negative 393 00:06:39.82000:06:39.830 now the enthalpy for a pure element is 00:06:43.34000:06:43.350 always zero thus the enthalpy of 00:06:47.96000:06:47.970 reaction is equal to the entropy of 00:06:50.51000:06:50.520 formation for co2 why because we have 00:06:54.98000:06:54.990 only pure elements on the left side 00:06:57.11000:06:57.120 which have an enthalpy of formation of 00:06:58.58000:06:58.590 zero and this is the only compound so 00:07:01.55000:07:01.560 the enthalpy of that reaction will equal 00:07:03.53000:07:03.540 the enthalpy of formation so as you can 00:07:06.29000:07:06.300 see the enthalpy of formation is the 00:07:08.69000:07:08.700 same as the enthalpy of reaction for any 00:07:11.72000:07:11.730 substance that is produced from its 00:07:14.38000:07:14.390 elements in a natural state now how 00:07:21.89000:07:21.900 would you write the reaction that's 00:07:24.68000:07:24.690 associated with the enthalpy of 00:07:25.94000:07:25.950 formation for water and ethanol so we're 00:07:30.83000:07:30.840 going to put water on the right side the 00:07:35.39000:07:35.400 elements that make up water are hydrogen 00:07:38.18000:07:38.190 and oxygen both of which are diatomic so 00:07:43.97000:07:43.980 we need a 1 in front of h2o we can't 00:07:46.67000:07:46.680 have anything else so the number of 00:07:49.43000:07:49.440 hydrogen atoms is balanced but we have 00:07:51.80000:07:51.810 one oxygen atom on the right side but 00:07:53.60000:07:53.610 two on the left so 1/2 is simply 1/2 so 00:08:00.23000:08:00.240 we need 1/2 in front of o2 so now is 00:08:02.96000:08:02.970 balanced the enthalpy for this reaction 00:08:05.24000:08:05.250 is the heat of formation of water which 00:08:10.04000:08:10.050 is approximately about negative 286 now 00:08:17.18000:08:17.190 how would you write the reaction for 00:08:20.53000:08:20.540 ethanol c2h5oh so we're going to have 1 00:08:26.42000:08:26.430 ethanol on the right side the elements 00:08:29.12000:08:29.130 that make up ethanol are carbon hydrogen 00:08:31.57000:08:31.580 and oxygen so we have 6 hydrogen atoms 00:08:38.48000:08:38.490 on the right side 6 divided by 2 is 3 00:08:40.85000:08:40.860 and only one oxygen atom so we need a 00:08:43.79000:08:43.800 1/2 and the enthalpy change of this 00:08:46.91000:08:46.920 reaction is equal to the enthalpy 00:08:49.91000:08:49.920 information for 00:08:51.47000:08:51.480 which is negative 278 and let's go ahead 00:08:58.22000:08:58.230 and add the other one for co2 which we 00:09:02.00000:09:02.010 wrote already and so that's negative 393 00:09:09.22000:09:09.230 now here's a question for you how can we 00:09:12.14000:09:12.150 use Hess's law to estimate the enthalpy 00:09:18.14000:09:18.150 change for the combustion of ethanol 00:09:22.60000:09:22.610 basically this reaction which we did 00:09:25.04000:09:25.050 already but we can use Hess along to get 00:09:27.77000:09:27.780 the answer so how can we use Hess's law 00:09:30.49000:09:30.500 to calculate the answer that we had 00:09:32.75000:09:32.760 previously so what we need to do is we 00:09:39.83000:09:39.840 need to adjust these three equations in 00:09:43.69000:09:43.700 such a way that they will add to the 00:09:47.03000:09:47.040 equation that we have on the bottom so 00:09:50.84000:09:50.850 notice that we have one water molecule 00:09:54.47000:09:54.480 on the right side so we're going to 00:09:57.14000:09:57.150 multiply this reaction by three and we 00:10:02.89000:10:02.900 have two co2 molecules on the right so 00:10:06.65000:10:06.660 we're going to multiply this reaction by 00:10:08.81000:10:08.820 two because we want to co2 is on the 00:10:11.69000:10:11.700 right and ethanol is on the left side so 00:10:15.95000:10:15.960 therefore we need to reverse the third 00:10:19.61000:10:19.620 reaction so for h2o we wanted three h2o 00:10:29.36000:10:29.370 molecules on the right side so we need 00:10:39.40000:10:39.410 three h2 molecules here and this is 00:10:42.48900:10:42.499 going to be 3 over 2 O 2 molecules now 00:10:47.19900:10:47.209 if you multiply the reaction by 3 the 00:10:51.48900:10:51.499 enthalpy you got to multiply the 00:10:54.18900:10:54.199 enthalpy by 3 so it was negative 286 so 00:10:57.30900:10:57.319 it's now negative 286 times 3 which is 00:11:03.02900:11:03.039 negative 858 that's the enthalpy of 00:11:06.18900:11:06.199 change for this reaction now the next 00:11:09.54900:11:09.559 one was carbon plus o2 which turned into 00:11:13.59900:11:13.609 co2 and because we had 2 CL 2 molecules 00:11:18.27900:11:18.289 on the right side we need to multiply 00:11:20.43900:11:20.449 this reaction went to the enthalpy 00:11:25.05900:11:25.069 change was negative 393 but we got to 00:11:28.56900:11:28.579 multiply that by 2 and so this is going 00:11:32.85900:11:32.869 to be a negative 786 now for the 00:11:37.56900:11:37.579 reaction with ethanol we have to reverse 00:11:40.14900:11:40.159 it so we need a thneed a left side and 00:11:44.40900:11:44.419 on the right side we have the elements 00:11:46.74900:11:46.759 that make up ethanol that's carbon 00:11:49.19900:11:49.209 hydrogen and oxygen 00:11:59.72000:11:59.730 the entropy of formation for ethanol was 00:12:03.42000:12:03.430 negative 278 but because we have to 00:12:05.85000:12:05.860 reverse it it's not going to be positive 00:12:08.63000:12:08.640 278 so what we're going to do is we're 00:12:11.82000:12:11.830 going to add these three reactions so if 00:12:16.41000:12:16.420 we add them notice that H 2 cancels we 00:12:20.34000:12:20.350 have 3 h2 on the left and 3 on the right 00:12:22.47000:12:22.480 so they completely cancel in addition 00:12:26.07000:12:26.080 the two carbon atoms cancel they're on 00:12:28.56000:12:28.570 opposite sides and that's about it on 00:12:34.08000:12:34.090 the left side we have ethanol which is 00:12:37.13000:12:37.140 c2h5oh so everything on the left is 00:12:40.77000:12:40.780 simply going to drop them now let's 00:12:43.83000:12:43.840 focus on the oxygen atoms so on the left 00:12:47.04000:12:47.050 side we have 2 plus 1.5 oxygen atoms 00:12:50.01000:12:50.020 that's 3.5 on the left but we have 1/2 00:12:53.19000:12:53.200 on the right if we subtract half on both 00:12:56.16000:12:56.170 sides this will cancel and on the left 00:12:59.49000:12:59.500 side instead of having 3.5 we'll have 3 00:13:01.64000:13:01.650 because 3 point 5 minus point 5 is 3 so 00:13:05.10000:13:05.110 we're going to have 3 o2 on the left 00:13:08.64000:13:08.650 side on the right side these two would 00:13:11.76000:13:11.770 simply fall down and it's going to be 2 00:13:14.82000:13:14.830 CL 2 + 3 H 2 on the right side so now if 00:13:20.39000:13:20.400 we add reactions 1 2 & 3 and since we 00:13:24.78000:13:24.790 came up with this reaction according to 00:13:26.82000:13:26.830 houses law the enthalpy change of that 00:13:30.48000:13:30.490 reaction can be calculated by adding the 00:13:34.59000:13:34.600 enthalpy change of the individual 00:13:36.24000:13:36.250 reactions that we use to get this 00:13:38.31000:13:38.320 reaction so we just got to add negative 00:13:41.49000:13:41.500 8 58 plus negative 7 86 plus 278 and we 00:13:47.37000:13:47.380 get the same answer that we had in a 00:13:49.20000:13:49.210 last example which is negative 13 66 so 00:13:53.70000:13:53.710 you have two ways of calculating the 00:13:55.86000:13:55.870 enthalpy change of a reaction using 00:13:58.50000:13:58.510 products minus reactants or even use 00:14:00.72000:14:00.730 intensive law 00:14:04.96000:14:04.970 ammonia reacts with oxygen gas to 00:14:10.77000:14:10.780 produce nitrogen gas and water so before 00:14:18.31000:14:18.320 we calculate the enthalpy change for 00:14:20.41000:14:20.420 this reaction 00:14:22.08000:14:22.090 make sure you balance it pause the video 00:14:24.94000:14:24.950 and go ahead and balance it so we can 00:14:32.98000:14:32.990 start out by putting the tool in front 00:14:34.78000:14:34.790 of nh3 so we can balance the Nitra 00:14:37.42000:14:37.430 numbers and that would mean that we have 00:14:40.00000:14:40.010 six hydrogen atoms on the left side 2 00:14:42.10000:14:42.110 times 3 or 6 which means that we need to 00:14:44.38000:14:44.390 put a 3 in front of h2o but now that we 00:14:49.63000:14:49.640 have three oxygen atoms on the right to 00:14:53.23000:14:53.240 balance it we need 3 over 2 in front of 00:14:55.69000:14:55.700 o2 so we can have three oxygen atoms on 00:14:57.97000:14:57.980 both sides 3 over 2 times 2 is string to 00:15:02.89000:15:02.900 get rid of the fraction multiply 00:15:04.42000:15:04.430 everything by 2 so we're going to double 00:15:07.53000:15:07.540 everything so let's start by putting a 4 00:15:11.43000:15:11.440 so if we have 4 nitrogens on the left we 00:15:14.26000:15:14.270 need a 2 in front of n2 to make it 4 and 00:15:16.90000:15:16.910 that is that we have 12 hydrogen's on 00:15:19.57000:15:19.580 the left side 4 times 3 is 12 00:15:21.10000:15:21.110 12 divided by 2 is 6 and now we have 6 00:15:24.70000:15:24.710 oxygen atoms on the right 6 over 2 is 3 00:15:27.64000:15:27.650 so now the reaction is balanced now 00:15:32.29000:15:32.300 let's say if you're given the enthalpy 00:15:35.83000:15:35.840 information for nh3 let's say it's 00:15:38.77000:15:38.780 negative 46 and for h2o which is 00:15:43.02000:15:43.030 negative 286 use miss-information 00:15:47.92000:15:47.930 calculate the enthalpy change of the 00:15:50.32000:15:50.330 reaction so to find it it's simply going 00:15:55.90000:15:55.910 to be the sum of the products minus the 00:16:00.73000:16:00.740 sum of the reactants 00:16:04.40000:16:04.410 so on the product side we have 2 n 2 00:16:09.59000:16:09.600 plus 6 h2o and on the reactant side we 00:16:16.04000:16:16.050 have nh3 and OH - so the enthalpy 00:16:23.57000:16:23.580 information for any pure element that 00:16:26.06000:16:26.070 includes nitrogen and oxygen are 0 so 00:16:30.46000:16:30.470 this is going to be 0 plus 6 times the 00:16:34.04000:16:34.050 value for h2o 00:16:35.33000:16:35.340 that's negative 286 minus 4 times the 00:16:42.53000:16:42.540 value for NH stream which is negative 46 00:16:45.47000:16:45.480 and 402 is 0 so 6 times negative 2 00:16:53.03000:16:53.040 eight-six 00:16:53.84000:16:53.850 that's about negative 17 16 and negative 00:17:01.16000:17:01.170 4 times negative 46 that's positive 184 00:17:11.24000:17:11.250 so if we add these two numbers we can 00:17:13.37000:17:13.380 get the enthalpy change of the reaction 00:17:15.07900:17:15.089 and so the final answer should be 00:17:18.55000:17:18.560 negative 15 32 and typically the units 00:17:22.76000:17:22.770 is kilojoules per mole 00:17:37.27000:17:37.280 consider the following reactions we're 00:17:43.78000:17:43.790 going to use Hess's law to calculate the 00:17:46.99000:17:47.000 enthalpy change of another reaction 00:17:52.56000:17:52.570 let's say the enthalpy change for this 00:17:54.73000:17:54.740 reaction is 1215 00:18:10.50000:18:10.510 and the enthalpy change for the second 00:18:12.78000:18:12.790 reaction we're going to say it's 00:18:16.11000:18:16.120 positive 283 so using these two 00:18:22.29000:18:22.300 reactions calculate the enthalpy change 00:18:25.70000:18:25.710 for the combustion of methane methane is 00:18:31.86000:18:31.870 ch4 and it's going to react with oxygen 00:18:35.46000:18:35.470 to produce water and carbon dioxide 00:18:52.88000:18:52.890 so what's the enthalpy change for this 00:18:55.01000:18:55.020 reaction 00:19:04.32000:19:04.330 feel free to pause the video and see if 00:19:07.17000:19:07.180 you can get the right answer 00:19:09.47000:19:09.480 so you need to know what to focus on and 00:19:13.47000:19:13.480 what not to focus on I wouldn't focus on 00:19:16.74000:19:16.750 CEO because it's found in reaction 1 & 2 00:19:20.49000:19:20.500 and I wouldn't focus on o2 because it's 00:19:23.79000:19:23.800 found in both reactions now if you focus 00:19:28.41000:19:28.420 on ch4 and co2 which is found in only 00:19:32.67000:19:32.680 one of those two reactions it's going to 00:19:35.55000:19:35.560 make the problem a lot easier so let's 00:19:37.95000:19:37.960 start by focusing on ch4 so if we look 00:19:42.09000:19:42.100 at the reaction that we want to 00:19:45.30000:19:45.310 calculate the enthalpy change for notice 00:19:48.03000:19:48.040 that we have one ch4 on the left side 00:19:51.26000:19:51.270 but here we have two ch4 on the right so 00:19:55.08000:19:55.090 that means that we need to reverse to 00:19:56.85000:19:56.860 reaction 1 and multiply it by 1/2 so I'm 00:20:01.80000:20:01.810 going to write R for reverse and then 00:20:04.47000:20:04.480 times 1/2 so if we reverse it ch4 is now 00:20:09.54000:20:09.550 on the left and if we divided by 2 we're 00:20:11.64000:20:11.650 going to have 1 ch4 plus 3 over 2 for o2 00:20:21.09000:20:21.100 and then half of 4 is 2 so we're going 00:20:24.42000:20:24.430 to have 2 h2o molecules and 1 co 00:20:30.00000:20:30.010 molecule so how is that going to effect 00:20:34.35000:20:34.360 the enthalpy change of the reaction 00:20:36.92000:20:36.930 since we reverse it the positive 12:15 00:20:40.44000:20:40.450 becomes negative 12:15 and since we 00:20:43.05000:20:43.060 divided by 2 we need to divide 12:15 by 00:20:45.84000:20:45.850 2 so this should be negative six hundred 00:20:49.89000:20:49.900 and seven point five so now let's focus 00:20:54.36000:20:54.370 on the second reaction let's focus on 00:20:57.15000:20:57.160 co2 we have one co2 molecule on the 00:21:00.33000:21:00.340 right side here it's on the left so all 00:21:02.73000:21:02.740 we need to do is we need to reverse the 00:21:05.16000:21:05.170 second reaction so it's one half O 2 and 00:21:11.19000:21:11.200 the left plus co yield co2 00:21:15.96000:21:15.970 so the enthalpy change for that is going 00:21:18.72000:21:18.730 to be negative 283 so now let's add the 00:21:22.91900:21:22.929 two reactions so let's see what cancels 00:21:26.94000:21:26.950 first notice that CL cancel so we have 00:21:30.84000:21:30.850 one on the left one on the right and so 00:21:34.25900:21:34.269 this is going to fall down so we have CH 00:21:36.33000:21:36.340 4 and notice that we have oxygen 00:21:40.95000:21:40.960 molecules only on the left so we got to 00:21:42.81000:21:42.820 add 1 over 2 plus 3 over 2 is 4 over 2 4 00:21:48.21000:21:48.220 divided by 2 is 2 so this is going to be 00:21:51.24000:21:51.250 202 on the left side and on the right 00:21:53.85000:21:53.860 side all we have left over is the two 00:21:56.70000:21:56.710 water molecules and carbon dioxide so we 00:22:03.11900:22:03.129 have the same reaction as this one 00:22:07.10000:22:07.110 therefore to find the enthalpy change of 00:22:09.77900:22:09.789 that reaction accordance it has as long 00:22:11.58000:22:11.590 we simply need to add these two numbers 00:22:14.81000:22:14.820 so negative 6 so 7.5 plus negative 283 00:22:24.79900:22:24.809 this is equal to negative 890 point five 00:22:30.02900:22:30.039 and so now you know how to use houses 00:22:33.81000:22:33.820 lotsa estimate the enthalpy change of a 00:22:36.06000:22:36.070 reaction another way in which you can 00:22:40.87900:22:40.889 measure the enthalpy change of a 00:22:43.37900:22:43.389 reaction is using calorimetry so here's 00:22:47.78900:22:47.799 an example problem 30 grams of sodium 00:22:50.94000:22:50.950 hydroxide was dissolved in 80 00:22:54.14900:22:54.159 milliliters of water the temperature 00:22:57.13900:22:57.149 increased from 27 to 70 degrees Celsius 00:23:02.11900:23:02.129 estimate the enthalpy change for the 00:23:04.68000:23:04.690 dissolution of NaOH so how can we do 00:23:08.85000:23:08.860 this how can we find the enthalpy change 00:23:11.90900:23:11.919 of the reaction the first thing we need 00:23:14.85000:23:14.860 to do is find out how much heat energy 00:23:17.93000:23:17.940 was absorbed by the water and we can do 00:23:22.11000:23:22.120 this using Q equals M cap MC delta T 00:23:27.70000:23:27.710 anytime you want to find out how much 00:23:29.84000:23:29.850 heat was absorbed or released by 00:23:31.58000:23:31.590 substance if you know the specific heat 00:23:33.44000:23:33.450 capacity of that substance and if a 00:23:35.90000:23:35.910 temperature change occurs you use this 00:23:37.34000:23:37.350 equation now the energy released by the 00:23:42.89000:23:42.900 reaction is equal to the energy absorbed 00:23:45.59000:23:45.600 by water so Q of the reaction in this 00:23:48.83000:23:48.840 case the dissolution of sodium hydroxide 00:23:51.44000:23:51.450 is equal to the amount of heat energy 00:23:53.51000:23:53.520 absorbed by the water now the amount of 00:23:56.96000:23:56.970 heat absorbed by the water is an 00:23:59.21000:23:59.220 endothermic process because the 00:24:01.25000:24:01.260 temperature of the water rose it 00:24:02.57000:24:02.580 increased now because of reaction 00:24:05.00000:24:05.010 released energy it's exothermic for the 00:24:08.24000:24:08.250 reaction so the dissolution of sodium 00:24:13.01000:24:13.020 hydroxide is an exothermic process 00:24:14.65000:24:14.660 because it caused the surrounding water 00:24:17.90000:24:17.910 molecules to increase in temperature now 00:24:21.41000:24:21.420 to find the enthalpy change to the 00:24:22.88000:24:22.890 reaction it's going to be Q divided by n 00:24:28.99000:24:29.000 now typically enthalpy is represented in 00:24:32.63000:24:32.640 units of kilojoules per mole and in this 00:24:36.59000:24:36.600 equation Q is going to be in joules so 00:24:39.20000:24:39.210 you have to convert it to kilojoules and 00:24:40.79000:24:40.800 simply divided by moles and you can 00:24:44.03000:24:44.040 estimate the enthalpy change for that 00:24:46.31000:24:46.32000:24:53.31000:24:53.320 so let's calculate the amount of energy 00:24:57.36000:24:57.370 absorbed by the water so the mass of the 00:25:04.14000:25:04.150 water is 80 keep in mind the density of 00:25:07.41000:25:07.420 water is one gram per milliliter that 00:25:09.93000:25:09.940 means one milliliter of water has a mass 00:25:12.33000:25:12.340 of 1 gram therefore 80 milliliters of 00:25:14.34000:25:14.350 water has a mass of 80 grams the 00:25:18.36000:25:18.370 specific heat capacity for water is 00:25:21.20000:25:21.210 4.184 and the units are joules per gram 00:25:25.50000:25:25.510 per Celsius and 80 has units of grams so 00:25:31.44000:25:31.450 as we can see grams cancel you need to 00:25:34.47000:25:34.480 understand what the specific heat 00:25:36.21000:25:36.220 capacity is what it represents it is the 00:25:39.33000:25:39.340 amount of energy needed to heat up one 00:25:43.83000:25:43.840 gram of a substance by one degree 00:25:45.45000:25:45.460 Celsius so in the case of water it takes 00:25:49.13000:25:49.140 4.184 joules of water I mean it takes 00:25:53.91000:25:53.920 four point money for joules of heat 00:25:55.56000:25:55.570 energy to heat up one gram of water by 00:25:59.22000:25:59.230 one degree Celsius by the way sometimes 00:26:04.43000:26:04.440 the heat capacity might be used in 00:26:09.02000:26:09.030 calories instead of joules so the heat 00:26:12.45000:26:12.460 capacity of water is also one calorie 00:26:14.73000:26:14.740 per gram per celsius one calorie is 00:26:18.62000:26:18.630 4.184 joules 00:26:20.19000:26:20.200 it's another unit of energy specifically 00:26:22.74000:26:22.750 heat energy a capital calorie is equal 00:26:29.04000:26:29.050 to a thousand lowercase calories so 00:26:33.36000:26:33.370 these are some other units to know 00:26:40.10000:26:40.110 the temperature change for this reaction 00:26:42.28900:26:42.299 is the final minus the initial 00:26:43.94000:26:43.950 temperature so 70 minus 27 that's a 00:26:49.31000:26:49.320 temperature change of 40 degrees Celsius 00:26:53.86000:26:53.870 so the unit Celsius will cancel 00:26:57.13000:26:57.140 therefore Q is going to be in joules so 00:27:02.32000:27:02.330 80 times 4.1 84 times 43 that's about 14 00:27:12.11000:27:12.120 thousand three hundred ninety three 00:27:16.02900:27:16.039 jewels so that's how much heat energy 00:27:19.58000:27:19.590 was absorbed by the water so the amount 00:27:23.12000:27:23.130 of energy that was released by the 00:27:24.56000:27:24.570 reaction is negative fourteen thousand 00:27:29.33000:27:29.340 393 joules keep mine it's an atomic for 00:27:35.96000:27:35.970 the water because the water absorb 00:27:38.96000:27:38.970 energy increase in its temperature and 00:27:41.06000:27:41.070 therefore it's kinetic energy but it's 00:27:44.06000:27:44.070 negative for the reaction it's positive 00:27:46.43000:27:46.440 for water since its endo but it's 00:27:48.28900:27:48.299 negative for the reaction because it's 00:27:50.24000:27:50.250 XO the reaction release heat energy 00:27:59.59000:27:59.600 now we need to convert jewels into 00:28:02.11000:28:02.120 kilojoules now keep in mind one 00:28:07.69000:28:07.700 kilojoule is equal to a thousand joules 00:28:12.06000:28:12.070 so we need the unit joules to be on the 00:28:14.53000:28:14.540 bottom so that these units will cancel 00:28:16.99000:28:17.000 and so the unit kilojoules going to go 00:28:20.08000:28:20.090 on top so to convert Joules to 00:28:22.87000:28:22.880 kilojoules you need to divide by a 00:28:24.43000:28:24.440 thousand so it's negative fourteen point 00:28:27.16000:28:27.170 three nine three kilojoules now in order 00:28:33.13000:28:33.140 to estimate the enthalpy change of the 00:28:34.66000:28:34.670 reaction we need to calculate the moles 00:28:36.55000:28:36.560 of sodium hydroxide right now we have 00:28:39.43000:28:39.440 the grams so we have 30 grams of NaOH to 00:28:45.52000:28:45.530 convert it to moles we're going to find 00:28:47.11000:28:47.120 the molar mass so we need to use the 00:28:51.52000:28:51.530 periodic table the atomic mass of na is 00:28:54.66000:28:54.670 approximately about 23 and for oxygen at 00:28:58.18000:28:58.190 16 for hydrogen its 1 if you add these 00:29:01.21000:29:01.220 three numbers you should get about 40 00:29:02.98000:29:02.990 the unit for molar mass is grams per 00:29:06.22000:29:06.230 mole so this is 40 grams per mole what 00:29:10.09000:29:10.100 that means is that one mole of sodium 00:29:13.39000:29:13.400 hydroxide has a mass of 40 grams so the 00:29:23.20000:29:23.210 unit grams of NaOH cancel 30/40 we can 00:29:29.50000:29:29.510 cancel the zero so this is basically 3 00:29:31.57000:29:31.580 over 4 3/4 is 0.75 so we have 27 5 moles 00:29:37.81000:29:37.820 of NaOH so now we can calculate or 00:29:42.58000:29:42.590 estimate the enthalpy change of the 00:29:45.04000:29:45.050 reaction for the dissolution of sodium 00:29:47.77000:29:47.780 hydroxide as it dissolves into the 00:29:51.22000:29:51.230 sodium cation and the hydroxide ion 00:30:01.26000:30:01.270 you 00:30:03.49000:30:03.500 and now the Enfamil change of the 00:30:05.98000:30:05.990 reaction as we mentioned before it's q 00:30:09.94000:30:09.950 of the reaction divided by n so we just 00:30:13.33000:30:13.340 gotta divide the kilojoules by the 00:30:15.13000:30:15.140 number of moles so it's negative 00:30:17.79900:30:17.809 fourteen point three nine three 00:30:20.19000:30:20.200 kilojoules divided by 0.75 moles 00:30:34.02000:30:34.030 so you should get 00:30:42.67000:30:42.680 negative 19.2 kilojoules per mole as the 00:30:48.91000:30:48.920 final answer for this particular example 00:30:51.66000:30:51.670 so that's how you can estimate the 00:30:53.77000:30:53.78000:30:56.05000:30:56.060 calorimetry how would you solve this 00:31:00.34000:31:00.350 problem the enthalpy of combustion for 00:31:03.73000:31:03.740 benzene is negative 30 to 70 kilojoules 00:31:07.87000:31:07.880 per mole how much heat energy will be 00:31:11.59000:31:11.600 released if 30 grams of benzene is 00:31:14.08000:31:14.090 burned in air let's focus on part a so 00:31:18.30000:31:18.310 we need to write a reaction a thermo 00:31:21.34000:31:21.350 chemical equation a balance reaction 00:31:23.38000:31:23.390 that is associated with the enthalpy 00:31:25.60000:31:25.610 change of the reaction as well so let's 00:31:28.63000:31:28.640 start with benzene which we know is c6h6 00:31:35.64000:31:35.650 now it's burned in air so it's going to 00:31:39.25000:31:39.260 react with the oxygen molecules that's 00:31:41.32000:31:41.330 in the air and in any combustion 00:31:43.87000:31:43.880 reaction we're always going to get cl2 00:31:46.51000:31:46.520 and water as the products for the 00:31:49.21000:31:49.220 reaction so we need to balance the 00:31:51.61000:31:51.620 reaction so we have six carbon atoms on 00:31:55.84000:31:55.850 the left side we need to put a six in 00:31:57.82000:31:57.830 front of co2 and since we have six 00:32:01.00000:32:01.010 hydrogen atoms on the left six divided 00:32:03.82000:32:03.830 by two is three and if we count the 00:32:07.36000:32:07.370 oxygen atoms on the right we have 12 six 00:32:10.51000:32:10.520 times two is twelve and three from water 00:32:13.57000:32:13.580 so that's 15 15 divided by two is 15 00:32:18.13000:32:18.140 over two so at this point the reaction 00:32:20.98000:32:20.990 is balanced but we have a fraction with 00:32:24.82000:32:24.830 a denominator of two therefore we want 00:32:27.70000:32:27.710 to multiply everything by two so let's 00:32:33.01000:32:33.020 start by putting a 2 in front of benzene 00:32:35.74000:32:35.750 so we now need 12 carbons instead of six 00:32:39.24000:32:39.250 now we have 12 hydrogen's on the left 00:32:42.01000:32:42.020 side so we need a 6 in front of h2o so 00:32:45.79000:32:45.800 at this point we have 24 oxygen atoms 00:32:48.88000:32:48.890 from the co2 and 6 from water that's 30 00:32:51.73000:32:51.740 30 divided by 2 is 15 00:32:54.71000:32:54.720 so now the reaction is balanced by the 00:32:59.00900:32:59.019 way the enthalpy change for benzene the 00:33:03.64900:33:03.659 3270 00:33:05.21000:33:05.220 doesn't correspond to this reaction 00:33:07.49000:33:07.500 rather it corresponds to this reaction 00:33:22.63000:33:22.640 this is the amount of heat energy that's 00:33:25.29900:33:25.309 released when one mole of benzene is 00:33:29.91900:33:29.929 burning in this the kilojoules per a 00:33:32.28900:33:32.299 single mom out two moles 00:33:34.00000:33:34.010 so therefore the enthalpy change for the 00:33:37.87000:33:37.880 second reaction this is negative thirty 00:33:40.24000:33:40.250 to seventy so for this reaction since is 00:33:43.69000:33:43.700 multiplied by two it's going to be twice 00:33:45.91000:33:45.920 as much thirty to seventy times 2 is 00:33:51.03000:33:51.040 negative sixty five forty 00:34:08.12900:34:08.139 so now that we have the energy that's 00:34:11.81900:34:11.829 associated with this reaction we can 00:34:14.78900:34:14.799 answer the question in part a how much 00:34:17.12900:34:17.139 heat energy will be released if 30 grams 00:34:19.82900:34:19.839 of benzene is burned in air so what we 00:34:23.60900:34:23.619 need to do is convert grams to 00:34:25.85900:34:25.869 kilojoules so let's start by converting 00:34:30.41900:34:30.429 grams to moles so we need to find the 00:34:38.03900:34:38.049 molar mass of benzene 00:34:40.75900:34:40.769 so for c6h6 00:34:44.38900:34:44.399 the atomic mass of carbon is about 12 00:34:47.48900:34:47.499 but there's six of them so we're going 00:34:49.16900:34:49.179 to multiply 12 by 6 then for hydrogen 00:34:51.74900:34:51.759 it's about one but there's six of them 00:34:53.60900:34:53.619 as well 12 times 6 is 72 plus 6 that's 00:34:57.80900:34:57.819 78 so there's 78 grams of benzene in a 00:35:04.97900:35:04.989 single mole of benzene now in this 00:35:12.42000:35:12.430 thermo chemical equation it relates the 00:35:15.05900:35:15.069 number of kilojoules per mole so if 2 00:35:19.55900:35:19.569 moles of benzene reacts negative 65 40 00:35:24.35900:35:24.369 kilojoules of heat energy will be 00:35:26.06900:35:26.079 released if 15 moles of old to v 00:35:29.22000:35:29.230 ax6gwsf0 q1o in a substance you can 00:35:44.57900:35:44.589 convert from moles to kilojoules using 00:35:48.10900:35:48.119 that connection so since we have benzene 00:35:52.94000:35:52.950 we need to use 2 2 so 2 moles of benzene 00:35:56.64000:35:56.650 will release this many kilojoules of 00:35:58.89000:35:58.900 heat energy so let's write that in the 00:36:00.90000:36:00.910 next fraction so we have two moles of 00:36:04.40000:36:04.410 c6h6 00:36:09.70000:36:09.710 and that's going to release negative 65 00:36:16.59900:36:16.609 40 kilojoules of heat so the unit grams 00:36:22.32900:36:22.339 cancel and also moles cancel so now we 00:36:27.97000:36:27.980 can calculate the answer 30 divided by 00:36:31.66000:36:31.670 78 times 65 40 divided by 2 is negative 00:36:41.28000:36:41.290 12 00:36:43.07900:36:43.089 58 kilojoules that's how much heat 00:36:47.17000:36:47.180 energy will be released if all of the 30 00:36:50.47000:36:50.480 grams of benzene is burned in air 00:37:00.05000:37:00.060 so now let's focus on Part B how many 00:37:03.32000:37:03.330 grams of co2 will be produced if 8800 00:37:08.12000:37:08.130 kilojoules of heat energy was released 00:37:10.97000:37:10.980 from the reaction well let's rewrite the 00:37:14.42000:37:14.430 reaction because I erased it 00:37:38.90000:37:38.910 and the energy change was 65 40 but- 00:37:47.44000:37:47.450 so this time we need to convert from 00:37:50.87000:37:50.880 kilojoules to grams of co2 so let's 00:37:55.49000:37:55.500 start with what we have 00:37:56.53900:37:56.549 that's 8800 kilojoules over 1 now the 00:38:03.04900:38:03.059 connection between kilojoules and co2 is 00:38:05.98000:38:05.990 that if 12 moles of co2 is produced 65 00:38:11.53900:38:11.549 40 kilojoules of heat energy will be 00:38:13.57900:38:13.589 released so we can put the 65 40 00:38:17.45000:38:17.460 kilojoules on the bottom so that the 00:38:19.73000:38:19.740 unit kilojoules will cancel and 12 moles 00:38:24.07900:38:24.089 of CL 2 correspond to that amount of 00:38:27.79900:38:27.809 energy so make sure you use the right 00:38:32.24000:38:32.250 coefficient that correlates to the 00:38:34.94000:38:34.950 substance that you're looking for so now 00:38:37.57900:38:37.589 all we need to do is convert moles of 00:38:40.78900:38:40.799 co2 into grams so we need to find the 00:38:43.81900:38:43.829 molar mass the atomic mass of carbon is 00:38:46.78900:38:46.799 12 and for oxygen is 16 but times 2 2 00:38:50.96000:38:50.970 times 16 is 32 plus 12 that's 44 00:38:54.94000:38:54.950 therefore one mole of carbon dioxide has 00:38:59.29900:38:59.309 a mass of 44 grams so now the unit moles 00:39:04.43000:39:04.440 of co2 cancel so 80 800 times 12 divided 00:39:10.06900:39:10.079 by 65 40 times 44 is equal to seven 00:39:17.39000:39:17.400 hundred and ten grams 0.5 or point four 00:39:22.49000:39:22.500 six so I'm going to round up to 17 so 00:39:26.07900:39:26.089 710 grams of co2 will be produced if 00:39:31.00000:39:31.010 8,800 of kilojoules of heat energy was 00:39:34.13000:39:34.140 released from the reaction so now you 00:39:36.58900:39:36.599 know how to convert from grams to 00:39:38.83900:39:38.849 kilojoules and kilojoules to grams using 00:39:42.04900:39:42.059 a thermal chemical equation 00:39:47.60000:39:47.610 now you need to be familiar with phase 00:39:51.15000:39:51.160 changes whenever a substance goes from 00:39:56.28000:39:56.290 the liquid state to the solid state what 00:39:58.38000:39:58.390 is the process called this process is 00:40:02.07000:40:02.080 known as freezing freezing is it an 00:40:08.34000:40:08.350 endothermic process or an exothermic 00:40:11.01000:40:11.020 process do you have to add heat energy 00:40:14.64000:40:14.650 or remove the energy in order to convert 00:40:18.09000:40:18.100 let's say liquid water in size you need 00:40:20.13000:40:20.140 to remove heat energy because you need 00:40:21.93000:40:21.940 to bring down the temperature so because 00:40:24.90000:40:24.910 you're removing heat energy freezing is 00:40:27.14000:40:27.150 an exothermic process now the reverse we 00:40:33.12000:40:33.130 learn from a solid to a liquid is called 00:40:36.24000:40:36.250 melting if you want to melt ice you need 00:40:39.15000:40:39.160 to add heat to it if you put ice on a 00:40:41.40000:40:41.410 stove and if you turn up the heat it's 00:40:45.75000:40:45.760 going to melt so melton is an 00:40:47.85000:40:47.860 endothermic process which means that the 00:40:50.25000:40:50.260 enthalpy change is positive now what is 00:40:53.25000:40:53.260 the process called when the liquid 00:40:54.80000:40:54.810 converts into a gas and also when a gas 00:40:58.35000:40:58.360 converts to a liquid and when a solid 00:41:02.04000:41:02.050 goes to a gas and when the gas goes to a 00:41:04.44000:41:04.450 solid feel free to pause the video and 00:41:07.85000:41:07.860 identify the names of these physical 00:41:10.26000:41:10.270 processes and also determine the sign of 00:41:13.35000:41:13.360 the enthalpy change is it positive or is 00:41:15.24000:41:15.250 it negative going from a liquid to a gas 00:41:17.96000:41:17.970 this is called vaporization and gas the 00:41:24.90000:41:24.910 liquid is known as condensation 00:41:31.79000:41:31.800 now I'm going from a solid to a gas this 00:41:36.23000:41:36.240 is called sublimation 00:41:39.76000:41:39.770 now what do you think gas the solid is 00:41:42.38000:41:42.390 called what is that called if a gas goes 00:41:45.29000:41:45.300 directly into a solid this is called 00:41:49.06000:41:49.070 deposition so vaporization is in 00:41:56.51000:41:56.520 endothermic or exothermic it turns out 00:42:00.17000:42:00.180 that to vaporize liquid water into steam 00:42:03.50000:42:03.510 you got to add heat to it so this is an 00:42:05.99000:42:06.000 endothermic process it's positive so 00:42:10.33000:42:10.340 condensation which is the reverse 00:42:12.38000:42:12.390 that's exothermic you need to remove 00:42:15.05000:42:15.060 heat energy from steam if you wish to 00:42:17.84000:42:17.850 condense it back to liquid water 00:42:19.78000:42:19.790 sublimation going directly from a solid 00:42:22.73000:42:22.740 to a gas that isn't happening with water 00:42:24.53000:42:24.540 at least not under normal pressure 00:42:27.08000:42:27.090 conditions but at a standard pressure of 00:42:31.22000:42:31.230 1 atm at sea level 00:42:32.54000:42:32.550 carbon dioxide Sublime's directly from a 00:42:35.36000:42:35.370 solid to a gas that's why it's called 00:42:37.31000:42:37.320 dry ice because it doesn't liquefy under 00:42:40.07000:42:40.080 room temperature conditions under normal 00:42:41.66000:42:41.670 pressure conditions it goes directly 00:42:44.66000:42:44.670 from a solid to a gas so it's dry ice 00:42:46.31000:42:46.320 and as it goes from a solid to a gas 00:42:49.91000:42:49.920 that's a endothermic process it has to 00:42:52.19000:42:52.200 absorb heat energy to do that deposition 00:42:56.93000:42:56.940 is exothermic because it's the reverse 00:43:01.27000:43:01.280 so let's see if we can summarize what 00:43:03.71000:43:03.720 we've just learned as you go from a 00:43:05.63000:43:05.640 solid to a liquid which is a melting and 00:43:08.33000:43:08.340 then a liquid to a gas which is 00:43:11.00000:43:11.010 vaporization go in in this direction the 00:43:15.32000:43:15.330 enthalpy change is positive the process 00:43:17.30000:43:17.310 is endothermic now if you go backwards 00:43:20.15000:43:20.160 from a gas to a liquid to a solid it's 00:43:24.23000:43:24.240 an exothermic process energy has to be 00:43:26.39000:43:26.400 released 00:43:30.98000:43:30.990 so solid to liquid we said it's melting 00:43:34.14000:43:34.150 liquid to gas and vaporization and going 00:43:37.38000:43:37.390 directly from a solid to a gas this is a 00:43:40.46000:43:40.470 sublimation now gas the liquid that's 00:43:44.70000:43:44.710 condensation liquid to solid is freezing 00:43:47.70000:43:47.710 and gas directly to solid is deposition 00:43:52.71000:43:52.720 all of those three processes are 00:43:55.08000:43:55.090 exothermic now let's work on a few other 00:44:01.77000:44:01.780 problems related to thermal chemistry 00:44:04.73000:44:04.740 how can you calculate the amount of 00:44:07.17000:44:07.180 energy required to heat 50 grams of 00:44:10.32000:44:10.330 water from 20 to 60 degrees Celsius to 00:44:15.08000:44:15.090 calculate the energy absorbed or release 00:44:17.54000:44:17.550 when given a temperature change you can 00:44:21.42000:44:21.430 use an equation that we used earlier Q 00:44:23.97000:44:23.980 is equal to M cap the mass of water is 00:44:29.19000:44:29.200 50 the heat capacity is 4.1 84 and a 00:44:34.26000:44:34.270 temperature change final minus initial 00:44:37.47000:44:37.480 that change is 40 so 50 times 40 that's 00:44:43.23000:44:43.240 2,000 times 4.184 this is going to be 00:44:49.91000:44:49.920 8300 68 joules of heat energy so now 00:44:56.04000:44:56.050 what about a phase change how can we 00:44:58.50000:44:58.51000:45:00.96000:45:00.970 when let's say water turns into ice or 00:45:04.53000:45:04.540 if it vaporizes into steam how can we do 00:45:06.96000:45:06.970 that so let's say if we have 80 grams of 00:45:12.57000:45:12.580 ice how much energy is required to melt 00:45:17.37000:45:17.380 80 grams of ice into water and let's say 00:45:22.05000:45:22.060 the temperature of the ice is at zero 00:45:23.82000:45:23.830 degrees Celsius 00:45:29.38000:45:29.390 and the temperature of the liquid water 00:45:32.25900:45:32.269 is also at zero degrees Celsius so 00:45:36.83900:45:36.849 before the summit arises all of the ice 00:45:39.51900:45:39.529 has to melt into liquid water while ice 00:45:41.79900:45:41.809 melts into liquid water the temperature 00:45:43.93000:45:43.940 will remain constant at zero degrees 00:45:45.99900:45:46.009 Celsius so we have a phase change 00:45:51.22000:45:51.230 problem as opposed to a temperature 00:45:53.58900:45:53.599 change problem and so we need the heat 00:45:56.07900:45:56.089 of fusion to calculate the amount of 00:45:59.55900:45:59.569 joules that's going to be released or 00:46:01.96000:46:01.970 kilojoules instead of using an equation 00:46:05.89000:46:05.900 it's better to convert to get the answer 00:46:08.24900:46:08.259 the heat of fusion for water is about 00:46:12.84900:46:12.859 six kilojoules per mole so if you're 00:46:17.41000:46:17.420 going from solid to liquid or liquid to 00:46:19.42000:46:19.430 solid you can use the heat fusion from 00:46:22.85900:46:22.869 solid to liquid which is melting that's 00:46:26.10900:46:26.119 an endothermic process so it's going to 00:46:28.83900:46:28.849 be positive six kilojoules per mole if 00:46:31.47900:46:31.489 it was freezing which is exothermic it 00:46:33.27900:46:33.289 would be negative six kilojoules per 00:46:34.74900:46:34.759 mole so let's start with 80 grams of ice 00:46:38.27900:46:38.289 which has the formula h2o and let's 00:46:42.60900:46:42.619 convert that into moles so the molar 00:46:45.81900:46:45.829 mass of water is 16 for the atomic mass 00:46:50.68000:46:50.690 for oxygen plus two for the two hydrogen 00:46:56.76900:46:56.779 atoms which is 18 so there's 18 grams of 00:47:00.16000:47:00.170 water per one mole of h2o so these units 00:47:07.39000:47:07.400 will disappear and now we can convert 00:47:09.57900:47:09.589 moles into kilojoules so for every mole 00:47:13.96000:47:13.970 of ice that that melts to liquid water 00:47:18.56900:47:18.579 six kilojoules of heat energy will be 00:47:21.33900:47:21.349 released so it's going to be 80 times 6 00:47:27.13000:47:27.140 divided by 18 which is about twenty six 00:47:32.07900:47:32.089 point seven kilojoules so that's how 00:47:37.53900:47:37.549 much heat energy is required to melt 80 00:47:40.50900:47:40.519 grams of ice 00:47:45.18000:47:45.190 what about this one how much energy is 00:47:48.27900:47:48.289 required to heat 50 grams of water from 00:47:50.62000:47:50.630 20 degrees Celsius to steam and 170 and 00:47:56.03900:47:56.049 the heat capacity of liquid water is 00:48:02.30900:48:02.319 4.184 the heat capacity of steam or 00:48:07.83900:48:07.849 water in the gas density we're going to 00:48:10.72000:48:10.730 say it's about 2 2 joules per gram per 00:48:13.79900:48:13.809 Celsius the heat of vaporization for 00:48:19.26900:48:19.279 water 00:48:20.64000:48:20.650 it's about 41 kilojoules per mole so 00:48:24.48000:48:24.490 using this information how can you 00:48:26.52900:48:26.539 calculate the amount of energy that's 00:48:28.96000:48:28.970 required to heat water from 20 degrees 00:48:31.59900:48:31.609 Celsius to steam at 170 so you need to 00:48:34.99000:48:35.000 realize that this is a multiple part 00:48:37.21000:48:37.220 problem there's many steps so let's say 00:48:43.05900:48:43.069 if we were to draw a number line water 00:48:46.34900:48:46.359 boils at 100 the current temperature is 00:48:51.16000:48:51.170 20 and we need to get to 170 so we got 00:48:56.79900:48:56.809 to find out how much heat energy is 00:48:58.35900:48:58.369 required to heat liquid water from 20 to 00:49:01.48000:49:01.490 100 degrees Celsius that's Q 1 Q 2 00:49:05.58900:49:05.599 that's the amount of energy required to 00:49:07.45000:49:07.460 vaporize liquid water into steam at 100 00:49:10.50900:49:10.519 so that's a phase change Q 3 is the 00:49:13.77900:49:13.789 temperature change that's how much 00:49:15.37000:49:15.380 energy is required to heat up steam from 00:49:18.43000:49:18.440 100 to 170 so you got to break this 00:49:20.20000:49:20.210 problem into three parts 00:49:26.01000:49:26.020 so let's calculate q1 so because it's 00:49:31.47000:49:31.480 the time to change we're going to use 00:49:32.55000:49:32.560 the equation M time the mass of water is 00:49:35.67000:49:35.680 50 the heat capacity of liquid water is 00:49:38.91000:49:38.920 4.1 84 and the temperature change of 00:49:42.30000:49:42.310 liquid water is from 20 to 100 past 100 00:49:45.90000:49:45.910 it's a no longer liquid so it has to 00:49:49.17000:49:49.180 increase by a degree Celsius that's 100 00:49:52.10900:49:52.119 minus 20 so 50 times 80 that's 4,000 00:49:56.58000:49:56.590 times four point ninety four so it 00:50:00.45000:50:00.460 requires sixteen thousand seven hundred 00:50:03.48000:50:03.490 thirty six joules to heat liquid water 00:50:06.09000:50:06.100 from 20 to 100 degrees Celsius now let's 00:50:11.88000:50:11.890 focus on the second part q2 which is a 00:50:15.00000:50:15.010 phase change so to calculate the energy 00:50:18.48000:50:18.490 required to vaporize water from 100 00:50:21.63000:50:21.640 degrees Celsius to steam at 100 we need 00:50:24.87000:50:24.880 to do a conversion so let's start with 00:50:27.41000:50:27.420 50 grams of water and let's convert it 00:50:32.04000:50:32.050 to moles so we know that the molar mass 00:50:34.80000:50:34.810 of h2o is about 18 and the heat of 00:50:40.58000:50:40.590 vaporization for water is about 41 00:50:43.53000:50:43.540 kilojoules per one mole so grams will 00:50:48.66000:50:48.670 cancel and the unit moles cancel as well 00:50:51.23000:50:51.240 so it's going to be 15 divided by 18 00:50:55.01000:50:55.020 times 41 now you should get one hundred 00:50:59.88000:50:59.890 thirteen point nine kilojoules but 00:51:02.88000:51:02.890 notice that the unit here is in joules 00:51:04.77000:51:04.780 so we're going to have to add q1 q2 and 00:51:06.69000:51:06.700 q3 in order to add them they must share 00:51:09.78000:51:09.790 the same unit so we need to convert 00:51:11.73000:51:11.740 kilojoules into joules so therefore we 00:51:16.08000:51:16.090 can add an extra step we can multiply 00:51:18.99000:51:19.000 this by a thousand joules per kilogram 00:51:29.08000:51:29.090 and the unit kilojoules will cancel so 00:51:33.81000:51:33.820 therefore this is going to be equal to 00:51:37.80000:51:37.810 one hundred thirteen thousand eight 00:51:42.10000:51:42.110 hundred and eighty nine joules so now we 00:51:46.90000:51:46.910 need to find q3 which is a temperature 00:51:49.36000:51:49.370 change it's the amount of energy that's 00:51:51.79000:51:51.800 required to heat up steam from 100 to 00:51:54.01000:51:54.020 170 so we're going to use the equation 00:51:57.37000:51:57.380 MCAT again so the mass is 50 the heat 00:52:01.18000:52:01.190 capacity for steam is about half of that 00:52:03.79000:52:03.800 for liquid water it's about two joules 00:52:06.31000:52:06.320 per gram for Celsius the temperature 00:52:08.59000:52:08.600 change final minus initial that's 170 00:52:10.81000:52:10.820 minus 100 00:52:11.71000:52:11.720 so that's 70 50 times 70 is 35 hundred 00:52:17.68000:52:17.690 times 2 so that's about seven thousand 00:52:20.95000:52:20.960 joules so notice that most of the energy 00:52:26.59000:52:26.600 that's required is to vaporize liquid 00:52:29.41000:52:29.420 water into steam that's where you got to 00:52:32.29000:52:32.300 put most of the energy in it takes a lot 00:52:36.94000:52:36.950 of energy to vaporize liquid water to 00:52:39.58000:52:39.590 steam after that it doesn't take much 00:52:41.17000:52:41.180 energy to heat up steam so now let's 00:52:44.86000:52:44.870 find the total energy which is the sum 00:52:48.46000:52:48.470 of q1 q2 and q3 so seven thousand plus 00:52:55.81000:52:55.820 sixteen thousand 736 plus one hundred 00:52:59.29000:52:59.300 thirteen thousand eight hundred eighty 00:53:00.88000:53:00.890 nine the total energy is about 137 00:53:05.62000:53:05.630 thousand 625 so that's how many joules 00:53:11.20000:53:11.210 of heat energy is required to heat 50 00:53:13.93000:53:13.940 grams of water from 20 degrees Celsius 00:53:16.39000:53:16.400 to one hundred seventy and if you want 00:53:18.31000:53:18.320 to convert this to kilojoules divided by 00:53:20.50000:53:20.510 thousand so this is one hundred thirty 00:53:22.66000:53:22.670 seven point six kilojoules now what if 00:53:29.11000:53:29.120 you want to calculate the amount of heat 00:53:31.96000:53:31.970 energy that's required to heat up 80 00:53:34.93000:53:34.940 grams of ice from negative thirty 00:53:40.36000:53:40.37000:53:41.82900:53:41.839 to steam at 300 degrees Celsius what do 00:53:50.31900:53:50.329 you need to do well let's draw the 00:53:52.35900:53:52.369 heating curve for water so first we need 00:53:57.51900:53:57.529 to heat up ice and then ice is going to 00:54:00.57900:54:00.589 melt to liquid water then we need to 00:54:02.79900:54:02.809 heat up liquid water and then that's 00:54:04.66000:54:04.670 going to turn into steam and then when 00:54:06.91000:54:06.920 you say heat up steam so on the y-axis 00:54:10.42000:54:10.430 we have temperature on the x-axis we 00:54:12.72900:54:12.739 have the amount of heat energy that 00:54:15.75900:54:15.769 we're adding to it so this is the solid 00:54:19.23900:54:19.249 phase which is ice this is the liquid 00:54:21.21900:54:21.229 phase water and it's esteem on the 00:54:27.40000:54:27.410 horizontal line we have the solid and 00:54:29.85900:54:29.869 the liquid phase coexisting 00:54:31.42000:54:31.430 so the temperature that corresponds to 00:54:33.33900:54:33.349 it is zero degree Celsius on the second 00:54:36.18900:54:36.199 horizontal line we have the liquid and 00:54:37.92900:54:37.939 the gas phase coexisting 00:54:40.94900:54:40.959 so let me write into the right s for 00:54:43.32900:54:43.339 steam I'm going to write G for gas 00:54:48.29900:54:48.309 so there's five Q's that we need to 00:54:51.37000:54:51.380 calculate in this problem we're starting 00:54:54.24900:54:54.259 at negative 30 so we need to calculate 00:54:56.55900:54:56.569 q1 00:54:57.37000:54:57.380 that's for ice as we heat it from 00:54:59.46900:54:59.479 negative 30 it's a zero so you need the 00:55:01.98900:55:01.999 specific heat capacity fries which is 00:55:03.84900:55:03.859 very similar to steam it's about 2 00:55:06.13000:55:06.140 joules per gram per Celsius and then we 00:55:10.26900:55:10.279 need to calculate Q 2 that's the that's 00:55:12.93900:55:12.949 a phase change problem that's the energy 00:55:15.13000:55:15.140 that's required to melt ice into liquid 00:55:18.78900:55:18.799 water and then Q 3 is a temperature 00:55:21.18900:55:21.199 change problem where we need to use Q 00:55:23.55900:55:23.569 equals MCAT and we need to heat up 00:55:25.87000:55:25.880 liquid water from 0 to 100 degrees 00:55:30.16000:55:30.170 Celsius Q 4 is a phase change problem 00:55:34.35900:55:34.369 who need to do a conversion then we're 00:55:38.14000:55:38.150 going to vaporize liquid water into 00:55:40.12000:55:40.130 steam and Q 5 that's a temperature 00:55:43.35900:55:43.369 change we need to use MCAT again and 00:55:45.42900:55:45.439 we're going to heat up steam all the way 00:55:48.81900:55:48.829 to 300 degree Celsius so to solve this 00:55:51.96900:55:51.979 problem you need to find Q 1 Q 2 Q 3 Q 00:55:55.54900:55:55.559 4q5 make sure they have the same unit 00:55:58.27900:55:58.289 either joules or kilojoules and then add 00:56:01.18900:56:01.199 them so we're not going to actually do 00:56:04.42900:56:04.439 this example but I just want to give you 00:56:07.74900:56:07.759 the guideline of how to do it or how to 00:56:10.45900:56:10.469 set it up I just want to help you to see 00:56:12.79900:56:12.809 the process of solving it by the way 00:56:16.00900:56:16.019 keeping this in mind as you add heat to 00:56:19.51900:56:19.529 ice whenever the temperature increases 00:56:22.68900:56:22.699 that's in these three segments for q1 q3 00:56:25.75900:56:25.769 and q5 the kinetic energy is increasing 00:56:29.19900:56:29.209 so as you add Heat as the temperature 00:56:31.96900:56:31.979 goes up the kinetic energy of the 00:56:33.38000:56:33.390 molecules increases now during the 00:56:37.51900:56:37.529 horizontal part of the graph that's Q 2 00:56:39.85900:56:39.869 and Q 4 as you add heat the solid and Q 00:56:44.35900:56:44.369 2 is melted into a liquid and so because 00:56:48.10900:56:48.119 the temperature is constant you're not 00:56:49.37000:56:49.380 increasing the kinetic energy what's 00:56:51.76900:56:51.779 happening is you're increasing the 00:56:53.59900:56:53.609 potential energy of the water molecules 00:56:56.23900:56:56.249 as you melt them from liquid from solid 00:56:58.96900:56:58.979 to liquid the same is true for Q 4 as 00:57:01.24900:57:01.259 you vaporize liquid water and cysteine 00:57:04.90000:57:04.910 your increase in the potential energy of 00:57:07.77900:57:07.789 the system 00:57:23.75000:57:23.760 seventy grams of metal at 200 degrees 00:57:27.44000:57:27.450 Celsius was dropped in the bucket 00:57:29.81000:57:29.820 containing 40 grams of water at 20 00:57:32.27000:57:32.28000:57:33.14000:57:33.150 the temperature increased at 32 degrees 00:57:35.84000:57:35.850 Celsius calculate the specific heat 00:57:38.18000:57:38.190 capacity of the metal so now we're going 00:57:40.52000:57:40.530 to work on some heat transfer problems 00:57:42.02000:57:42.030 this is one of them and how can we solve 00:57:45.44000:57:45.450 this so first let's understand what's 00:57:47.90000:57:47.910 happening so let's say if we have a 00:57:51.71000:57:51.720 bucket that has water and we add a metal 00:57:56.21000:57:56.220 to it so initially the temperature of 00:57:59.59900:57:59.609 the metal is 200 degrees Celsius and the 00:58:03.41000:58:03.420 temperature of the water is 20 so heat 00:58:06.47000:58:06.480 is going to flow from hot to cold so 00:58:09.98000:58:09.990 heat is going to flow out of the metal 00:58:12.97000:58:12.980 into liquid water so because he is being 00:58:18.85900:58:18.869 released from the metal it's exothermic 00:58:21.80000:58:21.810 from the metal the temperature of the 00:58:23.24000:58:23.250 metal is going to go down now since heat 00:58:29.03000:58:29.040 is being absorbed by the liquid water 00:58:30.74000:58:30.750 molecules the temperature of the water 00:58:32.45000:58:32.460 is going to go up eventually equilibrium 00:58:37.22000:58:37.230 will be established and the temperature 00:58:44.90000:58:44.910 of the metal will decrease to a point 00:58:48.43000:58:48.440 where water and let me rephrase that 00:58:53.78000:58:53.790 so the temperature of the metal is going 00:58:55.25000:58:55.260 to decrease and the temperature of the 00:58:57.08000:58:57.090 water is going to increase at some point 00:58:59.27000:58:59.280 at some temperature these two will be 00:59:02.63000:59:02.640 the same they're going to meet at some 00:59:03.95000:59:03.960 point and that point is at 32 degrees 00:59:07.07000:59:07.080 Celsius once the temperature of the 00:59:12.41000:59:12.420 metal is the same as the temperature of 00:59:16.25000:59:16.260 the water at that point equilibrium has 00:59:19.40000:59:19.410 been established the temperature will no 00:59:21.62000:59:21.630 longer change it's going to remain at 32 00:59:23.68000:59:23.690 the amount of heat that leaves the metal 00:59:27.41000:59:27.420 is going to be equal to the amount of 00:59:29.84000:59:29.850 heat that goes back to the metal from 00:59:32.39000:59:32.400 the water so at that point the 00:59:36.53000:59:36.540 temperature 00:59:37.19000:59:37.200 I can change that's the final 00:59:38.36000:59:38.370 temperature which is the same for both 00:59:42.16000:59:42.170 so now let's see if we can calculate the 00:59:46.34000:59:46.350 specific heat capacity of a metal so the 00:59:48.89000:59:48.900 amount of heat energy that's released by 00:59:51.68000:59:51.690 the metal is equal to the amount of heat 00:59:53.99000:59:54.000 energy that's absorbed by the water so 00:59:58.61000:59:58.620 whenever you have a heat transfer 01:00:00.68001:00:00.690 problem where you have to set QA equal 01:00:03.35001:00:03.360 to QB you need the one side to be 01:00:05.87001:00:05.880 negative for the stored because on one 01:00:08.72001:00:08.730 side the temperature is increasing and 01:00:10.61001:00:10.620 on the other side the temperature is 01:00:11.84001:00:11.850 decreasing so to make sure the math 01:00:14.38001:00:14.390 works out you have to have a negative 01:00:17.12001:00:17.130 sign on one side doesn't matter which 01:00:18.74001:00:18.750 side you put it on as long as you have a 01:00:20.87001:00:20.880 negative sign it's going to work so 01:00:24.56001:00:24.570 let's go ahead and solve it so Q equals 01:00:29.24001:00:29.250 M cap for both sides on the left side 01:00:32.78001:00:32.790 we're dealing with a metal the mass of 01:00:34.97001:00:34.980 the metal is 70 we're looking for C the 01:00:37.37001:00:37.380 heat capacity of the metal or the 01:00:39.26001:00:39.270 specific heat capacity delta T is the 01:00:42.35001:00:42.360 final temperature minus the initial 01:00:43.76001:00:43.770 temperature the final temperature is 32 01:00:46.40001:00:46.410 for both samples the initial temperature 01:00:48.74001:00:48.750 for the metal is 200 the mass of the 01:00:52.40001:00:52.410 water is 40 the heat capacity is 4.1 84 01:00:55.16001:00:55.170 and the temperature change its final 01:00:58.37001:00:58.380 minus the initial which is 20 so we have 01:01:02.33001:01:02.340 negative 70 C times 32 minus 200 that's 01:01:08.99001:01:09.000 negative 168 forty times 4.184 that's 01:01:20.17001:01:20.180 167 point three six and 32 minus twenty 01:01:24.53001:01:24.540 is 12 positive 12 01:01:27.98001:01:27.990 notice that the two negative signs 01:01:30.26001:01:30.270 cancel and it makes the left side 01:01:33.35001:01:33.360 positive which is equal to the right 01:01:35.57001:01:35.580 side which is also positive that's why 01:01:37.91001:01:37.920 this negative sign is important so 70 01:01:43.13001:01:43.140 times 168 that's eleven seventy well 01:01:49.66001:01:49.670 11700 01:01:50.72001:01:50.730 sixty times C and twelve times 160 7.36 01:02:03.05001:02:03.060 is about two thousand eight point three 01:02:07.55001:02:07.560 two so to solve for C we need to divide 01:02:10.25001:02:10.260 both sides by eleven thousand seven 01:02:12.50001:02:12.510 sixty so two thousand eight point three 01:02:15.13001:02:15.140 2/11 seven sixty the answer is point one 01:02:21.38001:02:21.390 seven that's the heat capacity for the 01:02:24.05001:02:24.060 model notice that it's very low the 01:02:28.22001:02:28.230 metal had a very large temperature 01:02:30.71001:02:30.720 change let's compare the two so on the 01:02:35.81001:02:35.820 left side we have the metal and on the 01:02:40.16001:02:40.170 right side water so the heat capacity of 01:02:44.93001:02:44.940 the metal we calculated to be about 01:02:47.27001:02:47.280 point seventeen for water it's four 01:02:49.64001:02:49.650 point one eight four now the temperature 01:02:55.34001:02:55.350 for the metal and went from 200 to 01:02:58.03001:02:58.040 thirty-two that's a large temperature 01:03:00.71001:03:00.720 change but for water it increased from 01:03:04.28001:03:04.290 twenty to thirty two so here the 01:03:07.84901:03:07.859 temperature change is only twelve and on 01:03:11.69001:03:11.700 this side the temperature change is 168 01:03:17.32001:03:17.330 so notice that a substance with a very 01:03:21.10901:03:21.119 high heat capacity typically experience 01:03:24.47001:03:24.480 a very small temperature change for the 01:03:27.14001:03:27.150 same amount of energy a substance with a 01:03:29.51001:03:29.520 very low heat capacity will experience a 01:03:32.39001:03:32.400 very high temperature change metals 01:03:34.43001:03:34.440 usually have very low heat capacities 01:03:36.80001:03:36.810 which means that they can conduct heat 01:03:40.40001:03:40.410 very well if you put a metal on a stove 01:03:42.85901:03:42.869 the temperature of the metal is going to 01:03:44.78001:03:44.790 increase rapidly almost instantaneously 01:03:49.39001:03:49.400 metals are good conductors of heat now 01:03:52.49001:03:52.500 let's say if you pour water in a metal 01:03:56.03001:03:56.040 bucket and heat it up the temperature of 01:03:58.46001:03:58.470 the water is not going to increase 01:03:59.51001:03:59.520 drastically quickly it's going to take 01:04:03.14001:04:03.150 some time 01:04:04.10001:04:04.110 because it takes a lot more energy for 01:04:06.86001:04:06.870 water to increase its temperature by one 01:04:09.11001:04:09.120 degree Celsius compared to a metal so 01:04:12.53001:04:12.540 make sure you keep this concept in mind 01:04:14.35001:04:14.360 substances that have a very high 01:04:16.25001:04:16.260 specific heat capacity are usually 01:04:20.51001:04:20.520 associated with a small temperature 01:04:21.95001:04:21.960 change it takes a lot of energy to raise 01:04:24.29001:04:24.300 the temperature so water can store more 01:04:27.80001:04:27.810 heat energy than the metal because it 01:04:29.39001:04:29.400 has a higher heat capacity substances 01:04:32.03001:04:32.040 with a very low heat capacity can't 01:04:36.14001:04:36.150 store a lot of heat energy they can 01:04:38.06001:04:38.070 transfer heat pretty well they're good 01:04:39.59001:04:39.600 conductors of heat but they don't store 01:04:41.75001:04:41.760 heat energy very well and so if you add 01:04:44.36001:04:44.370 heat to it the temperature of such 01:04:46.46001:04:46.470 substances will increase greatly
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