WEBVTT 01-01 00:00:00.180 --> 00:00:04.276 Welcome to Week five, from phases to chemical reactions. 01-02 00:00:04.456 --> 00:00:08.595 This is our independent lesson for Friday, November sixth. 01-03 00:00:08.775 --> 00:00:12.956 We will meet onsite again on Friday, November thirteenth. 01-04 00:00:13.136 --> 00:00:17.403 Please begin with the short review in section R of the worksheet. 01-05 00:00:17.583 --> 00:00:23.044 Your recent questions showed that we should separate two ideas before moving on. 01-06 00:00:23.224 --> 00:00:32.205 Bulk phase equilibrium concerns a substance distributed between macroscopic phases, such as a solid and its vapor. 01-07 00:00:32.385 --> 00:00:36.375 Adsorption concerns uptake associated with a surface. 01-08 00:00:36.555 --> 00:00:43.104 The amount adsorbed depends on an adsorption relation and the available surface capacity. 01-09 00:00:43.284 --> 00:00:47.252 A vapor pressure alone does not describe that uptake. 01-10 00:00:47.432 --> 00:00:55.961 Today, we will use the same habit of combining an equilibrium relation with conservation, but the species can also react. 01-11 00:00:56.141 --> 00:01:01.709 Our main example is a synthetic dissociation, A two forming two A atoms. 01-12 00:01:01.889 --> 00:01:05.431 We will follow one worksheet and use Lab twelve. 01-13 00:01:05.611 --> 00:01:10.923 Keep the worksheet beside the video so every pause has a clear place to work. 02-01 00:01:13.453 --> 00:01:18.560 Before we discuss reactions, revisit pressure in a closed, rigid vessel. 02-02 00:01:18.740 --> 00:01:21.428 The temperature and the volume are fixed. 02-03 00:01:21.608 --> 00:01:27.923 The ideal gas equation connects the amount of each gas to its own partial pressure. 02-04 00:01:28.103 --> 00:01:31.623 Total pressure is the sum of those partial pressures. 02-05 00:01:31.803 --> 00:01:38.928 If all of the volatile solid has disappeared, calculate the volatile gas pressure from its available amount. 02-06 00:01:39.108 --> 00:01:46.255 Do not continue forcing its pressure to equal the saturation value after that phase is depleted. 02-07 00:01:46.435 --> 00:01:50.851 The assumption of an ideal gas is a separate modeling choice. 02-08 00:01:51.031 --> 00:01:55.489 Losing the solid phase does not make a real gas ideal. 02-09 00:01:55.669 --> 00:01:59.104 Now pause for four minutes and complete R one. 02-10 00:01:59.284 --> 00:02:12.297 Use the values printed in the worksheet, calculate the volatile partial pressure and total pressure, then state which pressure changes when inert gas is added at fixed temperature and volume. 02-11 00:02:12.477 --> 00:02:15.443 Keep the volatile inventory unchanged. 02-12 00:02:15.623 --> 00:02:17.287 Record a unit check. 02-13 00:02:17.467 --> 00:02:20.901 After your attempt, continue to the next question. 02-14 00:02:21.081 --> 00:02:23.961 We will check both review answers together. 03-01 00:02:26.491 --> 00:02:30.673 The next question concerns adsorption in a finite vessel. 03-02 00:02:30.853 --> 00:02:37.295 Conservation tells us that the feed is divided between free gas and adsorbed material. 03-03 00:02:37.475 --> 00:02:43.427 The Langmuir equation supplies another condition that the proposed state must satisfy. 03-04 00:02:43.607 --> 00:02:52.738 Choosing two amounts that add to the feed is insufficient if the adsorbed amount disagrees with the isotherm at the same pressure. 03-05 00:02:52.918 --> 00:02:57.227 In R two, the pressure and the model parameters are supplied. 03-06 00:02:57.407 --> 00:03:02.186 First calculate the gas amount from pressure, volume and temperature. 03-07 00:03:02.366 --> 00:03:09.918 Independently calculate the loading from the Langmuir equation, then multiply that loading by the adsorbent mass. 03-08 00:03:10.098 --> 00:03:15.154 Only after these two calculations should you compare their sum with the feed. 03-09 00:03:15.334 --> 00:03:21.265 Use units of mole per kilogram for the loading and kilograms for the adsorbent mass. 03-10 00:03:21.445 --> 00:03:24.111 Their product is an amount in moles. 03-11 00:03:24.291 --> 00:03:27.299 Pause for four minutes and complete R two. 03-12 00:03:27.479 --> 00:03:33.261 If a number seems unreasonable, write which equation or conversion you need to check. 03-13 00:03:33.441 --> 00:03:36.043 Continue when the time budget ends. 04-01 00:03:38.573 --> 00:03:40.237 Check R one first. 04-02 00:03:40.417 --> 00:03:51.041 The gas capacity at the sublimation pressure is about four thousandths of a mole, which exceeds the available zero point zero zero two moles. 04-03 00:03:51.221 --> 00:03:53.312 All of the solid disappears. 04-04 00:03:53.492 --> 00:03:58.996 The volatile partial pressure is about four point nine eight nine kilopascals. 04-05 00:03:59.176 --> 00:04:12.851 Adding the specified inert gas at fixed temperature and volume raises total pressure to about nine point nine seven seven kilopascals, while the volatile partial pressure stays unchanged. 04-06 00:04:13.031 --> 00:04:14.759 Now check R two. 04-07 00:04:14.939 --> 00:04:26.253 At the proposed ten kilopascals, the gas contains about four thousandths of a mole, and the isotherm requires zero point zero one moles on the surface. 04-08 00:04:26.433 --> 00:04:28.374 Their sum exceeds the feed. 04-09 00:04:28.554 --> 00:04:33.930 That proposed pressure cannot satisfy this model and inventory together. 04-10 00:04:34.110 --> 00:04:39.870 The checked equilibrium pressure is about six point zero eight zero kilopascals. 04-11 00:04:40.050 --> 00:04:49.799 It gives about zero point zero zero two four four moles of gas and zero point zero zero seven five six moles adsorbed. 04-12 00:04:49.979 --> 00:04:51.857 The sum matches the feed. 04-13 00:04:52.037 --> 00:04:58.245 We will use the same combination of equilibrium and conservation checks for reactions. 05-01 00:05:00.775 --> 00:05:03.463 Here is the complete route for this week. 05-02 00:05:03.643 --> 00:05:07.931 Use the six-page Week five worksheet as your working record. 05-03 00:05:08.111 --> 00:05:12.505 Keep Lab twelve open in a second tab when we reach section D. 05-04 00:05:12.685 --> 00:05:17.208 The short Lab guide is available when you need help finding a control. 05-05 00:05:17.388 --> 00:05:20.588 The worked solutions are for checking after an attempt. 05-06 00:05:20.768 --> 00:05:25.824 There is no additional required preparation or follow-up assignment. 05-07 00:05:26.004 --> 00:05:31.913 The planned activities use one hundred and twenty-eight minutes, with a separate ten-minute break. 05-08 00:05:32.093 --> 00:05:38.429 Add the recording duration shown on the video page for a total of approximately three hours. 05-09 00:05:38.609 --> 00:05:40.807 Your actual time may differ. 05-10 00:05:40.987 --> 00:05:44.016 Please use the suggested stop budgets. 05-11 00:05:44.196 --> 00:05:49.551 If you remain stuck at the end of a pause, record the specific step and continue. 05-12 00:05:49.731 --> 00:05:54.189 Bring that question to our onsite class on November thirteenth. 05-13 00:05:54.369 --> 00:05:57.847 You do not need to repeat the whole worksheet neatly. 05-14 00:05:58.027 --> 00:06:04.128 A first attempt, a correction and a short explanation provide a useful record of your learning. 06-01 00:06:06.658 --> 00:06:11.010 Our reaction is A two gas in equilibrium with two A gas. 06-02 00:06:11.190 --> 00:06:14.326 We begin with one mole of A two and no A. 06-03 00:06:14.506 --> 00:06:18.474 No inert gas is present in the three required cases. 06-04 00:06:18.654 --> 00:06:26.718 A two contains two A atoms, so the reaction conserves atoms even though the number of gas molecules changes. 06-05 00:06:26.898 --> 00:06:33.682 Reaction extent tells us how far the reaction has proceeded on the written stoichiometric basis. 06-06 00:06:33.862 --> 00:06:37.574 Consuming one mole of A two produces two moles of A. 06-07 00:06:37.754 --> 00:06:45.370 The general expression is the initial amount plus the stoichiometric coefficient times the reaction extent. 06-08 00:06:45.550 --> 00:06:50.777 Reactant coefficients are negative and product coefficients are positive. 06-09 00:06:50.957 --> 00:06:58.253 For this particular feed, it is convenient to define f as the reaction extent divided by one mole. 06-10 00:06:58.433 --> 00:07:00.865 The fraction f is dimensionless. 06-11 00:07:01.045 --> 00:07:05.994 The amounts are one minus f moles of A two and two f moles of A. 06-12 00:07:06.174 --> 00:07:10.398 Nonnegative amounts require f between zero and one. 06-13 00:07:10.578 --> 00:07:16.697 These bounds come from the available material, before we use any equilibrium constant. 07-01 00:07:19.227 --> 00:07:22.576 To describe equilibrium, we need activities. 07-02 00:07:22.756 --> 00:07:33.163 For today's ideal gas model, the activity of a species is its mole fraction multiplied by total pressure, divided by the standard pressure. 07-03 00:07:33.343 --> 00:07:35.924 Both pressures must use the same units. 07-04 00:07:36.104 --> 00:07:40.157 Their ratio is dimensionless, and so is the activity. 07-05 00:07:40.337 --> 00:07:45.713 We use a standard pressure of one hundred kilopascals throughout this lesson. 07-06 00:07:45.893 --> 00:07:48.091 This is a reference convention. 07-07 00:07:48.271 --> 00:07:52.943 The actual total pressure is a separate input that we can change. 07-08 00:07:53.123 --> 00:08:01.059 For A two dissociating into two A, the quotient is the activity of A squared, divided by the activity of A two. 07-09 00:08:01.239 --> 00:08:05.036 The exponents come directly from the reaction coefficients. 07-10 00:08:05.216 --> 00:08:09.909 Write this activity expression before substituting mole fractions. 07-11 00:08:10.089 --> 00:08:14.612 That step makes it easier to see how pressure enters the calculation. 07-12 00:08:14.792 --> 00:08:20.488 Later we will compare one hundred and four hundred kilopascals at the same temperature. 07-13 00:08:20.668 --> 00:08:25.703 Our standard pressure stays at one hundred kilopascals in both cases. 08-01 00:08:28.233 --> 00:08:32.265 Please pause for ten minutes and complete A one through A three. 08-02 00:08:32.445 --> 00:08:34.322 Begin with a mole table. 08-03 00:08:34.502 --> 00:08:39.003 Write the amounts of A two, A and the total gas in terms of f. 08-04 00:08:39.183 --> 00:08:44.773 Divide each species amount by the total amount to obtain its mole fraction. 08-05 00:08:44.953 --> 00:08:47.193 Check that the fractions sum to one. 08-06 00:08:47.373 --> 00:08:53.922 Substitute those fractions into the activity definition and then into the reaction quotient. 08-07 00:08:54.102 --> 00:08:57.899 Try to simplify the expression without looking ahead. 08-08 00:08:58.079 --> 00:09:02.239 Keep the actual-pressure to standard-pressure ratio visible. 08-09 00:09:02.419 --> 00:09:11.742 Evaluate the quotient at a trial fraction of zero point five, first at a pressure ratio of one and then at a ratio of four. 08-10 00:09:11.922 --> 00:09:17.213 Finally, check the total number of A atoms using the stoichiometric counts. 08-11 00:09:17.393 --> 00:09:25.175 The total molecular amount can change, but the atom inventory must remain equal to the initial inventory. 08-12 00:09:25.355 --> 00:09:34.145 If the algebra becomes difficult, retain the unsimplified expression and substitute one trial value of f to test it. 08-13 00:09:34.325 --> 00:09:37.119 That is a useful intermediate result. 08-14 00:09:37.299 --> 00:09:42.675 At the end of ten minutes, record any unfinished step and continue to the check. 08-15 00:09:42.855 --> 00:09:45.415 The next slide shows the derivation. 09-01 00:09:47.945 --> 00:09:51.359 The total gas amount is one plus f moles. 09-02 00:09:51.539 --> 00:09:57.128 Therefore, the mole fraction of A two is one minus f, divided by one plus f. 09-03 00:09:57.308 --> 00:10:01.660 The mole fraction of A is two f, divided by one plus f. 09-04 00:10:01.840 --> 00:10:11.120 Substituting these into the quotient gives four f squared divided by one minus f squared, multiplied by the pressure ratio. 09-05 00:10:11.300 --> 00:10:19.961 One pressure factor remains because the product side has one more mole of gas per mole of reaction than the reactant side. 09-06 00:10:20.141 --> 00:10:27.587 If your expression has two pressure factors or none, return to the activity exponents and cancel them explicitly. 09-07 00:10:27.767 --> 00:10:32.204 The atom check is twice the amount of A two plus the amount of A. 09-08 00:10:32.384 --> 00:10:37.675 Substitution gives two moles of conserved A atoms for every feasible f. 09-09 00:10:37.855 --> 00:10:43.359 Notice that molecular moles and atom moles describe different inventories. 09-10 00:10:43.539 --> 00:10:56.253 At a trial fraction of zero point five, the quotient is one point three three three three at a pressure ratio of one, and five point three three three three at a pressure ratio of four. 09-11 00:10:56.433 --> 00:11:01.447 These are trial states, so we have not yet established equilibrium. 09-12 00:11:01.627 --> 00:11:06.533 The quotient changes with f, while the atom balance remains satisfied. 10-01 00:11:09.063 --> 00:11:12.669 How can the quotient tell us which direction is favored? 10-02 00:11:12.849 --> 00:11:23.771 At fixed temperature and pressure, the reaction Gibbs energy is the standard reaction Gibbs energy plus R T times the natural logarithm of Q. 10-03 00:11:23.951 --> 00:11:30.287 An equivalent form is R T times the natural logarithm of Q divided by K. 10-04 00:11:30.467 --> 00:11:33.646 A negative value favors forward reaction. 10-05 00:11:33.826 --> 00:11:37.111 A positive value favors reverse reaction. 10-06 00:11:37.291 --> 00:11:43.755 At an interior equilibrium, the reaction Gibbs energy is zero and Q equals K. 10-07 00:11:43.935 --> 00:11:47.199 These statements refer to the reaction as written. 10-08 00:11:47.379 --> 00:11:51.838 Reversing the reaction reverses the corresponding sign convention. 10-09 00:11:52.018 --> 00:12:01.725 For B one, use the supplied standard reaction Gibbs energy of minus four thousand joules per mole at three hundred and fifty kelvin. 10-10 00:12:01.905 --> 00:12:05.361 Test Q equal to one and Q equal to ten. 10-11 00:12:05.541 --> 00:12:11.514 Pause for six minutes, calculate the signs and write the favored direction for each state. 10-12 00:12:11.694 --> 00:12:13.998 Use the natural logarithm. 10-13 00:12:14.178 --> 00:12:19.213 We will return to these answers after calculating K by two routes. 11-01 00:12:21.743 --> 00:12:27.737 The first route starts with standard reaction Gibbs energy at the temperature of interest. 11-02 00:12:27.917 --> 00:12:35.704 The equation is natural log K equals minus standard reaction Gibbs energy divided by R T. 11-03 00:12:35.884 --> 00:12:43.223 For this example, the supplied standard reaction Gibbs energy applies at three hundred and fifty kelvin. 11-04 00:12:43.403 --> 00:12:47.733 It is minus four kilojoules per mole of reaction as written. 11-05 00:12:47.913 --> 00:12:56.063 To use the gas constant in joules per mole per kelvin, convert that energy to minus four thousand joules per mole. 11-06 00:12:56.243 --> 00:13:00.232 Keep the two minus signs separate until you substitute. 11-07 00:13:00.412 --> 00:13:08.732 A negative standard reaction Gibbs energy gives a positive natural logarithm of K, so K should exceed one. 11-08 00:13:08.912 --> 00:13:12.837 This sign prediction is a quick check on the arithmetic. 11-09 00:13:13.017 --> 00:13:20.740 The final equilibrium constant is dimensionless because it is defined using dimensionless activities. 11-10 00:13:20.920 --> 00:13:29.482 A larger K describes a stronger product preference in this convention, but it does not by itself specify the conversion. 11-11 00:13:29.662 --> 00:13:32.691 The feed and constraints are also needed. 12-01 00:13:35.221 --> 00:13:44.139 The second route constructs standard reaction Gibbs energy from standard reaction enthalpy and entropy at the same temperature. 12-02 00:13:44.319 --> 00:13:49.524 For the reference case, the enthalpy is fifty kilojoules per mole of reaction. 12-03 00:13:49.704 --> 00:13:56.979 The entropy is about one hundred and fifty-four point two eight five seven joules per mole per kelvin. 12-04 00:13:57.159 --> 00:14:03.175 Convert either the energy or entropy so that the subtraction uses consistent units. 12-05 00:14:03.355 --> 00:14:10.843 In joules, calculate fifty thousand minus three hundred and fifty multiplied by the entropy value. 12-06 00:14:11.023 --> 00:14:15.780 Then use that Gibbs energy in the equation for natural log K. 12-07 00:14:15.960 --> 00:14:21.635 These two routes should agree because the synthetic inputs were constructed consistently. 12-08 00:14:21.815 --> 00:14:31.647 In an experimental or literature calculation, agreement also depends on using the same phases, standard states and reaction basis. 12-09 00:14:31.827 --> 00:14:38.525 If the routes disagree, check the inputs before interpreting the discrepancy as physical behavior. 12-10 00:14:38.705 --> 00:14:46.172 For today's comparison, keep all supplied properties at the reference temperature of three hundred and fifty kelvin. 13-01 00:14:48.702 --> 00:14:52.585 Pause for ten minutes and complete B two through B four. 13-02 00:14:52.765 --> 00:14:59.143 Calculate natural log K and then K from the directly supplied standard Gibbs energy. 13-03 00:14:59.323 --> 00:15:07.317 Next, calculate standard Gibbs energy from enthalpy and entropy, and use that value to obtain a second K. 13-04 00:15:07.497 --> 00:15:12.723 Show the energy conversion on your worksheet so someone else can follow your units. 13-05 00:15:12.903 --> 00:15:18.493 Keep several digits during the calculation and round only the reported result. 13-06 00:15:18.673 --> 00:15:24.177 Compare the two routes and write why agreement is expected for this synthetic case. 13-07 00:15:24.357 --> 00:15:31.909 If you obtain a reciprocal value or a very small constant, revisit the minus sign in the exponential. 13-08 00:15:32.089 --> 00:15:38.958 If your two Gibbs energies differ by a factor of a thousand, check the joule and kilojoule conversion. 13-09 00:15:39.138 --> 00:15:43.383 You can use a calculator or a spreadsheet for the arithmetic. 13-10 00:15:43.563 --> 00:15:50.198 The objective is to make the route from the supplied properties to the equilibrium constant visible. 13-11 00:15:50.378 --> 00:15:53.941 Continue to the numerical check after your attempt. 14-01 00:15:56.471 --> 00:16:04.961 Both routes give a standard reaction Gibbs energy of minus four thousand joules per mole at three hundred and fifty kelvin. 14-02 00:16:05.141 --> 00:16:13.653 Dividing its negative by R T gives natural log K equal to about one point three seven four five four. 14-03 00:16:13.833 --> 00:16:19.423 Exponentiating gives K equal to about three point nine five three two six. 14-04 00:16:19.603 --> 00:16:30.781 The enthalpy-entropy calculation reaches the same result because fifty thousand minus fifty-four thousand equals minus four thousand joules per mole. 14-05 00:16:30.961 --> 00:16:33.180 Return briefly to B one. 14-06 00:16:33.360 --> 00:16:44.432 When Q is one, its natural logarithm is zero, so the reaction Gibbs energy is minus four kilojoules per mole and forward reaction is favored. 14-07 00:16:44.612 --> 00:16:47.298 When Q is ten, Q exceeds K. 14-08 00:16:47.478 --> 00:16:53.025 The logarithm of Q over K is positive, so reverse reaction is favored. 14-09 00:16:53.205 --> 00:16:56.640 At equilibrium, Q equals this finite K. 14-10 00:16:56.820 --> 00:17:02.452 The standard reaction Gibbs energy remains minus four kilojoules per mole. 14-11 00:17:02.632 --> 00:17:07.005 It is the actual reaction Gibbs energy that becomes zero. 15-01 00:17:09.535 --> 00:17:12.714 Now use K to calculate composition. 15-02 00:17:12.894 --> 00:17:17.779 At equilibrium, set the quotient expression from section A equal to K. 15-03 00:17:17.959 --> 00:17:29.474 If r denotes the actual-pressure to standard-pressure ratio, the equation is K equals four f squared times r, divided by one minus f squared. 15-04 00:17:29.654 --> 00:17:35.286 Rearranging gives f squared equal to K divided by K plus four r. 15-05 00:17:35.466 --> 00:17:43.508 Take the nonnegative root because our initial feed allows only forward extent between zero and one mole. 15-06 00:17:43.688 --> 00:17:48.531 For the reference pressure of one hundred kilopascals, r equals one. 15-07 00:17:48.711 --> 00:17:52.850 Pause for fifteen minutes and complete C one through C three. 15-08 00:17:53.030 --> 00:18:00.852 Calculate f, convert it to reaction extent, and calculate both species amounts and their mole fractions. 15-09 00:18:01.032 --> 00:18:03.443 Then verify the atom balance. 15-10 00:18:03.623 --> 00:18:08.935 Reconstruct Q from the calculated mole fractions and the pressure ratio. 15-11 00:18:09.115 --> 00:18:13.083 This second calculation should return K within rounding. 15-12 00:18:13.263 --> 00:18:17.956 Please do not use the rearranged equation as your only check. 15-13 00:18:18.136 --> 00:18:22.424 The purpose is to test the final composition independently. 16-01 00:18:24.954 --> 00:18:29.733 For the reference case, f is about zero point seven zero five. 16-02 00:18:29.913 --> 00:18:35.140 The reaction extent is therefore about zero point seven zero five moles. 16-03 00:18:35.320 --> 00:18:45.074 The remaining A two amount is about zero point two nine five moles, and the A amount is about one point four one zero moles. 16-04 00:18:45.254 --> 00:18:49.115 Their sum is about one point seven zero five moles. 16-05 00:18:49.295 --> 00:18:58.596 Dividing by this total gives mole fractions of about zero point one seven three for A two and zero point eight two seven for A. 16-06 00:18:58.776 --> 00:19:04.600 These fractions differ from the species amounts, even though both describe the same state. 16-07 00:19:04.780 --> 00:19:10.071 Use the more precise on-screen values when checking numerical closure. 16-08 00:19:10.251 --> 00:19:13.344 The atom check gives two moles of A atoms. 16-09 00:19:13.524 --> 00:19:23.679 At this pressure, the activities equal the mole fractions numerically, so the activity quotient returns about three point nine five three two six. 16-10 00:19:23.859 --> 00:19:30.472 K greater than one has produced substantial dissociation, while a finite amount of A two remains. 17-01 00:19:33.002 --> 00:19:40.874 Next, increase total pressure to four hundred kilopascals while keeping temperature at three hundred and fifty kelvin. 17-02 00:19:41.054 --> 00:19:43.998 Keep the same feed and standard pressure. 17-03 00:19:44.178 --> 00:19:49.469 Before calculating, predict whether dissociation should increase or decrease. 17-04 00:19:49.649 --> 00:19:54.428 The reaction produces more gas molecules in the forward direction. 17-05 00:19:54.608 --> 00:19:59.429 At a fixed composition, increasing pressure therefore increases Q. 17-06 00:19:59.609 --> 00:20:06.265 Use that observation to predict which way the system must react to recover Q equal to K. 17-07 00:20:06.445 --> 00:20:11.202 Then calculate the new equilibrium using r equal to four. 17-08 00:20:11.382 --> 00:20:14.348 Pause for eight minutes and complete C four. 17-09 00:20:14.528 --> 00:20:18.645 Report the new extent, species amounts and mole fractions. 17-10 00:20:18.825 --> 00:20:26.057 Write one sentence explaining what happens to K and one sentence explaining what happens to composition. 17-11 00:20:26.237 --> 00:20:29.842 Use the temperature condition in your explanation. 17-12 00:20:30.022 --> 00:20:38.300 It is possible for equilibrium composition to change while the standard equilibrium constant remains the same. 17-13 00:20:38.480 --> 00:20:44.837 This controlled comparison will help you distinguish those two quantities when you open the Lab. 18-01 00:20:47.367 --> 00:20:58.012 At four hundred kilopascals, K remains about three point nine five three two six because temperature and the standard-state convention are unchanged. 18-02 00:20:58.192 --> 00:21:04.251 The equilibrium extent decreases to about zero point four four five one moles. 18-03 00:21:04.431 --> 00:21:14.710 The A two amount becomes about zero point five five four nine moles, and the A amount becomes about zero point eight nine zero two moles. 18-04 00:21:14.890 --> 00:21:25.147 The A mole fraction is about zero point six one six zero three, and the A two mole fraction is about zero point three eight three nine seven. 18-05 00:21:25.327 --> 00:21:30.042 The higher-pressure mixture therefore contains less dissociated A. 18-06 00:21:30.222 --> 00:21:38.755 You can explain this directly from the quotient: at a fixed f, multiplying pressure by four multiplies Q by four. 18-07 00:21:38.935 --> 00:21:43.351 Reverse reaction then reduces f until Q returns to K. 18-08 00:21:43.531 --> 00:21:50.144 This is more precise than applying a pressure slogan without specifying the reaction or constraint. 18-09 00:21:50.324 --> 00:21:55.572 Keep the baseline and pressure-only cases beside each other in your worksheet. 18-10 00:21:55.752 --> 00:22:01.555 The atom total should be two in both cases, even though the molecular totals differ. 18-11 00:22:01.735 --> 00:22:07.900 We will now reproduce these states using the same ideal gas model in Lab twelve. 19-01 00:22:10.430 --> 00:22:15.124 Open Lab twelve using the link on this slide or the video page. 19-02 00:22:15.304 --> 00:22:19.186 Its title is Chemical equilibrium and routes to K. 19-03 00:22:19.366 --> 00:22:22.033 Complete D one before continuing. 19-04 00:22:22.213 --> 00:22:27.013 Reset the synthetic example so the feed and property values are known. 19-05 00:22:27.193 --> 00:22:38.542 Check that evaluation temperature and reference temperature are three hundred and fifty kelvin, and that total pressure and standard pressure are one hundred kilopascals. 19-06 00:22:38.722 --> 00:22:41.858 Set the reaction heat-capacity change to zero. 19-07 00:22:42.038 --> 00:22:49.270 The default value is thirty, so this is an intentional change for today's constant-enthalpy comparison. 19-08 00:22:49.450 --> 00:22:51.776 Keep the inert amount at zero. 19-09 00:22:51.956 --> 00:22:59.209 For the first run, select the direct standard Gibbs energy as the K source for extent and driving force. 19-10 00:22:59.389 --> 00:23:03.101 The direct data apply at three hundred and fifty kelvin. 19-11 00:23:03.281 --> 00:23:06.310 Check the species table and the source note. 19-12 00:23:06.490 --> 00:23:09.477 Pause for five minutes to prepare these inputs. 19-13 00:23:09.657 --> 00:23:14.244 If you need help finding a control, use the two-page Lab guide. 19-14 00:23:14.424 --> 00:23:18.221 Resume for the demonstration when your inputs are ready. 20-01 00:23:20.751 --> 00:23:25.914 Here is the actual Lab twelve calculator with the reference inputs prepared. 20-02 00:23:26.094 --> 00:23:34.499 In the thermal-properties section, the heat-capacity change is zero, and the reference enthalpy and entropy match the worksheet. 20-03 00:23:34.679 --> 00:23:46.672 In the reaction-extent section, the total pressure is one hundred kilopascals, the inert amount is zero, and the selected K source is direct standard Gibbs energy. 20-04 00:23:46.852 --> 00:23:51.161 Now use Compare K methods to calculate the reference case. 20-05 00:23:51.341 --> 00:23:58.957 The methods table shows the agreement of the direct Gibbs and enthalpy-entropy routes at the reference temperature. 20-06 00:23:59.137 --> 00:24:08.360 The formation-energy sum is another route available in this synthetic example, but we do not need an additional worksheet task for it. 20-07 00:24:08.540 --> 00:24:12.124 Scroll to From K to equilibrium composition. 20-08 00:24:12.304 --> 00:24:19.660 The extent is about zero point seven zero five zero three moles, matching the hand calculation. 20-09 00:24:19.840 --> 00:24:23.488 Check the species amounts and the atom-balance message. 20-10 00:24:23.668 --> 00:24:27.146 The observed-activities panel needs careful reading. 20-11 00:24:27.326 --> 00:24:34.747 Its supplied mole fractions, zero point two and zero point eight, are independent observations. 20-12 00:24:34.927 --> 00:24:42.202 They are not the initial feed and they are not the equilibrium composition found by the extent solver. 20-13 00:24:42.382 --> 00:24:52.064 Those observations give Q equal to three point two at this temperature and pressure, so they favor forward reaction relative to the selected K. 20-14 00:24:52.244 --> 00:24:55.252 Leave the equilibrium assertion unchecked. 20-15 00:24:55.432 --> 00:25:02.895 Checking that box would label the supplied observations as claimed equilibrium, but would not prove that claim. 20-16 00:25:03.075 --> 00:25:10.678 Return to the K-source control, select the constant-value enthalpy and entropy route, and recalculate. 20-17 00:25:10.858 --> 00:25:16.853 The extent remains the same because the two property routes agree at the reference temperature. 20-18 00:25:17.033 --> 00:25:26.569 Return to direct standard Gibbs energy and recalculate before saving the reference case as twelve A using Export complete study. 20-19 00:25:26.749 --> 00:25:33.597 Keep the downloaded JSON file in your Week five folder so you can restore it later. 21-01 00:25:36.127 --> 00:25:40.735 Now pause for twenty-two minutes and complete D two through D four. 21-02 00:25:40.915 --> 00:25:47.742 First reproduce the reference run and compare the displayed extent with your calculation in section C. 21-03 00:25:47.922 --> 00:25:52.423 Record the selected K source and the units beside your result. 21-04 00:25:52.603 --> 00:25:56.422 Check the enthalpy-entropy route at the reference temperature. 21-05 00:25:56.602 --> 00:26:03.172 Then change only total pressure to four hundred kilopascals and calculate case twelve B. 21-06 00:26:03.352 --> 00:26:08.319 Keep the temperature, standard pressure, feed and inert amount unchanged. 21-07 00:26:08.499 --> 00:26:11.955 Compare K and extent across the two cases. 21-08 00:26:12.135 --> 00:26:18.201 Save evidence of each result using the Lab export, together with your worksheet table. 21-09 00:26:18.381 --> 00:26:22.285 A short note identifying the changed input is enough. 21-10 00:26:22.465 --> 00:26:27.308 You do not need to copy every table or produce a separate report. 21-11 00:26:27.488 --> 00:26:34.016 Inspect the observed-activities label before using any displayed quotient as a solver check. 21-12 00:26:34.196 --> 00:26:40.766 If the application result differs from your hand calculation, compare the inputs one by one. 21-13 00:26:40.946 --> 00:26:46.834 At the end of the pause, record any remaining mismatch and continue to the result check. 22-01 00:26:49.364 --> 00:26:53.588 This is the pressure-case result in the actual calculator. 22-02 00:26:53.768 --> 00:27:04.408 At three hundred and fifty kelvin and four hundred kilopascals, the selected route still gives K equal to about three point nine five three two six. 22-03 00:27:04.588 --> 00:27:11.436 The extent is about zero point four four five one one moles, consistent with case twelve B. 22-04 00:27:11.616 --> 00:27:18.251 The species table lets you reconstruct both the atom balance and the equilibrium quotient. 22-05 00:27:18.431 --> 00:27:22.634 Now inspect the Gibbs-energy curve along reaction extent. 22-06 00:27:22.814 --> 00:27:31.219 Its minimum occurs at the calculated equilibrium extent for this fixed-temperature, fixed-pressure ideal gas model. 22-07 00:27:31.399 --> 00:27:38.596 The vertical axis is the change in total Gibbs energy from the reference state, divided by R T. 22-08 00:27:38.776 --> 00:27:40.568 Its units are moles. 22-09 00:27:40.748 --> 00:27:48.705 This plotted quantity is different from standard reaction Gibbs energy and from the actual reaction Gibbs energy. 22-10 00:27:48.885 --> 00:27:55.861 The actual reaction Gibbs energy is the slope of total Gibbs energy with respect to extent. 22-11 00:27:56.041 --> 00:28:00.862 The minimum tells us where the slope with respect to extent becomes zero. 22-12 00:28:01.042 --> 00:28:05.501 The allowed interval still comes from nonnegative species amounts. 22-13 00:28:05.681 --> 00:28:11.014 Lab twelve solves one balanced reaction among neutral ideal gases. 22-14 00:28:11.194 --> 00:28:20.368 It does not automatically calculate liquid splitting, multiple simultaneous reactions or a nonideal equation of state. 22-15 00:28:20.548 --> 00:28:27.716 These limits belong beside the numerical answer when you describe what the calculation establishes. 23-01 00:28:30.246 --> 00:28:33.147 Please take the planned ten-minute break now. 23-02 00:28:33.327 --> 00:28:40.026 Save your worksheet and make sure you can find the reference and pressure-case JSON files. 23-03 00:28:40.206 --> 00:28:44.216 You have completed the reference-property and pressure comparisons. 23-04 00:28:44.396 --> 00:28:50.434 When you return, we will change temperature and calculate a new equilibrium constant. 23-05 00:28:50.614 --> 00:28:53.963 Leave the optional extensions for another time. 23-06 00:28:54.143 --> 00:29:01.610 Pause the video, step away from the screen if you can, and resume at the temperature section after about ten minutes. 23-07 00:29:01.790 --> 00:29:04.734 The break is included in the lesson budget. 24-01 00:29:07.264 --> 00:29:10.378 We have changed pressure while holding K fixed. 24-02 00:29:10.558 --> 00:29:12.542 Next we change temperature. 24-03 00:29:12.722 --> 00:29:20.466 The standard equilibrium constant depends on temperature for a fixed reaction and standard-state convention. 24-04 00:29:20.646 --> 00:29:26.577 A Gibbs-energy value supplied at one temperature gives one point on that relationship. 24-05 00:29:26.757 --> 00:29:30.853 It does not automatically give a law for other temperatures. 24-06 00:29:31.033 --> 00:29:41.380 To estimate K at four hundred kelvin, we will use the reference constant at three hundred and fifty kelvin together with the standard reaction enthalpy. 24-07 00:29:41.560 --> 00:29:47.256 Our model assumes that the reaction enthalpy remains constant over this interval. 24-08 00:29:47.436 --> 00:29:52.961 This is an approximation with a stated range, rather than a new measured data set. 24-09 00:29:53.141 --> 00:29:58.730 Because the reaction enthalpy is positive, the dissociation is endothermic. 24-10 00:29:58.910 --> 00:30:06.590 Before using the equation, predict whether a higher temperature should increase or decrease the equilibrium constant. 24-11 00:30:06.770 --> 00:30:14.109 Keep pressure at one hundred kilopascals for the temperature case so the comparison isolates temperature. 24-12 00:30:14.289 --> 00:30:18.534 The initial feed and standard pressure also remain unchanged. 25-01 00:30:21.064 --> 00:30:31.283 The integrated van't Hoff equation relates the natural logarithm of the ratio of the two equilibrium constants to the reaction enthalpy. 25-02 00:30:31.463 --> 00:30:41.980 With constant enthalpy, it is the enthalpy divided by R, multiplied by one over the reference temperature minus one over the new temperature. 25-03 00:30:42.160 --> 00:30:49.696 Here the reference temperature is three hundred and fifty kelvin, and the new temperature is four hundred kelvin. 25-04 00:30:49.876 --> 00:30:54.292 Use fifty thousand joules per mole for the reaction enthalpy. 25-05 00:30:54.472 --> 00:31:01.725 Since the new temperature is higher, one over the reference temperature is larger than one over the new temperature. 25-06 00:31:01.905 --> 00:31:03.846 The difference is positive. 25-07 00:31:04.026 --> 00:31:10.298 The enthalpy is also positive, so the logarithm of the K ratio should be positive. 25-08 00:31:10.478 --> 00:31:12.761 That predicts an increase in K. 25-09 00:31:12.941 --> 00:31:19.746 To find the new natural logarithm of K, add this change to the reference natural logarithm. 25-10 00:31:19.926 --> 00:31:21.953 Then exponentiate once. 25-11 00:31:22.133 --> 00:31:28.170 Keep both temperatures in kelvin and check the sign before calculating the final value. 26-01 00:31:30.700 --> 00:31:34.626 Pause for fifteen minutes and complete E one through E three. 26-02 00:31:34.806 --> 00:31:44.427 Start with the constant-enthalpy van't Hoff equation and substitute the reference temperature, new temperature and enthalpy with their units. 26-03 00:31:44.607 --> 00:31:47.615 Calculate the logarithmic change first. 26-04 00:31:47.795 --> 00:31:54.323 Add it to the reference natural logarithm of K, then obtain K at four hundred kelvin. 26-05 00:31:54.503 --> 00:32:04.444 Cross-check the new Gibbs energy with enthalpy minus the new temperature times entropy, under the same zero heat-capacity-change assumption. 26-06 00:32:04.624 --> 00:32:10.896 Then use the new K in the equilibrium-extent expression at a pressure ratio of one. 26-07 00:32:11.076 --> 00:32:16.687 Calculate the species amounts and make the same two independent checks as before. 26-08 00:32:16.867 --> 00:32:24.888 The atom total should remain unchanged, and the quotient reconstructed from the final composition should match the new K. 26-09 00:32:25.068 --> 00:32:29.399 Keep enough intermediate digits to avoid a rounding mismatch. 26-10 00:32:29.579 --> 00:32:37.067 Write a short explanation connecting the positive reaction enthalpy to the direction of the temperature effect. 26-11 00:32:37.247 --> 00:32:43.220 If you need more time, mark the specific unfinished line and continue to the solution. 26-12 00:32:43.400 --> 00:32:49.310 You can return to that line during our onsite question period on November thirteenth. 27-01 00:32:51.840 --> 00:33:00.458 The logarithmic change from three hundred and fifty to four hundred kelvin is about two point one four seven seven two. 27-02 00:33:00.638 --> 00:33:06.740 Adding the reference natural logarithm gives about three point five two two two six. 27-03 00:33:06.920 --> 00:33:13.469 The new equilibrium constant is therefore about thirty-three point eight six zero nine three. 27-04 00:33:13.649 --> 00:33:25.382 The enthalpy-entropy cross-check gives standard reaction Gibbs energy of about minus eleven point seven one four three kilojoules per mole at four hundred kelvin. 27-05 00:33:25.562 --> 00:33:33.050 At one hundred kilopascals, this gives an extent of about zero point nine four five seven zero moles. 27-06 00:33:33.230 --> 00:33:43.732 The A two amount is about zero point zero five four three zero moles, and the A amount is about one point eight nine one four zero moles. 27-07 00:33:43.912 --> 00:33:53.597 The resulting mole fractions are about zero point nine seven two zero nine for A and zero point zero two seven nine one for A two. 27-08 00:33:53.777 --> 00:34:00.156 Heating has increased dissociation for this endothermic reaction under the stated model. 27-09 00:34:00.336 --> 00:34:04.283 The atom inventory still equals two moles of A atoms. 27-10 00:34:04.463 --> 00:34:13.380 Compare this case with the reference case, not with the four-hundred- kilopascal case, when attributing the difference to temperature alone. 27-11 00:34:13.560 --> 00:34:17.699 The increase in K is a property of the temperature model. 27-12 00:34:17.879 --> 00:34:22.551 The increase in extent also uses the specified feed and pressure. 28-01 00:34:25.081 --> 00:34:34.275 At four hundred kelvin, the direct Gibbs and formation-energy rows supplied at three hundred and fifty kelvin become unavailable. 28-02 00:34:34.455 --> 00:34:36.055 That is appropriate. 28-03 00:34:36.235 --> 00:34:42.358 Changing a temperature label would not create valid property data at a new temperature. 28-04 00:34:42.538 --> 00:34:45.674 Use the temperature-dependent route instead. 28-05 00:34:45.854 --> 00:34:55.149 For our required calculation, select van't Hoff with constant reaction enthalpy, and keep the heat-capacity change at zero. 28-06 00:34:55.329 --> 00:34:58.827 The Lab also offers a heat-capacity correction. 28-07 00:34:59.007 --> 00:35:05.685 With its default nonzero value, that is a different temperature model and can give a different K. 28-08 00:35:05.865 --> 00:35:10.895 You may explore it later, but it is outside this week's required work. 28-09 00:35:11.075 --> 00:35:16.750 The independent observed-composition panel has its own observation temperature. 28-10 00:35:16.930 --> 00:35:28.855 A three-hundred-and-fifty-kelvin observation does not become a four- hundred-kelvin observation simply because the main evaluation temperature changes. 28-11 00:35:29.035 --> 00:35:38.209 Its unavailable message does not prevent the equilibrium extent calculation from using an applicable thermodynamic route. 28-12 00:35:38.389 --> 00:35:43.359 Read each panel's temperature and label before comparing its output. 29-01 00:35:45.889 --> 00:35:50.156 Pause for nineteen minutes and complete F one through F three. 29-02 00:35:50.336 --> 00:35:57.440 Set evaluation temperature to four hundred kelvin and total pressure to one hundred kilopascals. 29-03 00:35:57.620 --> 00:36:04.703 Keep the reference temperature and reference K at their original three-hundred-and-fifty-kelvin values. 29-04 00:36:04.883 --> 00:36:13.401 Select van't Hoff with constant reaction enthalpy as the K source, and check that the heat-capacity change is zero. 29-05 00:36:13.581 --> 00:36:17.250 Run case twelve C and compare it with section E. 29-06 00:36:17.430 --> 00:36:19.478 Export the complete study. 29-07 00:36:19.658 --> 00:36:27.210 Next, use Restore calculator study JSON to reopen your saved twelve A reference file. 29-08 00:36:27.390 --> 00:36:33.321 The calculator should recover the original inputs and recalculate the reference result. 29-09 00:36:33.501 --> 00:36:37.789 Verify the restored temperature, pressure, K and extent. 29-10 00:36:37.969 --> 00:36:45.627 A restored filename alone is insufficient evidence that the numerical state is the one you intended. 29-11 00:36:45.807 --> 00:36:52.805 Finish the three-case table and keep one note about the pressure effect and one about the temperature effect. 29-12 00:36:52.985 --> 00:37:01.155 If restoring a file fails, retain your calculation evidence and record the message for our next onsite class. 30-01 00:37:03.685 --> 00:37:07.461 Here is the temperature-case setup in the actual Lab. 30-02 00:37:07.641 --> 00:37:19.669 The evaluation temperature is four hundred kelvin, pressure is one hundred kilopascals, and the selected extent route is van't Hoff with constant reaction enthalpy. 30-03 00:37:19.849 --> 00:37:22.431 The heat-capacity change is zero. 30-04 00:37:22.611 --> 00:37:31.699 After Compare K methods, the temperature-dependent route gives K equal to about thirty-three point eight six zero nine three. 30-05 00:37:31.879 --> 00:37:37.831 The direct Gibbs and formation rows remain tied to their supplied reference temperature. 30-06 00:37:38.011 --> 00:37:40.613 Scroll to the equilibrium composition. 30-07 00:37:40.793 --> 00:37:47.364 The extent is about zero point nine four five seven zero moles, agreeing with the worksheet. 30-08 00:37:47.544 --> 00:37:53.560 Export complete study saves the inputs and calculated evidence for case twelve C. 30-09 00:37:53.740 --> 00:38:01.676 To check reproducibility, choose the earlier twelve A file under Restore calculator study JSON. 30-10 00:38:01.856 --> 00:38:08.235 The restored controls return to three hundred and fifty kelvin and one hundred kilopascals. 30-11 00:38:08.415 --> 00:38:18.719 The recalculated K is about three point nine five three two six, with extent about zero point seven zero five zero three moles. 30-12 00:38:18.899 --> 00:38:21.501 Compare these with your reference row. 30-13 00:38:21.681 --> 00:38:25.244 Your final table now separates two changes. 30-14 00:38:25.424 --> 00:38:29.371 The pressure-only case changes extent while K stays fixed. 30-15 00:38:29.551 --> 00:38:32.985 The temperature-only case changes K and extent. 30-16 00:38:33.165 --> 00:38:37.005 All three cases conserve the same A-atom inventory. 30-17 00:38:37.185 --> 00:38:46.534 The JSON files let you reopen the exact chosen inputs, while your worksheet explains what those inputs and results mean. 31-01 00:38:49.064 --> 00:38:54.141 Before the final form, check the explanation you would give to another student. 31-02 00:38:54.321 --> 00:38:59.804 K follows from standard thermodynamic properties and the chosen temperature. 31-03 00:38:59.984 --> 00:39:03.824 Q follows from the activities of a particular state. 31-04 00:39:04.004 --> 00:39:11.343 Equilibrium extent combines the equilibrium condition with the feed and conservation constraints. 31-05 00:39:11.523 --> 00:39:16.041 These three ideas work together, but they answer different questions. 31-06 00:39:16.221 --> 00:39:20.125 The inert-gas question also depends on the constraint. 31-07 00:39:20.305 --> 00:39:29.095 In this Lab, adding inert at fixed temperature and total pressure dilutes the reacting species and can shift the equilibrium. 31-08 00:39:29.275 --> 00:39:43.183 In a rigid ideal gas vessel at fixed temperature and volume, adding inert leaves reacting-species partial pressures unchanged, so an initially equilibrated reacting mixture does not shift. 31-09 00:39:43.363 --> 00:39:47.075 Name the constraint whenever you use that conclusion. 31-10 00:39:47.255 --> 00:39:54.039 Choose one unresolved question from your worksheet and identify the step where your reasoning stops. 31-11 00:39:54.219 --> 00:39:59.765 We will use those questions when we meet onsite on Friday, November thirteenth. 31-12 00:39:59.945 --> 00:40:03.999 Further heat-capacity and custom-reaction work is optional. 32-01 00:40:06.529 --> 00:40:12.886 Use the final ten minutes for section G and the Session five exit ticket and feedback. 32-02 00:40:13.066 --> 00:40:17.354 Open the form using the link or QR code on this slide. 32-03 00:40:17.534 --> 00:40:21.353 Check that Session five is selected before submitting. 32-04 00:40:21.533 --> 00:40:27.655 Explain one result from the required cases using your own calculation evidence. 32-05 00:40:27.835 --> 00:40:36.454 Distinguish K from Q, then explain the pressure-only comparison with one numerical result or an independent check. 32-06 00:40:36.634 --> 00:40:38.810 Include the settings and units. 32-07 00:40:38.990 --> 00:40:42.382 Include the question you still want to discuss. 32-08 00:40:42.562 --> 00:40:51.714 In the feedback, report your actual study time and identify any instruction, calculation or control that caused difficulty. 32-09 00:40:51.894 --> 00:40:55.691 Specific descriptions help us improve the next lesson. 32-10 00:40:55.871 --> 00:40:58.986 You may answer the form in English or Thai. 32-11 00:40:59.166 --> 00:41:03.923 It is anonymous and does not identify individual attendance. 32-12 00:41:04.103 --> 00:41:08.455 Submit once and wait for the Response received confirmation. 32-13 00:41:08.635 --> 00:41:12.347 Keep your worksheet and exported study files together. 32-14 00:41:12.527 --> 00:41:15.535 This completes the required Week five work. 32-15 00:41:15.715 --> 00:41:25.128 Please follow the video and its activity stops for November sixth, and we will return to the classroom on Friday, November thirteenth.