There is no in-person class on 6 November. Begin with answers to your Week 4 questions, then use one synthetic reaction to compare Gibbs-energy, enthalpy-entropy and van’t Hoff calculations. Follow the worksheet and the actual Lab 12 demonstrations. Our next in-person class is 13 November 2026.
41:27 recording179.5 min with planned work and break32 slides with activity stopsEnglish male narration and captions
The measured recording is 41:27 (about 41.5 minutes). Add 128 minutes of calculations, Lab work and reflection, and a 10-minute break: approximately 179.5 minutes total.
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Slide 1 of 32Worksheet R
Week 5: from phases to reactions
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Transcript · slide 1
Welcome to Week five, from phases to chemical reactions. This is our independent lesson for Friday, November sixth. We will meet onsite again on Friday, November thirteenth. Please begin with the short review in section R of the worksheet. Your recent questions showed that we should separate two ideas before moving on. Bulk phase equilibrium concerns a substance distributed between macroscopic phases, such as a solid and its vapor. Adsorption concerns uptake associated with a surface. The amount adsorbed depends on an adsorption relation and the available surface capacity. A vapor pressure alone does not describe that uptake. Today, we will use the same habit of combining an equilibrium relation with conservation, but the species can also react. Our main example is a synthetic dissociation, A two forming two A atoms. We will follow one worksheet and use Lab twelve. Keep the worksheet beside the video so every pause has a clear place to work.
Use the provided synthetic A₂ ⇌ 2A feed, fixed standard pressure 100 kPa, ΔCp° = 0 and the stated K source. The observed Q panel is an independent supplied state, not the equilibrium-solver composition. Optional heat-capacity and custom-reaction extensions are outside the required work.
Finish once, then bring your question onsite
Complete worksheet G during the ten-minute stop at slide 32. Open the Session 5 exit ticket and feedback form. Explain the pressure-only comparison, distinguish K from Q, and give a numerical result or independent check with its settings and units.
Include your actual study minutes and the step where you need help. English or Thai is welcome. Submit once and wait for Response received. Keep your worksheet, corrections and exported studies. The form is anonymous and does not identify individual attendance. Player progress and Lab exports are not form submissions.
There is no extra required homework. Our next in-person class is Friday, 13 November 2026; bring one unresolved question.
Full narration transcript · includes the answer checks
Attempt the worksheet questions before reading their checks. This transcript follows the recording and its activity instructions.
Slide 1: Week 5: from phases to reactions
Recording 0:00 · Worksheet R
Welcome to Week five, from phases to chemical reactions. This is our independent lesson for Friday, November sixth. We will meet onsite again on Friday, November thirteenth. Please begin with the short review in section R of the worksheet. Your recent questions showed that we should separate two ideas before moving on. Bulk phase equilibrium concerns a substance distributed between macroscopic phases, such as a solid and its vapor. Adsorption concerns uptake associated with a surface. The amount adsorbed depends on an adsorption relation and the available surface capacity. A vapor pressure alone does not describe that uptake. Today, we will use the same habit of combining an equilibrium relation with conservation, but the species can also react. Our main example is a synthetic dissociation, A two forming two A atoms. We will follow one worksheet and use Lab twelve. Keep the worksheet beside the video so every pause has a clear place to work.
Your action: Follow slide 1; keep worksheet R beside you. Attempt questions before their checks.
Slide 2: Partial and total pressure at fixed volume
Recording 1:13 · Worksheet R1
Before we discuss reactions, revisit pressure in a closed, rigid vessel. The temperature and the volume are fixed. The ideal gas equation connects the amount of each gas to its own partial pressure. Total pressure is the sum of those partial pressures. If all of the volatile solid has disappeared, calculate the volatile gas pressure from its available amount. Do not continue forcing its pressure to equal the saturation value after that phase is depleted. The assumption of an ideal gas is a separate modeling choice. Losing the solid phase does not make a real gas ideal. Now pause for four minutes and complete R one. 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. Keep the volatile inventory unchanged. Record a unit check. After your attempt, continue to the next question. We will check both review answers together.
Your action: Pause for 4 minutes. Complete worksheet R1 before continuing. At the time limit, record the sticking point and continue.
Slide 3: A balance needs an independent equation
Recording 2:26 · Worksheet R2
The next question concerns adsorption in a finite vessel. Conservation tells us that the feed is divided between free gas and adsorbed material. The Langmuir equation supplies another condition that the proposed state must satisfy. Choosing two amounts that add to the feed is insufficient if the adsorbed amount disagrees with the isotherm at the same pressure. In R two, the pressure and the model parameters are supplied. First calculate the gas amount from pressure, volume and temperature. Independently calculate the loading from the Langmuir equation, then multiply that loading by the adsorbent mass. Only after these two calculations should you compare their sum with the feed. Use units of mole per kilogram for the loading and kilograms for the adsorbent mass. Their product is an amount in moles. Pause for four minutes and complete R two. If a number seems unreasonable, write which equation or conversion you need to check. Continue when the time budget ends.
Your action: Pause for 4 minutes. Complete worksheet R2 before continuing. At the time limit, record the sticking point and continue.
Slide 4: Two calculations, one physical state
Recording 3:38 · Worksheet R1–R2 check
Check R one first. The gas capacity at the sublimation pressure is about four thousandths of a mole, which exceeds the available zero point zero zero two moles. All of the solid disappears. The volatile partial pressure is about four point nine eight nine kilopascals. 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. Now check R two. 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. Their sum exceeds the feed. That proposed pressure cannot satisfy this model and inventory together. The checked equilibrium pressure is about six point zero eight zero kilopascals. It gives about zero point zero zero two four four moles of gas and zero point zero zero seven five six moles adsorbed. The sum matches the feed. We will use the same combination of equilibrium and conservation checks for reactions.
Your action: Follow slide 4; keep worksheet R1–R2 check beside you. Attempt questions before their checks.
Slide 5: One worksheet, one Lab, three hours
Recording 5:01 · Worksheet All sections
Here is the complete route for this week. Use the six-page Week five worksheet as your working record. Keep Lab twelve open in a second tab when we reach section D. The short Lab guide is available when you need help finding a control. The worked solutions are for checking after an attempt. There is no additional required preparation or follow-up assignment. The planned activities use one hundred and twenty-eight minutes, with a separate ten-minute break. Add the recording duration shown on the video page for a total of approximately three hours. Your actual time may differ. Please use the suggested stop budgets. If you remain stuck at the end of a pause, record the specific step and continue. Bring that question to our onsite class on November thirteenth. You do not need to repeat the whole worksheet neatly. A first attempt, a correction and a short explanation provide a useful record of your learning.
Your action: Follow slide 5; keep worksheet All sections beside you. Attempt questions before their checks.
Slide 6: Reaction extent keeps the atoms balanced
Recording 6:06 · Worksheet A1
Our reaction is A two gas in equilibrium with two A gas. We begin with one mole of A two and no A. No inert gas is present in the three required cases. A two contains two A atoms, so the reaction conserves atoms even though the number of gas molecules changes. Reaction extent tells us how far the reaction has proceeded on the written stoichiometric basis. Consuming one mole of A two produces two moles of A. The general expression is the initial amount plus the stoichiometric coefficient times the reaction extent. Reactant coefficients are negative and product coefficients are positive. For this particular feed, it is convenient to define f as the reaction extent divided by one mole. The fraction f is dimensionless. The amounts are one minus f moles of A two and two f moles of A. Nonnegative amounts require f between zero and one. These bounds come from the available material, before we use any equilibrium constant.
Your action: Follow slide 6; keep worksheet A1 beside you. Attempt questions before their checks.
Slide 7: Activities make Q dimensionless
Recording 7:19 · Worksheet A2
To describe equilibrium, we need activities. For today's ideal gas model, the activity of a species is its mole fraction multiplied by total pressure, divided by the standard pressure. Both pressures must use the same units. Their ratio is dimensionless, and so is the activity. We use a standard pressure of one hundred kilopascals throughout this lesson. This is a reference convention. The actual total pressure is a separate input that we can change. For A two dissociating into two A, the quotient is the activity of A squared, divided by the activity of A two. The exponents come directly from the reaction coefficients. Write this activity expression before substituting mole fractions. That step makes it easier to see how pressure enters the calculation. Later we will compare one hundred and four hundred kilopascals at the same temperature. Our standard pressure stays at one hundred kilopascals in both cases.
Your action: Follow slide 7; keep worksheet A2 beside you. Attempt questions before their checks.
Slide 8: Your turn: build Q from the mole table
Recording 8:28 · Worksheet A1–A3
Please pause for ten minutes and complete A one through A three. Begin with a mole table. Write the amounts of A two, A and the total gas in terms of f. Divide each species amount by the total amount to obtain its mole fraction. Check that the fractions sum to one. Substitute those fractions into the activity definition and then into the reaction quotient. Try to simplify the expression without looking ahead. Keep the actual-pressure to standard-pressure ratio visible. 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. Finally, check the total number of A atoms using the stoichiometric counts. The total molecular amount can change, but the atom inventory must remain equal to the initial inventory. If the algebra becomes difficult, retain the unsimplified expression and substitute one trial value of f to test it. That is a useful intermediate result. At the end of ten minutes, record any unfinished step and continue to the check. The next slide shows the derivation.
Your action: Pause for 10 minutes. Complete worksheet A1–A3 before continuing. At the time limit, record the sticking point and continue.
Slide 9: Check the quotient and the atom balance
Recording 9:48 · Worksheet A1–A3 check
The total gas amount is one plus f moles. Therefore, the mole fraction of A two is one minus f, divided by one plus f. The mole fraction of A is two f, divided by one plus f. Substituting these into the quotient gives four f squared divided by one minus f squared, multiplied by the pressure ratio. One pressure factor remains because the product side has one more mole of gas per mole of reaction than the reactant side. If your expression has two pressure factors or none, return to the activity exponents and cancel them explicitly. The atom check is twice the amount of A two plus the amount of A. Substitution gives two moles of conserved A atoms for every feasible f. Notice that molecular moles and atom moles describe different inventories. 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. These are trial states, so we have not yet established equilibrium. The quotient changes with f, while the atom balance remains satisfied.
Your action: Follow slide 9; keep worksheet A1–A3 check beside you. Attempt questions before their checks.
Slide 10: Which way does the reaction proceed?
Recording 11:09 · Worksheet B1
How can the quotient tell us which direction is favored? At fixed temperature and pressure, the reaction Gibbs energy is the standard reaction Gibbs energy plus R T times the natural logarithm of Q. An equivalent form is R T times the natural logarithm of Q divided by K. A negative value favors forward reaction. A positive value favors reverse reaction. At an interior equilibrium, the reaction Gibbs energy is zero and Q equals K. These statements refer to the reaction as written. Reversing the reaction reverses the corresponding sign convention. For B one, use the supplied standard reaction Gibbs energy of minus four thousand joules per mole at three hundred and fifty kelvin. Test Q equal to one and Q equal to ten. Pause for six minutes, calculate the signs and write the favored direction for each state. Use the natural logarithm. We will return to these answers after calculating K by two routes.
Your action: Pause for 6 minutes. Complete worksheet B1 before continuing. At the time limit, record the sticking point and continue.
Slide 11: Route 1: K from standard Gibbs energy
Recording 12:22 · Worksheet B2
The first route starts with standard reaction Gibbs energy at the temperature of interest. The equation is natural log K equals minus standard reaction Gibbs energy divided by R T. For this example, the supplied standard reaction Gibbs energy applies at three hundred and fifty kelvin. It is minus four kilojoules per mole of reaction as written. To use the gas constant in joules per mole per kelvin, convert that energy to minus four thousand joules per mole. Keep the two minus signs separate until you substitute. A negative standard reaction Gibbs energy gives a positive natural logarithm of K, so K should exceed one. This sign prediction is a quick check on the arithmetic. The final equilibrium constant is dimensionless because it is defined using dimensionless activities. A larger K describes a stronger product preference in this convention, but it does not by itself specify the conversion. The feed and constraints are also needed.
Your action: Follow slide 11; keep worksheet B2 beside you. Attempt questions before their checks.
Slide 12: Route 2: K from enthalpy and entropy
Recording 13:35 · Worksheet B3
The second route constructs standard reaction Gibbs energy from standard reaction enthalpy and entropy at the same temperature. For the reference case, the enthalpy is fifty kilojoules per mole of reaction. The entropy is about one hundred and fifty-four point two eight five seven joules per mole per kelvin. Convert either the energy or entropy so that the subtraction uses consistent units. In joules, calculate fifty thousand minus three hundred and fifty multiplied by the entropy value. Then use that Gibbs energy in the equation for natural log K. These two routes should agree because the synthetic inputs were constructed consistently. In an experimental or literature calculation, agreement also depends on using the same phases, standard states and reaction basis. If the routes disagree, check the inputs before interpreting the discrepancy as physical behavior. For today's comparison, keep all supplied properties at the reference temperature of three hundred and fifty kelvin.
Your action: Follow slide 12; keep worksheet B3 beside you. Attempt questions before their checks.
Slide 13: Your turn: compare the two K routes
Recording 14:49 · Worksheet B2–B4
Pause for ten minutes and complete B two through B four. Calculate natural log K and then K from the directly supplied standard Gibbs energy. Next, calculate standard Gibbs energy from enthalpy and entropy, and use that value to obtain a second K. Show the energy conversion on your worksheet so someone else can follow your units. Keep several digits during the calculation and round only the reported result. Compare the two routes and write why agreement is expected for this synthetic case. If you obtain a reciprocal value or a very small constant, revisit the minus sign in the exponential. If your two Gibbs energies differ by a factor of a thousand, check the joule and kilojoule conversion. You can use a calculator or a spreadsheet for the arithmetic. The objective is to make the route from the supplied properties to the equilibrium constant visible. Continue to the numerical check after your attempt.
Your action: Pause for 10 minutes. Complete worksheet B2–B4 before continuing. At the time limit, record the sticking point and continue.
Slide 14: Both routes give the same reference K
Recording 15:56 · Worksheet B1–B4 check
Both routes give a standard reaction Gibbs energy of minus four thousand joules per mole at three hundred and fifty kelvin. Dividing its negative by R T gives natural log K equal to about one point three seven four five four. Exponentiating gives K equal to about three point nine five three two six. The enthalpy-entropy calculation reaches the same result because fifty thousand minus fifty-four thousand equals minus four thousand joules per mole. Return briefly to B one. 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. When Q is ten, Q exceeds K. The logarithm of Q over K is positive, so reverse reaction is favored. At equilibrium, Q equals this finite K. The standard reaction Gibbs energy remains minus four kilojoules per mole. It is the actual reaction Gibbs energy that becomes zero.
Your action: Follow slide 14; keep worksheet B1–B4 check beside you. Attempt questions before their checks.
Slide 15: Your turn: solve for equilibrium extent
Recording 17:09 · Worksheet C1–C3
Now use K to calculate composition. At equilibrium, set the quotient expression from section A equal to K. 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. Rearranging gives f squared equal to K divided by K plus four r. Take the nonnegative root because our initial feed allows only forward extent between zero and one mole. For the reference pressure of one hundred kilopascals, r equals one. Pause for fifteen minutes and complete C one through C three. Calculate f, convert it to reaction extent, and calculate both species amounts and their mole fractions. Then verify the atom balance. Reconstruct Q from the calculated mole fractions and the pressure ratio. This second calculation should return K within rounding. Please do not use the rearranged equation as your only check. The purpose is to test the final composition independently.
Your action: Pause for 15 minutes. Complete worksheet C1–C3 before continuing. At the time limit, record the sticking point and continue.
Slide 16: The reference equilibrium satisfies Q=K
Recording 18:25 · Worksheet C1–C3 check
For the reference case, f is about zero point seven zero five. The reaction extent is therefore about zero point seven zero five moles. 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. Their sum is about one point seven zero five moles. 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. These fractions differ from the species amounts, even though both describe the same state. Use the more precise on-screen values when checking numerical closure. The atom check gives two moles of A atoms. At this pressure, the activities equal the mole fractions numerically, so the activity quotient returns about three point nine five three two six. K greater than one has produced substantial dissociation, while a finite amount of A two remains.
Your action: Follow slide 16; keep worksheet C1–C3 check beside you. Attempt questions before their checks.
Slide 17: Your turn: compress at the same temperature
Recording 19:33 · Worksheet C4
Next, increase total pressure to four hundred kilopascals while keeping temperature at three hundred and fifty kelvin. Keep the same feed and standard pressure. Before calculating, predict whether dissociation should increase or decrease. The reaction produces more gas molecules in the forward direction. At a fixed composition, increasing pressure therefore increases Q. Use that observation to predict which way the system must react to recover Q equal to K. Then calculate the new equilibrium using r equal to four. Pause for eight minutes and complete C four. Report the new extent, species amounts and mole fractions. Write one sentence explaining what happens to K and one sentence explaining what happens to composition. Use the temperature condition in your explanation. It is possible for equilibrium composition to change while the standard equilibrium constant remains the same. This controlled comparison will help you distinguish those two quantities when you open the Lab.
Your action: Pause for 8 minutes. Complete worksheet C4 before continuing. At the time limit, record the sticking point and continue.
Slide 18: Pressure changes composition at fixed K
Recording 20:47 · Worksheet C4 check
At four hundred kilopascals, K remains about three point nine five three two six because temperature and the standard-state convention are unchanged. The equilibrium extent decreases to about zero point four four five one moles. 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. 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. The higher-pressure mixture therefore contains less dissociated A. You can explain this directly from the quotient: at a fixed f, multiplying pressure by four multiplies Q by four. Reverse reaction then reduces f until Q returns to K. This is more precise than applying a pressure slogan without specifying the reaction or constraint. Keep the baseline and pressure-only cases beside each other in your worksheet. The atom total should be two in both cases, even though the molecular totals differ. We will now reproduce these states using the same ideal gas model in Lab twelve.
Your action: Follow slide 18; keep worksheet C4 check beside you. Attempt questions before their checks.
Slide 19: Lab 12: prepare a controlled comparison
Recording 22:10 · Worksheet D1
Open Lab twelve using the link on this slide or the video page. Its title is Chemical equilibrium and routes to K. Complete D one before continuing. Reset the synthetic example so the feed and property values are known. Check that evaluation temperature and reference temperature are three hundred and fifty kelvin, and that total pressure and standard pressure are one hundred kilopascals. Set the reaction heat-capacity change to zero. The default value is thirty, so this is an intentional change for today's constant-enthalpy comparison. Keep the inert amount at zero. For the first run, select the direct standard Gibbs energy as the K source for extent and driving force. The direct data apply at three hundred and fifty kelvin. Check the species table and the source note. Pause for five minutes to prepare these inputs. If you need help finding a control, use the two-page Lab guide. Resume for the demonstration when your inputs are ready.
Your action: Pause for 5 minutes. Complete worksheet D1 before continuing. At the time limit, record the sticking point and continue.
Slide 20: Lab 12 demonstration: the reference state
Recording 23:21 · Worksheet Watch after D1
Here is the actual Lab twelve calculator with the reference inputs prepared. In the thermal-properties section, the heat-capacity change is zero, and the reference enthalpy and entropy match the worksheet. 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. Now use Compare K methods to calculate the reference case. The methods table shows the agreement of the direct Gibbs and enthalpy-entropy routes at the reference temperature. The formation-energy sum is another route available in this synthetic example, but we do not need an additional worksheet task for it. Scroll to From K to equilibrium composition. The extent is about zero point seven zero five zero three moles, matching the hand calculation. Check the species amounts and the atom-balance message. The observed-activities panel needs careful reading. Its supplied mole fractions, zero point two and zero point eight, are independent observations. They are not the initial feed and they are not the equilibrium composition found by the extent solver. Those observations give Q equal to three point two at this temperature and pressure, so they favor forward reaction relative to the selected K. Leave the equilibrium assertion unchecked. Checking that box would label the supplied observations as claimed equilibrium, but would not prove that claim. Return to the K-source control, select the constant-value enthalpy and entropy route, and recalculate. The extent remains the same because the two property routes agree at the reference temperature. Return to direct standard Gibbs energy and recalculate before saving the reference case as twelve A using Export complete study. Keep the downloaded JSON file in your Week five folder so you can restore it later.
Your action: Follow slide 20; keep worksheet Watch after D1 beside you. Attempt questions before their checks.
Slide 21: Your turn: reproduce the pressure comparison
Recording 25:36 · Worksheet D2–D4
Now pause for twenty-two minutes and complete D two through D four. First reproduce the reference run and compare the displayed extent with your calculation in section C. Record the selected K source and the units beside your result. Check the enthalpy-entropy route at the reference temperature. Then change only total pressure to four hundred kilopascals and calculate case twelve B. Keep the temperature, standard pressure, feed and inert amount unchanged. Compare K and extent across the two cases. Save evidence of each result using the Lab export, together with your worksheet table. A short note identifying the changed input is enough. You do not need to copy every table or produce a separate report. Inspect the observed-activities label before using any displayed quotient as a solver check. If the application result differs from your hand calculation, compare the inputs one by one. At the end of the pause, record any remaining mismatch and continue to the result check.
Your action: Pause for 22 minutes. Complete worksheet D2–D4 before continuing. At the time limit, record the sticking point and continue.
Slide 22: What the Lab results establish
Recording 26:49 · Worksheet D2–D4 check
This is the pressure-case result in the actual calculator. 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. The extent is about zero point four four five one one moles, consistent with case twelve B. The species table lets you reconstruct both the atom balance and the equilibrium quotient. Now inspect the Gibbs-energy curve along reaction extent. Its minimum occurs at the calculated equilibrium extent for this fixed-temperature, fixed-pressure ideal gas model. The vertical axis is the change in total Gibbs energy from the reference state, divided by R T. Its units are moles. This plotted quantity is different from standard reaction Gibbs energy and from the actual reaction Gibbs energy. The actual reaction Gibbs energy is the slope of total Gibbs energy with respect to extent. The minimum tells us where the slope with respect to extent becomes zero. The allowed interval still comes from nonnegative species amounts. Lab twelve solves one balanced reaction among neutral ideal gases. It does not automatically calculate liquid splitting, multiple simultaneous reactions or a nonideal equation of state. These limits belong beside the numerical answer when you describe what the calculation establishes.
Your action: Follow slide 22; keep worksheet D2–D4 check beside you. Attempt questions before their checks.
Slide 23: Ten-minute break
Recording 28:30 · Worksheet Break
Please take the planned ten-minute break now. Save your worksheet and make sure you can find the reference and pressure-case JSON files. You have completed the reference-property and pressure comparisons. When you return, we will change temperature and calculate a new equilibrium constant. Leave the optional extensions for another time. Pause the video, step away from the screen if you can, and resume at the temperature section after about ten minutes. The break is included in the lesson budget.
Your action: Pause for 10 minutes. Save your work and take the planned break.
Slide 24: Temperature changes the equilibrium constant
Recording 29:07 · Worksheet E1
We have changed pressure while holding K fixed. Next we change temperature. The standard equilibrium constant depends on temperature for a fixed reaction and standard-state convention. A Gibbs-energy value supplied at one temperature gives one point on that relationship. It does not automatically give a law for other temperatures. 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. Our model assumes that the reaction enthalpy remains constant over this interval. This is an approximation with a stated range, rather than a new measured data set. Because the reaction enthalpy is positive, the dissociation is endothermic. Before using the equation, predict whether a higher temperature should increase or decrease the equilibrium constant. Keep pressure at one hundred kilopascals for the temperature case so the comparison isolates temperature. The initial feed and standard pressure also remain unchanged.
Your action: Follow slide 24; keep worksheet E1 beside you. Attempt questions before their checks.
Slide 25: Route 3: K at a new temperature
Recording 30:21 · Worksheet E1–E2
The integrated van't Hoff equation relates the natural logarithm of the ratio of the two equilibrium constants to the reaction enthalpy. With constant enthalpy, it is the enthalpy divided by R, multiplied by one over the reference temperature minus one over the new temperature. Here the reference temperature is three hundred and fifty kelvin, and the new temperature is four hundred kelvin. Use fifty thousand joules per mole for the reaction enthalpy. Since the new temperature is higher, one over the reference temperature is larger than one over the new temperature. The difference is positive. The enthalpy is also positive, so the logarithm of the K ratio should be positive. That predicts an increase in K. To find the new natural logarithm of K, add this change to the reference natural logarithm. Then exponentiate once. Keep both temperatures in kelvin and check the sign before calculating the final value.
Your action: Follow slide 25; keep worksheet E1–E2 beside you. Attempt questions before their checks.
Slide 26: Your turn: heat from 350 K to 400 K
Recording 31:31 · Worksheet E1–E3
Pause for fifteen minutes and complete E one through E three. Start with the constant-enthalpy van't Hoff equation and substitute the reference temperature, new temperature and enthalpy with their units. Calculate the logarithmic change first. Add it to the reference natural logarithm of K, then obtain K at four hundred kelvin. Cross-check the new Gibbs energy with enthalpy minus the new temperature times entropy, under the same zero heat-capacity-change assumption. Then use the new K in the equilibrium-extent expression at a pressure ratio of one. Calculate the species amounts and make the same two independent checks as before. The atom total should remain unchanged, and the quotient reconstructed from the final composition should match the new K. Keep enough intermediate digits to avoid a rounding mismatch. Write a short explanation connecting the positive reaction enthalpy to the direction of the temperature effect. If you need more time, mark the specific unfinished line and continue to the solution. You can return to that line during our onsite question period on November thirteenth.
Your action: Pause for 15 minutes. Complete worksheet E1–E3 before continuing. At the time limit, record the sticking point and continue.
Slide 27: The endothermic reaction dissociates further
Recording 32:52 · Worksheet E1–E3 check
The logarithmic change from three hundred and fifty to four hundred kelvin is about two point one four seven seven two. Adding the reference natural logarithm gives about three point five two two two six. The new equilibrium constant is therefore about thirty-three point eight six zero nine three. 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. At one hundred kilopascals, this gives an extent of about zero point nine four five seven zero moles. 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. 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. Heating has increased dissociation for this endothermic reaction under the stated model. The atom inventory still equals two moles of A atoms. Compare this case with the reference case, not with the four-hundred-kilopascal case, when attributing the difference to temperature alone. The increase in K is a property of the temperature model. The increase in extent also uses the specified feed and pressure.
Your action: Follow slide 27; keep worksheet E1–E3 check beside you. Attempt questions before their checks.
Slide 28: A 350 K Gibbs value is not a temperature law
Recording 34:25 · Worksheet F1 preparation
At four hundred kelvin, the direct Gibbs and formation-energy rows supplied at three hundred and fifty kelvin become unavailable. That is appropriate. Changing a temperature label would not create valid property data at a new temperature. Use the temperature-dependent route instead. For our required calculation, select van't Hoff with constant reaction enthalpy, and keep the heat-capacity change at zero. The Lab also offers a heat-capacity correction. With its default nonzero value, that is a different temperature model and can give a different K. You may explore it later, but it is outside this week's required work. The independent observed-composition panel has its own observation temperature. A three-hundred-and-fifty-kelvin observation does not become a four-hundred-kelvin observation simply because the main evaluation temperature changes. Its unavailable message does not prevent the equilibrium extent calculation from using an applicable thermodynamic route. Read each panel's temperature and label before comparing its output.
Your action: Follow slide 28; keep worksheet F1 preparation beside you. Attempt questions before their checks.
Slide 29: Your turn: save and reopen the temperature case
Recording 35:46 · Worksheet F1–F3
Pause for nineteen minutes and complete F one through F three. Set evaluation temperature to four hundred kelvin and total pressure to one hundred kilopascals. Keep the reference temperature and reference K at their original three-hundred-and-fifty-kelvin values. Select van't Hoff with constant reaction enthalpy as the K source, and check that the heat-capacity change is zero. Run case twelve C and compare it with section E. Export the complete study. Next, use Restore calculator study JSON to reopen your saved twelve A reference file. The calculator should recover the original inputs and recalculate the reference result. Verify the restored temperature, pressure, K and extent. A restored filename alone is insufficient evidence that the numerical state is the one you intended. Finish the three-case table and keep one note about the pressure effect and one about the temperature effect. If restoring a file fails, retain your calculation evidence and record the message for our next onsite class.
Your action: Pause for 19 minutes. Complete worksheet F1–F3 before continuing. At the time limit, record the sticking point and continue.
Slide 30: Three cases separate pressure and temperature
Recording 37:04 · Worksheet F1–F3 check
Here is the temperature-case setup in the actual Lab. 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. The heat-capacity change is zero. After Compare K methods, the temperature-dependent route gives K equal to about thirty-three point eight six zero nine three. The direct Gibbs and formation rows remain tied to their supplied reference temperature. Scroll to the equilibrium composition. The extent is about zero point nine four five seven zero moles, agreeing with the worksheet. Export complete study saves the inputs and calculated evidence for case twelve C. To check reproducibility, choose the earlier twelve A file under Restore calculator study JSON. The restored controls return to three hundred and fifty kelvin and one hundred kilopascals. The recalculated K is about three point nine five three two six, with extent about zero point seven zero five zero three moles. Compare these with your reference row. Your final table now separates two changes. The pressure-only case changes extent while K stays fixed. The temperature-only case changes K and extent. All three cases conserve the same A-atom inventory. The JSON files let you reopen the exact chosen inputs, while your worksheet explains what those inputs and results mean.
Your action: Follow slide 30; keep worksheet F1–F3 check beside you. Attempt questions before their checks.
Slide 31: Bring one unresolved question to our next class
Recording 38:49 · Worksheet G preparation
Before the final form, check the explanation you would give to another student. K follows from standard thermodynamic properties and the chosen temperature. Q follows from the activities of a particular state. Equilibrium extent combines the equilibrium condition with the feed and conservation constraints. These three ideas work together, but they answer different questions. The inert-gas question also depends on the constraint. In this Lab, adding inert at fixed temperature and total pressure dilutes the reacting species and can shift the equilibrium. 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. Name the constraint whenever you use that conclusion. Choose one unresolved question from your worksheet and identify the step where your reasoning stops. We will use those questions when we meet onsite on Friday, November thirteenth. Further heat-capacity and custom-reaction work is optional.
Your action: Follow slide 31; keep worksheet G preparation beside you. Attempt questions before their checks.
Slide 32: Exit ticket, feedback and saved work
Recording 40:06 · Worksheet G
Use the final ten minutes for section G and the Session five exit ticket and feedback. Open the form using the link or QR code on this slide. Check that Session five is selected before submitting. Explain one result from the required cases using your own calculation evidence. Distinguish K from Q, then explain the pressure-only comparison with one numerical result or an independent check. Include the settings and units. Include the question you still want to discuss. In the feedback, report your actual study time and identify any instruction, calculation or control that caused difficulty. Specific descriptions help us improve the next lesson. You may answer the form in English or Thai. It is anonymous and does not identify individual attendance. Submit once and wait for the Response received confirmation. Keep your worksheet and exported study files together. This completes the required Week five work. Please follow the video and its activity stops for November sixth, and we will return to the classroom on Friday, November thirteenth.
Your action: Pause for 10 minutes. Complete worksheet G before continuing. At the time limit, record the sticking point and continue.