LearnChem · Part III — Why do some things react and others just sit there?
Downhill is not enough. A reaction also needs a push to start. Module 15 drew a cut-off across the fast tail of the speeds and did not name it. This module names it and prices it.
“Why does a match need striking? Why doesn't wood catch fire by itself?”
Module 15 left you a box of marbles nobody can slow down. Temperature was the average energy of that jostling. The marbles did not all move at one speed. At a single temperature they carried a whole spread of speeds. A few sat far out in the fast tail. Module 15 drew a line across that tail. It counted the share beyond the line. That line had no name there. Here it gets one. It is the reason a match needs striking.
A log sits in a fireplace for years and nothing happens. Module 12 measured the heat that burning wood gives out. It is a great deal of heat. Module 14's referee then said which way is downhill. Burning is downhill, and the air is already there. So the accounting is settled. The log still sits there. Years of sitting there change nothing at all. Then someone strikes a match, and the wood burns. The match gives back far less heat than the fire does. So the match is not paying for the burning. Something is being got past before the burning can start. This section asks what that something is.
A log sits in air for years and does not catch fire. Module 12 showed that burning it gives out a great deal of heat. Commit — what is stopping it?
Commit before you look. First answers are counted anonymously, never named.
Burning wood ends with new bonds. Those bonds hold more tightly than the old ones. Getting there means pulling the old bonds apart first. Pulling a bond apart costs energy. That cost is paid first. The new bonding energy only comes back later. So a collision that is too gentle leaves the old bonds intact. The activation energy is the energy a collision must carry before the old bonds will break. It is not the same thing as the heat of the reaction. Module 12's heat is the gap between the two ends. The activation energy is the cost of the crossing.
IUPAC's own definition starts somewhere else. It is built from how the rate constant k changes with temperature. The rate constant is the number that fixes how fast a reaction runs at one temperature. Section 3 comes back to it and measures it. So the activation energy is a number read off a graph of measurements. The hill is the picture that goes with it. The hill is not the definition.
When a match is struck, its head rubs against the strip on the box. Friction between the two heats one very small spot. In that spot a tiny amount of the phosphorus on the strip changes. It becomes a more reactive form. The more reactive form ignites in air. That ignition supplies a small pocket of heat in one place. The heat pushes the chemicals in the match head over their activation energy. The head flares, and the flare heats the wood. Now notice what the match did not do. It did not make the barrier any smaller. It paid the barrier, in one small place, for a moment. That is enough for a reaction of this kind. Burning gives out more than enough heat to pay for the next patch of wood. After the first patch, the fire pays its own way.
You may have met the idea of an ignition temperature. That temperature is real and it is measured. It is not the cause of anything, though. It is a name for the temperature at which the payment starts happening by itself. The barrier is what sits underneath it. A match does not bring the whole log to that temperature. It brings one small spot there, and the spot does the rest.
Quick check — a reaction gives out 20 kJ/mol. Going forward, the barrier is 60 kJ/mol. What is the barrier coming back?
Picture a hill between two valleys. In the reactions this section uses, the reactants sit higher and the products sit lower. Not every reaction is like that. Bench 3 will let you raise the far valley above the near one. A reaction has to go over the top either way. The height of the top is the activation energy.
The image breaks in four places. First, a walker looks at the hill and decides whether to climb it. A molecule decides nothing. A share of molecules simply arrive carrying enough energy, and module 15's spread of speeds fixes that share. Second, a walker climbs alone. A reaction needs two molecules to meet, so the hill is crossed by a collision. Third, the walker's hill is a real place with one path over it. A catalyst can open a second route, and the first picture has no room for that. Fourth, the picture puts the products lower than the reactants. A real reaction need not end lower than it started.
— The energy books settle whether a reaction can go at all, and the activation energy settles how fast it goes.
Section 1 gave you the bar. Now ask who is carrying enough energy to clear it. Warm a reaction by ten degrees and it can run several times faster. Module 15 measured what a ten-degree rise does to the speeds. It barely moves the average. The molecules are hardly faster than they were. So the average cannot be where the speed-up lives. Something else about the spread must be moving a great deal. Module 15 drew a cut-off across the fast tail and counted past it. That is the number to watch. The gate below asks you to say why.
A reaction runs several times faster when you warm it by ten degrees. Commit — where does that come from?
Module 15 measured how little the average speed moves for a ten-degree rise.
The average is the wrong place to look. Activation energies are large compared with the energy an ordinary collision carries. So the bar does not sit near the middle of the spread. It sits far out in the fast tail. Now think about what warming does out there. The middle of the spread shifts only slightly. The far tail is thin, and a slight shift makes a large difference to how thin it is. The share of collisions beyond a far-out line can change a great deal while the average hardly changes at all. That is worth measuring rather than being told. Bench 1 counts collisions and sorts them by the energy they carry. Set the bar where you like and watch what the share does.
Here is what the log panel should have shown you. The share of collisions clearing the bar falls as a straight line on that log scale. A straight line there means the share falls exponentially as the bar is raised. The slope of the line carries the temperature inside it. Set the dial to 300 K and then read the slope of that line. On one run of ours the slope handed back 294 K. That reading is low by about two per cent. Your own run will give a number a little different from ours. The slope is worked out from a pile of counted collisions. No two piles of collisions are ever quite the same.
Across every temperature we checked, the slope read low rather than high. The worst of them was low by about five per cent. A bigger pile of collisions does not make the slope steadily more accurate. What a bigger pile does buy is room higher up the bar slider. Let the bench count for longer and the high bars gather more collisions. Then the fit can use bars further up before the tail runs thin.
The bench was told nothing about exponentials, and no exponential appears anywhere in its code. It measured the energy along the line joining two centres, counted, and the shape came out of the counting. One warning belongs on the high end of the bar slider. A run of ours counted 41 432 collisions in all. Only seven of them cleared a bar of 20 kJ/mol. A share measured from seven collisions is luck rather than a measurement. Watch for the bench's own warning when the tail gets that thin. Read the slope from the low bars, where the counts are large.
Quick check — you raise the bar from 10 to 20 kJ/mol and leave the temperature alone. What happens to the share that clears it?
Module 15's image works again here. A class sits an exam and the marks spread out. Draw a cut-off somewhere near the top of that spread. Move the cut-off down a little and the number above it jumps. The bar in a reaction is that cut-off. The energy a collision carries is the mark.
A fair worry follows from all this. If only the fast collisions react, the fast molecules should run out. They do not. A molecule's speed is not a property it keeps. Every collision reshuffles it, about eight thousand million times a second, as module 15 worked out. A molecule is fast one moment and slow the next. So the group above the line is never the same molecules twice. It is refilled constantly. Only the shape of the spread holds still, and the temperature is what fixes the shape.
— Temperature works on a reaction through the thin far tail of the spread, not through the average.
A balanced equation is a stock-take. It says what goes in and what comes out. It makes the atoms on both sides tally. Balancing is about conservation. Conservation says nothing about how long anything takes. Yet the equation is the first thing anyone writes down. So the temptation is obvious. If a reactant appears once in the equation, surely doubling it doubles the speed. Put two molecules of it in the equation, and surely the effect becomes a squaring. That reasoning is used constantly, and it is worth testing. The gate below puts it to you.
An equation reads A + B → P. Commit — if you double how much B you start with, what happens to the speed?
There is a right answer here, and it is not about this particular reaction.
Two different things are being confused here. The equation is about amounts. Speed is a different question, and it needs its own statement. The rate law is the measured expression that gives a reaction's rate from the amounts present. The order in a reactant is the power that reactant's amount is raised to in the rate law. An order is not read off the equation. It is measured, by starting with more of one reactant and watching what the starting rate does. The rate constant k is the number in front. Your own textbook calls it a proportionality constant. That is a textbook definition rather than an IUPAC one, and it is the one this module uses.
There is a second word that is easy to mix up with order. Molecularity is the number of particles that take part in one single step of a reaction. Molecularity counts particles in a step, so it is always a whole number. It is one, two or three, and it is a theoretical count. Order is not a count of anything. Order is an exponent got from measurements, so it can be zero, a fraction or a negative number. The two agree only when the step you wrote down is the single step that sets the pace. Almost no equation you meet is a single step. Bench 2 below is where you can watch that happen. Your own textbook puts the warning plainly. NCERT's chapter on chemical kinetics says a rate law cannot be predicted from the balanced equation. It must be determined by experiment. The sentence is quoted in full under the table below.
Take the three mechanisms in order, and read the numbers the bench printed. The single-step mechanism measured an order of 1.0000 in A and 1.0000 in B. There the coefficients and the measured orders agree. The second mechanism has A breaking up first and B joining on afterwards. Its overall equation is A + B → P, and B is plainly in it. The bench measured the order in B as 0.0006, which is zero to the accuracy it can reach. Doubling B changes nothing, because the slow first step has already happened before B is needed. The third mechanism splits X2 in a fast balance and then reacts one piece of it. The bench measured the order in X2 as 0.5154 and the order in Y as 1.0017. A chemist would call that a half-order reaction.
Here is where the half comes from. The fast balance ties the amount of the loose piece to the square root of the amount of X2. So doubling X2 multiplies the amount of the loose piece by the square root of two. The measured order is 0.5154 rather than exactly 0.5. The balance is fast, but it is not infinitely fast. So the loose piece sits a little below its balance amount, and the reading drifts with concentration. It reads 0.522 when the starting amounts are small and 0.510 when they are large.
| the balanced equation | what the coefficients suggest | the rate law that was actually measured |
|---|---|---|
| CH3COOC2H5 + H2O → CH3COOH + C2H5OH | first order in the water | rate = k[ester] — order zero in the water |
| H2 + Br2 → 2 HBr | first order in the bromine | rate = k[H2][Br2]1/2 — order one half |
| CHCl3 + Cl2 → CCl4 + HCl | first order in the chlorine | rate = k[CHCl3][Cl2]1/2 — order one half |
| 2 N2O5 → 4 NO2 + O2 | second order | rate = k[N2O5] — order one |
| 2 O3 → 3 O2 | second order, and oxygen does not appear | rate = k[O3]2[O2]-1 — order minus one in the oxygen |
Quick check — a measured rate law reads rate = k[X]2[Y]0. You double the amount of Y. What happens to the rate?
The last row of that table is worth a second look. Ozone breaks down to oxygen, and the measured rate law has order minus one in oxygen. A negative order means that more of that substance makes the reaction slower. Oxygen is a product here, so the reaction slows itself as it goes along. Add extra oxygen at the start and it is slower from the beginning. Nothing in the balanced equation hints at that. An order of minus one is the fingerprint of a reaction that runs in more than one step. Oxygen takes part in one of those steps. This module does not have the machinery to draw those steps. The measurement stands either way.
— The balanced equation tells you what reacts, and only an experiment tells you how the speed depends on it.
Section 2 showed the shape. One exponent holds both the barrier and the temperature. Section 2 moved the barrier. This section moves the temperature instead. The Arrhenius equation ties the rate constant to temperature. Chemists use it everywhere. It has two parts. The first is a constant in front, written A. A is the overall scale. The second is the exponential factor. That factor is the share of collisions carrying enough energy. Take logarithms and it becomes a straight line. Plot ln k against 1/T, and the slope gives the activation energy. How far that fit can be trusted is this section's question. A fit comes with a goodness number, and the gate below is about it.
A line through some measurements has an R² of 0.999986. Commit — how far outside the measurements can you trust it?
R² says how closely the points used lie to the line.
R2 reports one thing. It says how closely the points the fit was given lie to the line. A low R2 tells you those points are scattered about the line. A high R2 tells you they are not. What R2 cannot tell you is anything about a point the fit never saw. It is a score on the homework the line was set, marked by the line itself. Step outside the range of the measurements and R2 has nothing to say. There is only one honest way to find out whether a line predicts. Keep some measurements back from it. Fit the line on what is left, then ask it for the measurements it never saw, and compare. Bench 4 is built to let you do exactly that.
Press “the three hottest” and the line is fitted to 40, 60 and 80 °C only. That fit covers the range 313 to 353 kelvin and gives an activation energy of 101.27 kJ/mol. The two cold points, 0 °C and 20 °C, were not used to make that line. The bench then asks the line for the rate constant at each of them. At 0 °C the prediction comes out 3.5 per cent below the measurement. At 20 °C it comes out 6.4 per cent below the measurement. Those are honest predictions, because the line never saw either point. Whether they are good ones is a question the next paragraph reopens. The scatter in these five measurements is larger than 3.5 per cent.
Now press “the three coldest” and the line is fitted to 0, 20 and 40 °C. That fit covers 273 to 313 kelvin and gives an activation energy of 102.90 kJ/mol. Its R2 is 0.999986, which is as close to a perfect straight line as measurements get. Three points is only one more than two. A line through three points can look nearly perfect and still be pinned to very little. The two hot points were not used to make it. Asked for the rate constant at 60 °C, this line comes out 53.8 per cent high. Asked for 80 °C, it comes out 4.9 per cent high. So the better-looking fit made the worse prediction, by a long way. R2 did not warn anybody, because R2 only ever looked inside the range.
It would be easy to blame extrapolation for that 53.8 per cent, and that would be half wrong. Fit all five points and look at how far each measurement sits from the line. Four of the five sit within about sixteen per cent of it. The 60 °C measurement sits 24.3 per cent below the line. That one point is the noisy one in this dataset. The cold fit was asked to predict a measurement that is itself well off the trend. Much of the 53.8 per cent belongs to that scatter rather than to the method.
You can watch the same point wreck a smaller fit. Fit only 40 °C and 60 °C and the activation energy comes out 83.930 kJ/mol. Fit only 60 °C and 80 °C and it comes out 121.623 kJ/mol. That is a spread of 37.7 kJ/mol out of the same five measurements. The five-point answer, 100.73 kJ/mol, sits between the two. A two-point activation energy is pinned to whatever scatter those two points happen to carry. Two points always lie perfectly on a line, so nothing inside the fit warns you. The honest statement is that a two-point answer is at the mercy of which two points. You cannot tell from the fit which pair you have got.
Quick check — a line is fitted to all five measurements. It is then used to state k at 40 °C. Is that a prediction?
Not always, and there is a measured case to prove it. NASA's Jet Propulsion Laboratory publishes recommended rate data for atmospheric chemistry. Two of its recommendations carry a negative fitted activation energy. The first is O(1D) + O2 → O(3P) + O2. In that reaction an oxygen molecule takes the extra energy off an excited oxygen atom. JPL tabulates the exponent as E/R, in kelvin, and for this reaction it is minus 55 K. Multiplied by the gas constant, that is an activation energy of minus 0.46 kJ/mol. JPL gives the recommendation for 104 to 424 kelvin. The second is O(1D) + N2 → O(3P) + N2. There E/R is minus 110 K. That is an activation energy of minus 0.9 kJ/mol. JPL gives this one for 104 to 673 kelvin.
A negative exponent means these reactions run more slowly as they get hotter. Do not read that as a hill of negative height. No source in this course's reading gave a verified reason for these two numbers, so this page offers none. What it can say is what a fitted activation energy actually is. It is a number got from the slope of a graph of measurements. It is not the height of a hill that anyone has measured. Notice one more thing about JPL's tables. Every recommendation has its temperature range printed beside it. That is the discipline this section has been arguing for.
— R2 marks the line's own homework, and only a held-back measurement can mark its predictions.
Everyone meets catalysts early. A catalyst makes a reaction go faster and is not used up doing it. Iron does it for ammonia in industry. A catalytic converter does it for a car's exhaust gases. Your own cells do it with enzymes, thousands of them. That much is usually remembered correctly. The question underneath is what else a catalyst changes. A reaction left alone stops somewhere, and module 14's referee decided where. Does a catalyst move that stopping point? Most people's first answer is yes. It is worth committing to an answer before you look. The gate is below.
You add a catalyst to a reaction and leave it until it stops changing. Commit — what is different about where it stopped?
Module 14's referee decided where a reaction stops. Think about what the catalyst did and did not touch.
A catalyst is a substance that increases the rate of a reaction without being used up in it. It works by opening a second route between the same two valleys. The new route is lower over the top. A route over a hill is a route in both directions. The same shortcut that makes the forward reaction faster makes the back reaction faster by the same factor. A reaction stops when the forward and back rate come into balance. Both rate constants are multiplied by the same factor, so their ratio is unchanged. That ratio is what fixes where the reaction stops. Module 14's referee is what fixes that ratio. The catalyst touched neither the ratio nor the referee. If a catalyst could shift a stopping point, you could add it, take out the product, remove the catalyst, and start again. That is a machine that never runs down. Nobody has built one.
Push the catalyst slider to the top and read the four numbers the bench prints. The forward rate constant is 1839.8521 times bigger than it was. The back rate constant is 1839.8521 times bigger than it was. That is the same number, to eight figures, and it is the whole point. Their ratio is the equilibrium constant, and the bench prints it as 90.9640412316. Slide the catalyst back to zero and the bench prints 90.9640412316 again. The two agree to twelve figures, so nothing about the balance point has moved. The final share of product is the same on both routes as well. Only one thing changed, and that is the time it took to get there. So the common answer is wrong, and it is wrong in a definite way. A catalyst does not give more product, and what it gives you instead is the same product sooner. There is one more thing the bench will not let you claim. The slider does not lower the barrier of the original route. The dashed hump stays where it was, and the catalyst draws a second route beside it.
Quick check — a catalyst doubles the forward rate constant. What happens to the rate constant for the reverse?
Real numbers are worth seeing next to that. Hydrogen peroxide breaks down slowly on its own, by 2 H2O2 → 2 H2O + O2. One published fit gives its activation energy as 55.3 ± 1.8 kJ/mol, over the range 297 to 338 kelvin. Your blood and your liver carry an enzyme called catalase that does the same reaction. A separate study of catalase gives an activation energy of 14 kJ/mol, over 283 to 318 kelvin. That is roughly a quarter of the barrier on the uncatalysed route. Read the comparison carefully, because it comes from two different studies. Different laboratories and different conditions sit behind those two numbers. Their temperature ranges only partly overlap. Nobody ran one controlled experiment with catalase in one flask and nothing in the other. So the direction of the effect is solid and the pair of numbers is not matched. Catalase is also among the fastest enzymes known, turning over about 4×107 molecules a second at each active site. That last figure comes from a secondary source, and this page flags it as one.
There is a trap waiting here. So far in this module, heat has made every reaction faster. Enzymes are where that stops being true. Warm an enzyme a little and it does work faster, for the reason section 2 gave. Warm it a lot and it stops working altogether. An enzyme is a large folded molecule, and its folded shape is what does the catalysis. Denaturation is the word for that shape coming apart. Heat denatures an enzyme. What is left has lost the arrangement the reaction needed, and it has lost the catalysis with it. So there is a temperature above which more heat makes an enzyme useless rather than faster. Where that temperature sits depends on the enzyme, and this page does not give a number for it.
— A catalyst changes how long a reaction takes and leaves where it ends up exactly as it was.
Cut an apple in half and leave it on the counter. It browns within the hour. A whole apple beside it does nothing for days. Put the cut half in the fridge and the browning almost stops. The same goes for milk, for meat and for a chopped onion. Two different things are going on in those examples. Cutting does one of them and cooling does the other. This module can put a number on one of the two. The other it can only point at and name. Commit to which is which before you read on.
A cut apple browns faster than a whole one, and a fridge slows both down. Commit — which of these is this module about?
You have two things to explain and one new tool.
Both matter, and they matter for different reasons. Cutting works on how much surface there is. A whole apple meets the air only at its skin. A cut one meets the air across the whole cut face. More surface means more places where two things can meet at all. Cutting also breaks cells open and puts their contents within reach of each other. None of that changes the barrier. None of it changes the share of collisions that clear the barrier. It changes how many collisions there are. Cold works on the other half of the story. Cooling does not change the barrier either. It changes the share of collisions carrying enough energy to clear it. That share is the thing section 4 gave you an equation for. So the fridge is the one this module can price.
Run the Arrhenius equation once, on the reaction that bench 4 has been using. Its activation energy from all five points is 100.73 kJ/mol. Take a room at 25 °C, which is 298.15 kelvin. Take a fridge at 4 °C, which is 277.15 kelvin. Start by dividing the activation energy by the gas constant. That is 100 730 joules per mole divided by 8.314 joules per kelvin per mole. The division gives 12 116 kelvin, and the joules and the moles cancel. Now take one over the colder temperature, which is one over 277.15 kelvin. From that subtract one over the warmer temperature, which is one over 298.15 kelvin. The colder temperature goes first, so the difference comes out positive. That difference is 2.541 × 10-4 per kelvin.
Now multiply the difference by the 12 116 kelvin of the first step. The kelvins cancel, and the product is 3.079, which is a plain number. The ratio of the two rate constants is e raised to the power 3.079. That comes out at 21.7. So this reaction runs 21.7 times faster in the room than in the fridge. The direction of that ratio is worth saying out loud. Cooling makes the share of clearing collisions smaller, so the ratio is bigger than one. The fridge has taken nothing away from the reaction's activation energy. That is still the 100.73 kJ/mol the fit gave. What the cold has taken away is the energy a small share of collisions needed. The equation takes any activation energy and any two temperatures. Tier 2 question 4 asks you to run it yourself.
There is one honest note to add before you do that. Section 2 said a reaction can run several times faster for a ten-degree rise. It did not say twice as fast, and the difference matters. The doubling rule is the one you are likeliest to have met. It is a rule of thumb and it is not a law. This reaction does not obey it. Warm this reaction from 25 °C to 35 °C and the rate constant goes up 3.74 times. The doubling rule belongs to a much smaller activation energy, near 53.59 kJ/mol. The rule of thumb is an activation energy in disguise.
Food is where this gets harder. The FAO publishes keeping times for fish at different temperatures. Salmon keeps 11.8 days at 0 °C and 3.0 days at 10 °C. Ten degrees of cooling has bought a factor of 3.9 in shelf life, which sounds like the equation above. The same FAO paper then says the equation does not work here. It says the Arrhenius equation has been shown not to be accurate for spoilage across a wide range of temperatures. Its exact words are in the source line below. That warning is worth taking seriously, and the reason is not hard to see. Spoiling is not one reaction with one barrier. It is many reactions at once, and it is bacteria growing and dividing as well. Living things have their own temperatures at which they stop. One activation energy cannot stand in for all of that.
Quick check — two reactions have activation energies of 50 and 100 kJ/mol. Which one does a fridge slow down more?
The United States Department of Agriculture names a range of temperatures food should not sit in. Its words are that bacteria grow most rapidly between 40 and 140 degrees Fahrenheit. In that range they can double in number in as little as 20 minutes. That range is about 4 to 60 °C. A fridge sits below that range and cooking takes food above it. That is what both are for. The same department's Agricultural Research Service has measured what cutting does on its own. Shredding and slicing raised microbial populations on cut cabbage, lettuce and onions by one to three powers of ten. That is ten to a thousand times more bacteria, immediately, with no change of temperature at all. So the cut apple and the fridge are two separate effects. The USDA has a number for each.
— Cutting adds places for collisions, and cooling thins the share of collisions that can clear the barrier.
New numbers every time. Press Deal again for a fresh set.
1. Module 12's heat, then this module's two barriers: (In kJ/mol.)
2. Module 13's shares, with module 15's cut-off: (Write it as a decimal or in the form 3.2e-5.)
3. (A number of times, to two decimal places.)
4.
These four questions use the five measurements bench 4 uses, and nothing else. Work them with a calculator, taking the gas constant as 8.314 J per kelvin per mole. Each answer is checked against a band, so a little rounding will not fail you.
| temperature | 0 °C | 20 °C | 40 °C | 60 °C | 80 °C |
|---|---|---|---|---|---|
| k / s-1 | 7.87×10-7 | 1.70×10-5 | 2.57×10-4 | 1.78×10-3 | 2.14×10-2 |
1. Use only the 0 °C and 20 °C measurements. Divide the warmer rate constant by the colder one, then take the natural logarithm of the result. Next, subtract one over the warmer temperature in kelvin from one over the colder temperature in kelvin. The activation energy is the gas constant times the first result, divided by the second. Give it in kJ/mol.
2. Now do the same with the 40 °C and 60 °C measurements only. Compare the answer with the one you got from the two coldest points. Both pairs came out of one experiment on one reaction.
3. Take the activation energy you just got from the 40 and 60 °C pair. Use it to predict the rate constant at 0 °C. Anchor the line on the 40 °C measurement and step down from there. Then compare your prediction with the measured value in the table. Give the answer as the number of times too big the prediction is, and not as a rate constant.
4. One last one, and it is the equation from section 6. Take a reaction with an activation energy of 100.73 kJ/mol. How many times faster is it in a room at 25 °C than in a fridge at 4 °C? Work in kelvin and give a plain number of times.
These three have no box to type into. Write a few sentences on each, or argue them out with somebody.
Start with the two questions at the top. Both were about the match and the log. Burning wood has a barrier in front of it, and a room never pays that barrier. So the log sits there for years. A struck match pays it in one small spot, for a moment. Section 6's box asked about the temperatures food should not sit in. Cold thins the share of collisions carrying enough energy, and this module can price that. It cannot price the cutting. A cut apple browns faster because cutting opens more surface to the air. Chopped food rots faster because the blade spreads bacteria through it.
You have also seen a line with an R2 of 0.999986 miss a held-back measurement by 53.8 per cent. Part of that miss was the measurement's own scatter. Bench 3's reaction stopped at 98.9 per cent product, not at 100. This module said only that a reaction stops when the forward and back rates come into balance. It did not say why that balance sits at 98.9 per cent rather than anywhere else. This module was about how fast. The next module asks why a reaction stops before it has used everything up.
Reactions need a push because bonds must break before better bonds can form.