Why do reactions stop before they finish?

Seal a bottle of fizzy drink and leave it. Gas comes out of the liquid at first. Then the pressure above the drink stops rising. Nothing seems to change after that. Open the same bottle and it goes flat in a day. So what is the sealed bottle doing while it sits? It looks still. Is it?

This is not a special property of fizzy drinks. Every reaction that can run backwards behaves this way. Module 16 gave every reaction a barrier. A molecule that crosses the barrier one way can cross it back. So the reverse reaction is always there. It is just slow at first. There is little product to feed it.

That is why the sealed bottle never goes flat. The cap does not trap the gas in some clever way. Gas leaves the liquid just as it would in the open. But the cap keeps the gas above the drink. So the return leg has something to work with. The two legs match, and the amounts hold.

So the balance belongs to the closed bottle, not to the gas. Take the lid off and the product leaves. The return leg has nothing to run on. The forward leg runs alone until the drink is flat. Keep the lid on and the product stays. Then the return leg catches up, and the drink stays fizzy.

— A sealed bottle stays fizzy because the return leg catches up with the leaving leg.

Picture one bridge with two lanes. Cars cross one way, and cars cross the other way. At first the whole town is on the left bank. So the right-hand lane is empty, and only one lane is busy. A reaction started from pure reactant is like that town. Only the forward direction can run. Here is the question. What happens to the two rates as time passes?

Look at the left panel first. The forward curve falls and the reverse curve rises. The shaded gap between them narrows. The readout gives the time for the gap to fall under one per cent of its start. It also counts how many times the rates crossed. That count comes out at zero. The two rates settle onto one common value. Neither one overshoots the other. So they never cross. The gap only keeps closing, until it has closed.

Now look at the right panel. The engine follows four thousand molecules one by one. It counts every molecule that changes from A to B. It counts every molecule that changes back. Both running totals keep climbing long after the amounts have flattened. The amounts hold steady because the two totals climb at the same pace. Nothing has stopped. Across this slider, the balance rate is not small either. From pure reactant, it is 1/(1 + K) of the starting forward rate. Here that runs from 80 per cent down to 20 per cent. Push K higher and that share becomes small. The first tier 3 exercise takes K to 100, where it is about one per cent.

So a balanced reaction is a busy bridge. Both lanes are full of traffic. The count on each bank stays the same. That is because the two flows are equal, not because they are zero. This is the bottle's dynamic equilibrium again. The forward and reverse reactions never stop. They only come to match.

— The amounts hold steady because the two rates match, not because either has stopped.

The two rates have come to match. It is tempting to think the two amounts match too. That would be half reactant and half product, with both lanes equally busy. Here is the question. When the flow is balanced, how much of the material is product?

Take K = 4 and work the balance out by hand. At balance, [B] divided by [A] equals 4, so [B] = 4 × [A]. The material is all A or B, so [A] + [B] = 1 of the whole. Put 4 × [A] in place of [B], which gives [A] + 4 × [A] = 1. That is 5 × [A] = 1, so [A] = 1/5 of the whole. Then [B] = 4 × 1/5 = 4/5 of the whole. Four fifths of the material is product, and one fifth is still reactant. The balance rate is 1/(1 + 4) = 1/5 of the starting forward rate, which is 20 per cent.

The table below shows the balance for five values of K. The same solver as bench 3 worked it out. Each row starts from pure A. Find the row for K = 4 and compare it with your hand result. The last column is the balance rate, as a share of the starting forward rate. That share is 1/(1 + K). It is not one of the two amounts. It is how hard both lanes are still running.

So the balance point is a number the reaction owns. It is the ratio of the two rate constants. Whether a flask reaches a balance at all depends on the lid. §1 showed that. Where that balance sits is the reaction's own number. Module 16 showed a catalyst multiplies both constants by the same factor. So a catalyst leaves K exactly where it was. It reaches the balance faster. It does not move it.

— The balance ratio is the ratio of the two rate constants, and nothing makes that a half.

Module 16 promised that the ratio of the two constants fixes the stopping point. That is a strong claim. Does the stopping point care where you began? Take three flasks of the same reaction, at the same temperature. Fill one with pure reactant. Fill one with pure product. Fill one with a mixture. Where does each one finish?

All three runs land on the dashed line. Read the bench source line before you trust that. Two rate constants went in, and three starting mixtures. Where the runs land is what came out. Nobody told the bench the answer. But it is still a simulation. The rules inside it are the rules this module claims. So it cannot prove those rules. Only a measurement can. The table below is measurement.

These five experiments were run with real gas at 25 °C. Two started from pure N₂O₄, two from pure NO₂, and one from a mixture. Read the last column. All five rows sit between 6.50 and 6.56 × 10^-3.

Here is where that column came from. For this reaction, K is [NO₂] squared, divided by [N₂O₄]. The square is there because the equation makes two NO₂ from one N₂O₄. Take experiment 3, which started from pure N₂O₄. Square the NO₂ value, which gives 0.0206 × 0.0206 = 0.0004244 (mol/L)². Divide that by 0.0647 mol/L, which gives 0.0004244 / 0.0647 = 0.00656 mol/L, or 6.56 × 10^-3 as the table writes it. Now take experiment 4, which started from pure NO₂. Square its NO₂ value, which gives 0.0141 × 0.0141 = 0.0001988 (mol/L)². Divide that by 0.0304 mol/L, which gives 0.0001988 / 0.0304 = 0.00654 mol/L, or 6.54 × 10^-3. One flask ran forwards and the other ran backwards, and they landed within rounding of each other. That is measurement, and it says what the simulation said.

So the simulation and the measurement agree. They are different kinds of evidence. The bench shows what the rules imply. The table shows what real gas did. The number they agree on is K. It belongs to the reaction and the temperature. It does not belong to any one flask. NCERT states the same rule in its Class 11 chapter 7, Equilibrium, on page 199.

— Start anywhere, and the reaction lands at the same ratio, because the balance condition never mentions the start.

Bench 2 showed every start heading for the same ratio. That gives a way to predict. Take the ratio a mixture has right now. Chemists call it Q, the reaction quotient. The reaction quotient is product over reactant for the mixture as it is, balanced or not. K is the same ratio at balance. Suppose a mixture has Q bigger than K. Which way will it run?

Take the bench's opening case and check it by hand. The reaction is A ⇌ B with K = 4. The start is 1.000 mol/L of A with no B. Q at the start is 0 divided by 1.000, which is 0. Since 0 is below 4, the reaction runs forwards. Call the amount that changes x. The change row is −x for A and +x for B. The equilibrium row is 1.000 − x for A and 0 + x for B. At balance the ratio equals K, so x / (1.000 − x) = 4. Multiply both sides by (1.000 − x), which gives x = 4.000 − 4x. Add 4x to both sides, which gives 5x = 4.000. Divide both sides by 5, which gives x = 0.800 mol/L. So [A] = 1.000 − 0.800 = 0.200 mol/L, and [B] = 0 + 0.800 = 0.800 mol/L. Check the ratio, 0.800 / 0.200 = 4, which is K.

Now switch the bench to N₂O₄ ⇌ 2 NO₂. It opens on experiment 1 from the table, with the measured K typed in. Read the equilibrium row it builds. Then read the balance columns of experiment 1 in the table. They agree within rounding. That is the prediction machine of this module. Give it K and a start, and it tells you where the mixture stops. The other two reactions have two products each. There the change row carries a product of two terms. The arithmetic is a quadratic, and the bench solves it the same way.

So Q against K says which way. The I, C, E table says how far. Together they are the module's prediction machine. Give them a reaction you have never met, its K, and a start. They tell you where it will stop. That is the tool. Try it on the fizzy bottle from §1. Its Q is the gas above the drink over the gas dissolved in it. Open the bottle and the gas above escapes. Q drops below K, so the exchange runs out of the liquid. That is the flat drink. Seal it again and Q climbs back to K. The last section asks the deeper question. Why does the balance sit where it does?

— Compare Q with K for the direction, then solve the I, C, E table for how far.

Module 14 gave every reaction a referee, called ΔG. ΔG is the free energy change. Its sign says whether the next step happens. Negative means the step goes. Zero means nothing pushes either way. A reaction that is downhill overall still stops early. So ΔG must reach zero before the reactants run out. Something in ΔG changes as product builds. What is it?

One step comes before K can go inside a logarithm. The K in §4 carries the unit mol/L. A logarithm can only take a plain number. So K is first divided by the agreed reference amount. That leaves a number with no unit. Chemists write that version K°. K° is the standard equilibrium constant, the unit-free form of K.

Set ΔG to zero and see what falls out. At balance, Q has become K. Written with no unit, that is K°. So 0 = ΔG° + RT ln K°. Take RT ln K° from both sides, which gives ΔG° = −RT ln K°. Here R is 8.314 J K^-1 mol^-1, and T is the temperature in kelvin. This is the line that ties module 14 to this module. A reaction stops early because mixing has made the next step cost nothing. The stopping point is where ΔG reaches zero. It is not where the two enthalpies balance. It is not where the reactants run out. Both terms of the referee, energy and entropy, are in it.

Here is that line with real numbers. NCERT's chapter on equilibrium gives ΔG° = 13.8 kJ/mol at 298 K, for glucose-1-phosphate turning into glucose-6-phosphate. Positive means the standard step is uphill, so K should be small. Multiply R by T, which gives 8.314 × 298 = 2478 J/mol. Turn ΔG° into joules, which gives 13.8 × 1000 = 13800 J/mol. Divide the second by the first, which gives 13800 / 2478 = 5.57, with no unit left. The definition puts a minus sign in front, so the exponent is −5.57. Then K° = e^−5.57 = 3.81 × 10^-3, which is the figure NCERT gets. Now ask what kind of step that was. It turned one number into another number. It did not measure anything.

IUPAC's Gold Book defines the standard equilibrium constant this way. K° is defined as e to the power of −ΔG°/RT. So ΔG° = −RT ln K° is a definition. It is not a discovery, and on its own it is not a prediction. The K this page trusts came from measurement, from the five flasks. The glucose number ran the other way, from ΔG° to K. That only counts as a prediction when the ΔG° was found without measuring any equilibrium. Where NCERT's 13.8 kJ/mol came from is not traced here. A real prediction needs an independent route to ΔG°. That route exists. Measure the heat of the reaction in a calorimeter. A calorimeter is a sealed vessel that catches the heat a reaction gives out. Get the entropies from heat capacity data. Neither step measures an equilibrium. Then the definition hands you K before any flask is opened. This page is not doing that. It only says the route exists.

— A reaction stops where mixing has made the next step cost nothing, and that is ΔG reaching zero.

You started with a bottle that never goes flat. You now know it is busy. Gas leaves and gas returns, at the same rate. That is dynamic equilibrium. The bridge carries the picture. Both lanes stay full while the count on each bank holds.

Then the numbers came. The balance ratio is K, the ratio of the two rate constants. It is not a half. It does not depend on the start, and five real flasks showed that. Q against K gives the direction. The I, C, E table gives the stopping point.

Underneath is module 14's referee. A reaction stops where ΔG reaches zero, and mixing is what brings it there. ΔG° = −RT ln K° is a definition, not a discovery. This module measured K. It borrowed a ΔG° only to show how the other route would run. So one question stays open. This page never predicted a K without opening a flask. Can a K be found before any flask is opened? That question stays on the table.