LearnChem · Part III — Why do some things react and others just sit there?
Module 13 left you holding two counts that pull opposite ways. One belongs to the reaction. The other belongs to the room around it. You can count the first. You cannot yet price the second. One line of arithmetic is missing.
“Why do some things happen all by themselves — rust, rot, a melting ice cube — while others never will, however long you wait?”
Last time you shuffled a deck and it never shuffled itself back. You counted the arrangements. You found that the world's total count only ever rises. That is the second law. But the total has two halves. One half is the reaction. The other half is the room around it. Ice melting gains freedom and swallows heat from the room. The two counts pull opposite ways. You cannot yet price the room's count. That is one line of arithmetic. This module supplies it.
A reaction gives out heat. The heat arrives in the room. The room's molecules now have more energy to share out. So the room's count of arrangements rises. Module 13 told you that much. It did not tell you by how much. That is the missing number. Without it you cannot add the two counts together. Without the sum you cannot use the second law at all. So the question is plain. You pour a known amount of heat into a room. How much does its count rise? Before the arithmetic, a guess.
You pour the same amount of heat into two rooms. One room is cold, near −20 °C. The other is hot, near 230 °C. Commit — in which room does the count of arrangements rise by more?
Commit before you look. First answers are counted anonymously, never named.
The same heat is worth more arrangements in a cold place. In a hot room the count is already huge. A fixed amount of heat adds a small fraction to it. In a cold room the count is small. The same heat adds a large fraction.
That much follows from counting. The exact shape does not. Getting from “it falls as the room warms” to the precise form needs more than this course can show. It comes out as one line.
ΔSsurroundings = −ΔH ⁄ T
Each piece has a plain meaning. ΔH is the heat the reaction takes in. The minus sign turns it round. Heat leaving the reaction is heat arriving in the room. T is the room's temperature in kelvin. ΔS is the change in the count of arrangements from module 13. Chemists call that count the entropy. Dividing by T is what makes cold rooms gain more.
Quick check — a reaction gives out 100 kJ into surroundings held at 300 K. By how much does the entropy of the surroundings change?
Two assumptions sit under that line. The surroundings are taken as very large. The heat arriving does not shift their temperature. And the change happens at a fixed pressure. That is what makes ΔH the heat rather than something else.
Now the two counts can be added. The total entropy change is the system's change plus the surroundings' change. Write the surroundings' change with the new line. The second law says the total must be greater than zero.
ΔStotal = ΔSsystem − ΔH ⁄ T > 0
Multiply the whole line by −T. Take it one term at a time. Multiplying ΔSsystem by −T gives −TΔSsystem. Multiplying −ΔH/T by −T cancels the T and leaves +ΔH. And T is always positive, so the minus sign turns the inequality round.
ΔH − TΔSsystem < 0
That is the whole derivation. Nothing new went in. The quantity on the left has a name. It is the change in the free energy. Chemists call it the Gibbs free energy change, written ΔG. It measures how far downhill a change is, once the room's share has been counted. So ΔG = ΔH − TΔS. A change goes on its own when ΔG is negative. Notice what happened to the room. It has been folded into the T. This is now a rule about the system alone. That is why chemists use it. You no longer have to think about the room.
— Heat is worth more in a cold room, and folding that in gives ΔG = ΔH − TΔS.
Picture a referee with two whistles. One whistle is in the energy hand. It blows for the heat the reaction lets go. That is ΔH. The other whistle is in the freedom hand. It blows for the arrangements gained, multiplied by the temperature. That is TΔS. The referee adds the two and reports which way is downhill. Module 13 left the cold pack as a puzzle. It goes, and it goes uphill in energy. Module 13 could not say how that is possible. Now the referee can be made to show its arithmetic.
An instant cold pack swallows 25.7 kJ for every mole that dissolves. It goes uphill in energy. It turns cold in seconds all the same. Commit — how big must TΔS be at 25 °C for that to work?
Commit before you look. First answers are counted anonymously, never named.
Both terms are in the same units once TΔS is worked out. ΔH is in kilojoules per mole. ΔS is in joules per kelvin per mole. Multiply ΔS by the temperature and the kelvins cancel. Divide by 1000 and you have kilojoules per mole.
The two whistles do not have to agree. Add them up instead. Play goes on when the sum comes out downhill. Temperature is the volume knob on the second whistle. Turn it up and TΔS grows. The energy whistle does not move with it. So temperature settles which of the two wins.
Module 13 said the arrangements beat the energy. Here is the price. Ammonium nitrate dissolving swallows 25.7 kJ per mole. That is ΔH, measured by calorimetry. Calorimetry means measuring heat by watching a temperature change. The dissolving is uphill. ΔS is +108.7 J/K per mole, worked out from tabulated entropies. Room temperature, 25 °C, is 298.15 K. So TΔS is 298.15 × 108.7 ÷ 1000, which is 32.41 kJ per mole. Now the referee adds. ΔG is 25.7 − 32.41, which comes to −6.71 kJ per mole. That is negative, so the pack goes. The margin is small.
Small margins flip. Cool the pack and TΔS shrinks. At 236.4 K, which is −36.7 °C, the two terms draw level. One caution on that number. At −36.7 °C the water in the pack would be ice, and nothing would dissolve at all. The flip temperature is what the arithmetic says. It is not a place you could take the pack. Watch for that whenever a flip temperature lands far from where the numbers were measured.
The cold itself comes from the room. The pack pulls heat out of your hand to pay the uphill 25.7 kJ.
Two honest flags. The same ΔH worked out a second way, from tabulated formation values, comes to +28.1 kJ per mole. That is about 2.5 kJ higher. This page uses the directly measured +25.7. Try the other value and see what it does. 28.1 − 32.41 comes to −4.3 kJ per mole. Still negative, so the pack still goes. The margin shrinks and the verdict holds. The second flag is smaller. Commercial packs are moving from ammonium nitrate to urea, because urea is less hazardous.
The bench names four cases, and they are worth having as a table.
| the energy term | the freedom term | it goes |
|---|---|---|
| gives out heat | gains freedom | at every temperature |
| takes in heat | loses freedom | at no temperature |
| gives out heat | loses freedom | only below a flip temperature |
| takes in heat | gains freedom | only above a flip temperature |
Only the bottom two rows have a flip temperature. In the top two the verdict is the same however hot it gets. The cold pack is a bottom row. It takes in heat and it gains freedom, so it goes above some temperature and not below it.
You can find that temperature. At the flip the two terms are equal, so ΔH = TΔS. Divide both sides by ΔS. The flip temperature is ΔH divided by ΔS. Put ΔH into joules first, because ΔS is in joules.
Quick check — a reaction takes in 60 kJ per mole. It also gains 150 J/K per mole of freedom. Above what temperature does it go?
Now load the iron preset. ΔH is −824.2 kJ per mole of rust. ΔS is −274.9 J/K per mole. Both come from measured tables. The bench works out ΔG and prints −742.24 kJ per mole. Chemists also list a ΔG for rust, and their figure is −742.2 kJ per mole.
Be careful what that agreement shows. It is not a second opinion. Table values of ΔG are made by this same arithmetic, from these same two measurements. So the two numbers had to match. The agreement shows the sum was done right. It does not show the rule is true. The real test comes in the next section. There, one number is reached by two roads that know nothing about each other.
The referee holds two whistles. One blows for the energy let go. One blows for the freedom gained. The referee adds them and reports which way is downhill. Temperature is the volume knob on the second whistle. Turn it up and freedom sounds louder.
Here the image breaks, in three places. A referee decides. ΔG decides nothing. It only reports which way is downhill. A referee's whistle also stops play at once. ΔG says nothing about how long the trip takes. And a diamond breaks the picture hardest. The referee waved play on a hundred years ago and nobody has moved.
— Two terms, one verdict, and temperature sets how loud the second one is.
Ice melts at 0 °C. It does not melt at −1 °C. Take a block at −1 °C and warm it by one degree. Nothing about the ice has changed. The same molecules sit in the same lattice. The same hydrogen bonds hold them. Melting swallows the same 6.01 kJ per mole at either temperature. Melting gains the same freedom at either temperature. Yet one degree separates a solid that stays from a solid that goes. Something must be different at 0 °C.
Ice melts at 0 °C and not at −1 °C. Melting swallows the same 6.01 kJ per mole either way. It gains the same freedom either way. Commit — so what is different about 0 °C?
Commit before you look. First answers are counted anonymously, never named.
The melting point is the temperature where the two terms draw level. ΔH of melting is +6.01 kJ per mole. It does not change much with temperature. ΔS of melting is fixed too. What moves is T, the multiplier on the freedom term. Below 273.15 K, TΔS is smaller than 6.01 kJ. The energy term wins and ice stays. Above it, TΔS is the bigger of the two. ΔG turns negative and ice goes. At exactly 273.15 K the two are equal. ΔG is zero. Ice and water sit together.
Nothing about the ice changed. The temperature changed. So a melting point is not a property of the solid alone. It is the point where the two terms meet. That is a definition, not a fact about ice.
The slope is the real news, and the slope has a name. It is the entropy of melting itself. Every extra kelvin multiplies ΔS by one more degree. So the line falls by ΔS for every kelvin you climb. For water that is 22 J per mole per kelvin. Ten degrees above the melting point, ΔG is only −220 J per mole. Ten degrees below, it is +220. That is a very small push. It is why ice and water sit together so easily in a glass. And it is why a little salt or a little pressure can move the line at all.
Now load naphthalene, the mothball. It melts at 80.2 °C, and its entropy of melting is 53.8 J/K per mole. Water's is 22.0. Melting a mothball buys more than twice the freedom that melting ice does. Hold that question. The rest of this section answers it, and it answers it with a table of boiling points.
The bench admits a circle. It worked out the entropy of melting from the melting point. So of course the line crosses zero at the melting point. That is not a prediction. It is the input coming back out.
The circle can be broken. The same number is reachable a second way, and that way knows nothing about melting points. Measure a substance's heat capacity, starting near absolute zero. Warm it step by step. Add up the entropy gained at each step. That is how chemists actually tabulate entropies. The two roads give one answer. Module 13 made the same move for R ln 2. R is the gas constant, 8.314 joules per kelvin per mole. Two independent routes to one number is what turns arithmetic into evidence.
The same line works for boiling. Take a liquid's heat of boiling. Divide it by its boiling point. That gives the entropy gained when it boils. Now do that for six ordinary liquids, all quite different from each other. Before you see the answers, commit.
Six liquids, six heats of boiling, six boiling points. Divide one by the other for each. Commit — will the six answers be scattered all over, or will most of them land near one number?
Commit before you look. First answers are counted anonymously, never named.
Here are the six, worked out.
| liquid | boiling point (K) | heat of boiling (kJ/mol) | entropy gained (J/K per mole) |
|---|---|---|---|
| diethyl ether | 307.7 | 26.52 | 86.2 |
| benzene | 353.3 | 30.72 | 87.0 |
| chloroform | 334.3 | 29.24 | 87.5 |
| methanol | 337.8 | 35.21 | 104.2 |
| water | 373.2 | 40.65 | 108.9 |
| ethanol | 351.5 | 38.56 | 109.7 |
Boiling points from the NIST Chemistry WebBook. So are the heats of boiling, for every liquid here except water. The WebBook's page for water carries no table of boiling enthalpies. Water's 40.65 kJ/mol comes from a secondary engineering source instead. It is cross-checked against an independently published entropy of 109.1 J/K per mole. The last column is worked out here, as the heat of boiling divided by the boiling point.
Three of them sit together near 87. The other three sit about 20 higher. So most of the six do cluster, and the three that do not are a set of their own.
Quick check — benzene, ether and chloroform each gain about 87 J/K per mole when they boil. Water gains 109, methanol 104 and ethanol 110. What in the liquid explains the second three?
The three that break the pattern are water, methanol and ethanol. They are the three that hold hands. Module 08 built that picture. A hydrogen bond is a strong grip between one molecule's hydrogen and a neighbour's oxygen. A liquid held by those grips is already tied down. So breaking it into a gas buys more freedom than usual.
That also answers naphthalene. Liquid water is still holding hands after the ice melts. Melting has not set the molecules loose. Melted naphthalene has no such grips, so its molecules really are loose. That is why melting a mothball buys 53.8 J/K per mole and melting ice buys only 22.0.
One honest note, because a careful reader will spot it. The three that break the pattern are also the three lightest in the table. So weight and hydrogen bonding point the same way here, and these six liquids cannot tell them apart. What settles it is that hydrogen bonding explains other things as well. Module 08 used it to explain why ice floats. It used it again to explain why water boils a hundred degrees later than its size says it should. Weight explains neither.
The pattern has a name. It is called Trouton's rule. Trouton's rule says that most liquids gain about the same entropy when they boil. Textbooks put the figure near 85 J/K per mole. The three well-behaved liquids in this table average 87.
That makes it a prediction machine. Take a liquid's heat of boiling. Put it into joules first, so 30.72 kJ becomes 30 720 J. Divide by 87. You get an estimate of the boiling point in kelvin.
Try it on benzene. 30 720 divided by 87 gives 353 K. The real answer is 353 K. Now try it on water. 40 650 divided by 87 gives 467 K, and water boils at 373 K. The machine is 94 K out. Water is one of the three that hold hands, and the rule does not cover those. So the machine works on liquids that do not hold hands. Knowing where it fails is part of owning it.
— A melting point is just the temperature where the two terms draw level.
ΔG has passed every test so far. Now it meets a living body. A cell runs thousands of reactions at once. Many of them are uphill. Building a protein from its pieces is uphill. Sticking a phosphate group onto a sugar is uphill. ΔG for each one is positive. The referee says no. Yet every one of them goes, in you, right now, without pause. Nothing in the ledger has been broken. So something in the way the question is asked must be incomplete.
Sticking a phosphate group onto glucose is uphill by about 13.8 kJ per mole. Your body does it millions of times a second. Commit — how?
Commit before you look. First answers are counted anonymously, never named.
Two reactions can be added only when they are really one reaction. Here they are. ATP is a small molecule a cell keeps ready. It carries a phosphate group, and it pays out energy when it lets that group go. The phosphate group leaves ATP and lands on glucose in a single step. Nothing is set free in between. One enzyme holds both partners while it happens. The enzyme is called hexokinase. Because the move is one step, one ΔG covers it. That ΔG is the sum of the two halves. Glucose plus phosphate costs +13.8 kJ per mole, and ATP giving up its phosphate pays −30.5. Add the two and the joined step comes to −16.7 kJ per mole. That is negative, so the referee says go. The uphill half never runs alone.
Set the switch to separate and the sum disappears. That is not a display fault. An uphill reaction is not helped by a downhill one it does not touch. Energy does not move across a cell from one reaction to another. There is no path for it to take. Coupling is a shared step, or it is nothing.
Two flags on the numbers. These are ΔG′ values at pH 7. That is the biochemists' standard state. A standard state is the fixed set of conditions a tabulated number was measured under. Change the conditions and the number changes. Everywhere else on this page the standard state is a different one. It means pure solids and liquids, and gases at 1 bar, all at 298.15 K. The second flag is the ATP figure itself. Its −30.5 kJ per mole is a midpoint, not a constant. Measured values run from −27.9 to −33.5, depending on the magnesium and the salt in the solution. Across that whole range the joined step stays between −14.1 and −19.7 kJ per mole. The verdict never changes.
The +13.8 figure is the weakest number on the page. The usual textbook value is 3.3 kcal per mole. The kilocalorie is an older unit of energy, and one kilocalorie is 4.184 kJ. So 3.3 kcal comes to 13.8 kJ. The one source this course could actually reach rounds it to about 3 kcal, which would give 12.6 kJ instead. The joined step would then be −17.9 kJ per mole. Still downhill. Treat the last digit as soft.
ATP comes in units. There is no half an ATP. A larger uphill job needs a whole number of them, joined to it. Metal extraction uses the same trick, and a later module takes it properly.
Quick check — a step is uphill by 50 kJ per mole. One ATP pays out about 30.5 kJ per mole. What is the smallest number of ATP that must be linked to it?
— An uphill step can be paid for, but only by a reaction it actually touches.
ΔG has priced a cold pack, a rusting nail, a melting point and a sugar in a cell. Every verdict came out right. So it is time to say where it stops. Here is a test you can run on something you have seen. A diamond in a ring is carbon. Graphite in a pencil is carbon. They are the same element in two arrangements. In diamond every carbon is bonded to four others, in a rigid frame. At room temperature graphite is the more stable of the two. Diamond sits 2.9 kJ per mole above it. So ΔG for diamond turning into graphite is negative. And yet there are diamond rings a hundred years old.
Graphite is the more stable form of carbon at room temperature. Diamond sits 2.9 kJ per mole above it, so ΔG for diamond turning into graphite is negative. Commit — why is a hundred-year-old ring still a diamond?
Commit before you look. First answers are counted anonymously, never named.
ΔG looks at the start and the end. It never looks at the road between them. Module 12's ledger was path-blind, and the route never entered its arithmetic. ΔG inherits that blindness and adds nothing to it. For diamond the road is blocked at room temperature. Every carbon is gripped by four neighbours. The whole neighbourhood has to move at once. Nothing in a warm room arranges that.
So the rate is not zero. It is unimaginably small, and a hundred years of it leaves nothing you could measure. ΔG cannot see any of this. It compares diamond with graphite, finds graphite lower, and reports downhill. That report is correct. Downhill is not the same as moving. Here is the rule, plainly. ΔG tells you which way is downhill. It tells you nothing about how long. The next module is the one that times it.
A balcony in Chennai, in warm wet sea air. An iron grille on it rusts within a few years. A gold bangle on the same balcony does not change in a lifetime. The referee can price the iron.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
Take it one mole of rust at a time. ΔG is −742.2 kJ. That is enormously downhill. Iron pays a large freedom bill on the way. The reaction swallows oxygen gas out of the air. A gas locked into a solid is a large loss of freedom. The measured figure is ΔS = −274.9 J/K per mole of rust. The energy payout covers it many times over.
Gold forming its oxide swallows oxygen gas too. So its freedom term says no, for the same reason. The difference is the energy term. Forming gold(III) oxide from the elements gives out only about 13 kJ per mole. The measured figure is −13.0 ± 2.4 kJ per mole, and it was measured once, in 1972. Iron's figure is 824. Gold's energy term is very nearly silent. One term says no and the other barely speaks. So the pair says no.
That argument is structural, not numerical. No standard Gibbs energy of formation for gold(III) oxide exists in any source this course could reach. Neither does a standard entropy for it. So this page puts no ΔG number on gold. A number would be better than an argument. The course will not invent one.
Gold(III) oxide is real. It can be made, and it falls apart when warmed to about 298 °C. That 298 °C is not a flip temperature. The oxide already sits above gold and oxygen at room temperature, and it survives anyway. That is this section's own lesson, met a second time.
One correction before the check. People say gold is unreactive because its oxide is very high in energy. It is not. The oxide sits about 13 kJ per mole below the elements.
Quick check — forming iron oxide from the elements pays out 824 kJ per mole. Forming gold oxide pays out about 13. Both swallow oxygen gas from the air. Which term stops gold?
Where this section's numbers come from. Iron, iron(III) oxide, oxygen, carbon and diamond are all in the standard thermodynamic appendices. This page used the ones in OpenStax Chemistry 2e and in the NIST Chemistry WebBook. The gold(III) oxide enthalpy is from a 1972 paper in the Journal of the Chemical Society, Faraday Transactions. It reports −13.0 ± 2.4 kJ/mol. No source reachable from here carries a Gibbs energy for gold(III) oxide, or a standard entropy for it. That is why the gold argument above carries no ΔG.
Module 13 said that NCERT folds the count and the verdict into one chapter. This course split them. This module is where the two paths rejoin. You now hold both halves, so the chapter should look familiar when you meet it.
— ΔG says which way is downhill. It never says how long the walk takes.
These check themselves. “New numbers” deals a fresh set each time, so there is nothing to memorise. Each stem says how exact to be.
1. Module 12's cycle, revisited: (To ±0.1 kJ.)
2. The new line: Find the entropy change of the surroundings. (To ±1 J/K.)
3. The verdict: Find ΔG, and watch the units, since ΔS is in joules and ΔH is in kilojoules. (To ±0.2 kJ/mol.)
4. The flip: Find the temperature at which the verdict changes. (To ±2 K.)
Lime is made by heating limestone until it gives up carbon dioxide. This is done everywhere, and has been for thousands of years.
CaCO3(s) → CaO(s) + CO2(g)
Below are the tabulated numbers you need. A standard formation enthalpy is the heat taken in when one mole of a substance is made from its elements. A standard entropy is the arrangement count of one mole of it, measured from near absolute zero upwards. The small ° on ΔH° and ΔS° means the figure is for the standard state.
The standard formation enthalpies, in kJ/mol, are −1207.6 for CaCO3, −634.9 for CaO and −393.5 for CO2. The standard entropies, in J/K per mole, are 91.7 for CaCO3, 38.1 for CaO and 213.8 for CO2. There are three parts.
Part 1. Find ΔH° and ΔS° for the reaction.
Part 2. Find the temperature at which the verdict flips. Give it in kelvin.
Part 3. Real lime kilns fire at 1000–1200 °C. That is far above the temperature you just found. Explain the gap. There is more than one reason, and one of them lies outside this module.
Part 1. ΔH° is the products minus the reactant. The two products give −634.9 plus −393.5, which is −1028.4. Now take away the reactant's value, and the reactant's value is −1207.6. Taking away a negative number means adding it. So the sum is −1028.4 plus 1207.6, and ΔH° = +179.2 kJ. The reaction is uphill in energy.
ΔS° goes the same way, and here the reactant's value is positive, so this time you really do subtract. The two products give 38.1 plus 213.8, which is 251.9. Take away 91.7. ΔS° = +160.2 J/K. A gas is set free, so the freedom gained is large.
Part 2. The verdict flips where ΔH = TΔS, so the flip temperature is ΔH divided by ΔS. Put ΔH into joules first. 179 200 divided by 160.2 gives 1119 K. That is 845 °C. Below it the stone stays stone. Above it, lime is downhill.
Part 3. Two reasons, and the first is about the arithmetic itself. ΔH° and ΔS° were measured at 25 °C. The flip temperature was then worked out as though both stayed the same all the way to 845 °C. They do not. Both drift slowly as a substance warms. So a flip temperature computed from room-temperature numbers, more than 800 degrees away, is an estimate and not a reading.
The second reason is speed. At 845 °C the two terms draw level and ΔG is zero. There is no push at all. A kiln needs margin above the crossing so the reaction runs at a useful rate. How much margin, and why the rate depends on it, is the next module's business.
A balcony in Chennai. An iron grille on it rusts. A gold bangle beside it does not. Same air, same years. Price the iron as far as the numbers go. Give both terms and the verdict, and say which term is doing the work. Then make the gold case from the two terms alone, without a ΔG value. Finally, say which of your two arguments is weaker, and name the single measurement that would settle it.
The iron case runs on numbers all the way. ΔG is −742.2 kJ per mole of rust. ΔS is −274.9 J/K per mole, a large loss because a gas is swallowed. The energy payout is far larger than the freedom loss it has to cover. Energy wins, and the verdict is strong.
The gold case has only one number, the enthalpy of about 13 kJ per mole. That is the energy term, and it is nearly silent. The freedom term is a loss, for the same reason as iron. A gas is swallowed. Nearly silent, plus a loss, comes to a loss. That is the whole argument. Notice what it is not. It is not a ΔG. No standard Gibbs energy of formation for gold(III) oxide was reachable. And the freedom loss has been argued by analogy with iron, not measured for gold.
So the gold argument is the weaker one. The measurement that would settle it is the standard entropy of gold(III) oxide. With that and the enthalpy, ΔG follows in one line.
You now have the referee with both whistles. Energy in one hand, freedom in the other, and one verdict for any change at any temperature. Two of the hook's three examples are settled. Rust goes because it is enormously downhill. Ice melts because the temperature crossed the line. Rot is the same kind of case. The diamond is the one this module cannot settle. Turning into graphite is downhill for a diamond, and no diamond has moved. What the referee cannot say is how long. The next module times the road.
Two terms set the direction. One is the heat moved. One is the freedom changed.