Topic 9.6 key terms
- three factors of dissolving
- Breaking the attractions holding the solid together (Δ H > 0, Δ S > 0), reorganizing the solvent around the ions (Δ H > 0, Δ S < 0), and forming solute–solvent (ion–dipole) attractions (Δ H < 0). They nearly cancel.
- hydration shell
- Water molecules gathered around a dissolved ion: oxygen ends toward cations, hydrogen ends toward anions. The shell keeps exchanging members.
Transcript
789 words · 2 parts
Free energy of dissolution: the cold pack that gets colder
Remember the lantern? The park calls its light a cool glow, with hardly any heat, because the energy leaves as light instead, so a favored reaction doesn’t have to make things hot, and some favored processes even get colder as they run.
Say you had an instant cold pack in your first-aid kit. In Unit 6 we called it endothermic, because it soaks up heat, and the question we skipped is why it goes at all. When you squeeze it, a pouch of water breaks inside and mixes with solid ammonium nitrate, and within seconds the whole pack starts getting cold.
Let’s go inside, and look at the water first. Every molecule is turning and jostling, and those flickering threads are hydrogen bonds, forming and breaking all the time. Here’s the ammonium nitrate crystal, ammonium and nitrate ions packed in a lattice. Each ion is a group of atoms held by covalent bonds, and those stay whole; only the attraction between the ions changes.
Watch this ammonium ion at the corner. Water molecules turn their oxygen ends, the partly negative ends, toward it, tug it off the crystal, and carry it away in a shell that keeps swapping members. And as more and more ions leave, the solution gets colder, because pulling the crystal apart costs more energy than the water gives back.
So let’s put numbers on it. For ammonium nitrate dissolving, delta H is plus twenty-six point two kilojoules per mole, which is why the pack gets cold, and delta S is plus one hundred eight joules per mole per kelvin. Let’s find delta G naught at 298 kelvin. First, joules to kilojoules, so one hundred eight becomes zero point one zero eight.
Now you finish it. Delta G naught is twenty-six point two minus two ninety-eight times zero point one zero eight, so what is it, in kilojoules per mole?
Negative, so dissolving is favored even though it’s uphill in heat. And from the first part of the lesson, K is e to the six thousand over RT, about eleven, so the dissolved ions are favored. That’s the whole cold pack in one line: the enthalpy goes the wrong way, and the entropy pays the bill the enthalpy can’t.
Three pulls inside a dissolving salt
But why are both of those positive? Chemists split dissolving into three pieces, and each one pushes delta H and delta S its own way. First, the ions come out of the crystal. Pulling apart ions that attract each other costs energy, so delta H goes up, and free ions have many more places to be, so delta S goes up too.
Second, the water makes room. Breaking some hydrogen bonds between water molecules costs energy, and the water that gathers around each ion has to stay pointing one way, with fewer ways to arrange itself, so delta S goes down. Third, the ions and the water attract each other, and forming those attractions releases energy, so this piece pushes delta H down.
For ammonium nitrate, the enthalpy costs win by a little, plus twenty-six, and the ions’ freedom wins the entropy by a lot, plus one hundred eight. Here’s why solubility is hard to predict. This is calcium chloride, the salt in the hot pack from Unit 6. Its delta H is about minus eighty-one, so it gets hot, and its delta S is about minus forty-five, so the entropy actually goes down.
The small calcium ion carries a two-plus charge, so it grips its water so tightly that the water loses more freedom than the ions gain. Its delta G naught is still about minus sixty-eight, so it dissolves easily, but this time the heat wins. And this is silver chloride. Its delta H is plus sixty-five point four, a much bigger cost, and its entropy gain is only about plus thirty-three, so delta G naught comes out plus fifty-five point six.
That makes K one point eight times ten to the minus ten, right in line with the measured Ksp of silver chloride. So only a tiny amount dissolves before ions are landing back on the crystal as fast as they leave it, and that balance is what a saturated solution looks like up close.
Put side by side, these three salts have three different winners. The pieces are big and nearly cancel, so what’s left over can go either way, and you can’t tell from a picture whether a salt dissolves; you need the numbers.
Picture two drawings of a salt dissolving, one with the ions spreading out and one with water turning to face the ions, and no numbers. Could drawings like that prove that delta S for dissolving is positive? Which of these descriptions of ammonium nitrate dissolving can’t be right?