Actual chemistry lecture this time!  The newcomers should probably at least try to attend; it's not relying on math from Keltham's earlier lectures.

This is the Periodic Table of Elements.  It goes by the number of electrons in the outermost electron orbital.

Keltham will spend some real time, this time, describing the electron orbitals, and how covalent bonds allow shared electrons to fill more of their shells; and how this lower-potential-energy state, where more electrons are closer in to stronger attractive charges of protons, makes molecules tightly bound together in a way that takes heat to pry apart.

Even when you could rearrange all the atoms in a reaction into a lower-energy state at the end, releasing heat along the way, you still need a high temperature so that the vibrations of heat will break apart some of the original molecules and let them randomly recombine, sometimes into the lower-energy state.  Ideally, you pick a reaction temperature that easily breaks apart the feed molecules, but doesn't so easily break apart the more tightly bound final forms.

The reaction they're trying for with sulfuric acid is:

Stage 1: burn sulfur to sulfur dioxide, this is exothermic but requires a high starting temperature

S + O2 => SO2

Stage 1.5: purify the sulfur dioxide?? Keltham doesn't actually remember how to do this, unfortunately

Stage 2: catalyze sulfur dioxide to sulfur trioxide, in the presence of oxygen, using Element-23 oxide as a catalyst

SO2 + V2O5 => SO3 + V2O4 (probably)
maybe SO2 + V2O4 => SO3 + V2O3 and then V2O3 + O2 = V2O5??

Stage 3: cool and dissolve the sulfur trioxide in sulfuric acid to form oleum

H2SO4 + SO3 => H2S2O7

Stage 4: react oleum with water to form concentrated sulfuric acid, twice as much as previously existed

H2S2O7 + H2O => 2 H2SO4

He's not really sure about the details, in the case of the sulfuric-acid synthesis pathway, but the reason the vanadium catalyst might help in the first place - is if it's relatively hard to tear apart an O2 molecule, for example, and the exoergic reaction of a single SO2 to SO3 isn't enough pull to tear apart a single O2 molecule - it could happen with two SO2 molecules at once, cooperating to tear apart an O2 molecule that's already heated and vibrating, but then it's less likely for all those molecules to be in the right positions.  But V2O5 could resist the tug more weakly, giving up an O molecule to a demanding SO2 molecule; and then giving up another O molecule and turning into V2O3; and then swallowing down a whole O2 molecule to go back to V2O5 again.  Possibly.  For example.

The main point of this process, as Keltham understands it, is that it's liable to produce relatively high-purity sulfuric acid at scale.

It will also - he suspects - be much easier to do, much much easier to do reliably, if they can learn to really use Prestidigitation.

Actual chemistry lecture this time!  The newcomers should probably at least try to attend; it's not relying on math from Keltham's earlier lectures.

This is the Periodic Table of Elements.  It goes by the number of electrons in the outermost electron orbital.

Keltham will spend much longer, this time, describing the electron orbitals; and how covalent bonds allow shared electrons to fill more of their shells; and how this lower-potential-energy state, where more electrons are closer in to stronger attractive charges of protons, makes molecules tightly bound together in a way that takes heat to pry apart.

Even when you could rearrange all the atoms in a reaction into a lower-energy state at the end, releasing heat along the way, you still need a high temperature so that the vibrations of heat will break apart some of the original molecules and let them randomly recombine, sometimes into the lower-energy state.  Ideally, you pick a reaction temperature that easily breaks apart the feed molecules, but doesn't so easily break apart the more tightly bound final forms.

The reaction they're trying for with sulfuric acid is:

Stage 1: burn sulfur to sulfur dioxide, this is exothermic but requires a high starting temperature

S + O2 => SO2

Stage 1.5: purify the sulfur dioxide?? Keltham doesn't actually remember how to do this, unfortunately

Stage 2: catalyze sulfur dioxide to sulfur trioxide, in the presence of oxygen, using Element-23 oxide as a catalyst

SO2 + V2O5 => SO3 + V2O4 (probably)
maybe SO2 + V2O4 => SO3 + V2O3 and then V2O3 + O2 = V2O5??

Stage 3: cool and dissolve the sulfur trioxide in sulfuric acid to form oleum

H2SO4 + SO3 => H2S2O7

Stage 4: react oleum with water to form concentrated sulfuric acid, twice as much as previously existed

H2S2O7 + H2O => 2 H2SO4

He's not really sure about the details, in the case of the sulfuric-acid synthesis pathway, but the reason the vanadium catalyst might help in the first place - is if it's relatively hard to tear apart an O2 molecule, for example, and the exoergic reaction of a single SO2 to SO3 isn't enough pull to tear apart a single O2 molecule - it could happen with two SO2 molecules at once, cooperating to tear apart an O2 molecule that's already heated and vibrating, but then it's less likely for all those molecules to be in the right positions.  But V2O5 could resist the tug more weakly, giving up an O molecule to a demanding SO2 molecule; and then giving up another O molecule and turning into V2O3; and then swallowing down a whole O2 molecule to go back to V2O5 again.  Possibly.  For example.

The main point of this process, as Keltham understands it, is that it's liable to produce relatively high-purity sulfuric acid at scale.

It will also - he suspects - be much easier to do, much much easier to do reliably, if they can learn to really use Prestidigitation.

Actual chemistry lecture this time!  The newcomers should probably at least try to attend; it's not relying on math from Keltham's earlier lectures.

This is the Periodic Table of Elements.  It's ordered by protons and organized by the behavior of electrons in the outermost electron orbital.

Keltham will spend much longer, this time, describing the electron orbitals; and how covalent bonds allow shared electrons to fill more of their shells; and how this lower-potential-energy state, where more electrons are closer in to stronger attractive charges of protons, makes molecules tightly bound together in a way that takes heat to pry apart.

Even when you could rearrange all the atoms in a reaction into a lower-energy state at the end, releasing heat along the way, you still need a high temperature so that the vibrations of heat will break apart some of the original molecules and let them randomly recombine, sometimes into the lower-energy state.  Ideally, you pick a reaction temperature that easily breaks apart the feed molecules, but doesn't so easily break apart the more tightly bound final forms.

The reaction they're trying for with sulfuric acid is:

Stage 1: burn sulfur to sulfur dioxide, this is exothermic but requires a high starting temperature

S + O2 => SO2

Stage 1.5: purify the sulfur dioxide?? Keltham doesn't actually remember how to do this, unfortunately

Stage 2: catalyze sulfur dioxide to sulfur trioxide, in the presence of oxygen, using Element-23 oxide as a catalyst

SO2 + V2O5 => SO3 + V2O4 (probably)
maybe SO2 + V2O4 => SO3 + V2O3 and then V2O3 + O2 = V2O5??

Stage 3: cool and dissolve the sulfur trioxide in sulfuric acid to form oleum

H2SO4 + SO3 => H2S2O7

Stage 4: react oleum with water to form concentrated sulfuric acid, twice as much as previously existed

H2S2O7 + H2O => 2 H2SO4

He's not really sure about the details, in the case of the sulfuric-acid synthesis pathway, but the reason the vanadium catalyst might help in the first place - is if it's relatively hard to tear apart an O2 molecule, for example, and the exoergic reaction of a single SO2 to SO3 isn't enough pull to tear apart a single O2 molecule - it could happen with two SO2 molecules at once, cooperating to tear apart an O2 molecule that's already heated and vibrating, but then it's less likely for all those molecules to be in the right positions.  But V2O5 could resist the tug more weakly, giving up an O molecule to a demanding SO2 molecule; and then giving up another O molecule and turning into V2O3; and then swallowing down a whole O2 molecule to go back to V2O5 again.  Possibly.  For example.

The main point of this process, as Keltham understands it, is that it's liable to produce relatively high-purity sulfuric acid at scale.

It will also - he suspects - be much easier to do, much much easier to do reliably, if they can learn to really use Prestidigitation.

Actual chemistry lecture this morning!  The newcomers should probably at least try to attend; it's not relying on math from Keltham's earlier lectures.

This is the Periodic Table of Elements.  It's ordered by protons and organized by the behavior of electrons in the outermost electron orbital.

Keltham will spend much longer, this time, describing the electron orbitals; and how covalent bonds allow shared electrons to fill more of their shells; and how this lower-potential-energy state, where more electrons are closer in to stronger attractive charges of protons, makes molecules tightly bound together in a way that takes heat to pry apart.

Even when you could rearrange all the atoms in a reaction into a lower-energy state at the end, releasing heat along the way, you still need a high temperature so that the vibrations of heat will break apart some of the original molecules and let them randomly recombine, sometimes into the lower-energy state.  Ideally, you pick a reaction temperature that easily breaks apart the feed molecules, but doesn't so easily break apart the more tightly bound final forms.

The reaction they're trying for with sulfuric acid is:

Stage 1: burn sulfur to sulfur dioxide, this is exothermic but requires a high starting temperature

S + O2 => SO2

Stage 1.5: purify the sulfur dioxide?? Keltham doesn't actually remember how to do this, unfortunately

Stage 2: catalyze sulfur dioxide to sulfur trioxide, in the presence of oxygen, using Element-23 oxide as a catalyst

SO2 + V2O5 => SO3 + V2O4 (probably)
maybe SO2 + V2O4 => SO3 + V2O3 and then V2O3 + O2 = V2O5??

Stage 3: cool and dissolve the sulfur trioxide in sulfuric acid to form oleum

H2SO4 + SO3 => H2S2O7

Stage 4: react oleum with water to form concentrated sulfuric acid, twice as much as previously existed

H2S2O7 + H2O => 2 H2SO4

He's not really sure about the details, in the case of the sulfuric-acid synthesis pathway, but the reason the vanadium catalyst might help in the first place - is if it's relatively hard to tear apart an O2 molecule, for example, and the exoergic reaction of a single SO2 to SO3 isn't enough pull to tear apart a single O2 molecule - it could happen with two SO2 molecules at once, cooperating to tear apart an O2 molecule that's already heated and vibrating, but then it's less likely for all those molecules to be in the right positions.  But V2O5 could resist the tug more weakly, giving up an O molecule to a demanding SO2 molecule; and then giving up another O molecule and turning into V2O3; and then swallowing down a whole O2 molecule to go back to V2O5 again.  Possibly.  For example.

The main point of this process, as Keltham understands it, is that it's liable to produce relatively high-purity sulfuric acid at scale.

It will also - he suspects - be much easier to do, much much easier to do reliably, if they can learn to really use Prestidigitation.

Actual chemistry lecture this morning!  The newcomers should probably at least try to attend; it's not relying on math from Keltham's earlier lectures.

This is the Periodic Table of Elements.  It's ordered by protons, and organized by the arrangement of electrons as they fill up the least energetic orbital positions.

Keltham will spend much longer, this time, describing the electron orbitals; and how covalent bonds allow shared electrons to fill more of their shells; and how this lower-potential-energy state, where more electrons are closer to stronger attractive charges of protons, makes molecules tightly bound together in a way that takes heat to pry apart.

Even when you could-in-principle rearrange all the atoms in a reaction into a lower-energy state at the end, releasing heat along the way, to get there, you still need a high temperature, so that the vibrations of heat will break apart some of the original molecules and let them randomly recombine, sometimes into the lower-energy state.  Ideally, you pick a reaction temperature that easily breaks apart the feed molecules, but doesn't so easily break apart the more tightly bound final forms.

The reaction they're trying for with sulfuric acid is:

Stage 1: burn sulfur to sulfur dioxide, this is exothermic but requires a high starting temperature

S + O2 => SO2

Stage 1.5: purify the sulfur dioxide?? Keltham doesn't actually remember how to do this, unfortunately

Stage 2: catalyze sulfur dioxide to sulfur trioxide, in the presence of oxygen, using Element-23 oxide as a catalyst

SO2 + V2O5 => SO3 + V2O4 (probably)
maybe SO2 + V2O4 => SO3 + V2O3 and then V2O3 + O2 = V2O5??

Stage 3: cool and dissolve the sulfur trioxide in sulfuric acid to form oleum

H2SO4 + SO3 => H2S2O7

Stage 4: react oleum with water to form concentrated sulfuric acid, twice as much as previously existed

H2S2O7 + H2O => 2 H2SO4

He's not really sure about the details, in the case of the sulfuric-acid synthesis pathway, but the reason the vanadium catalyst might help in the first place - is if it's relatively hard to tear apart an O2 molecule, for example, and the exoergic reaction of a single SO2 to SO3 isn't enough pull to tear apart a single O2 molecule - it could happen with two SO2 molecules at once, cooperating to tear apart an O2 molecule that's already heated and vibrating, but then it's less likely for all those molecules to be in the right positions.  But V2O5 could resist the tug more weakly, giving up an O molecule to a demanding SO2 molecule; and then giving up another O molecule and turning into V2O3; and then swallowing down a whole O2 molecule to go back to V2O5 again.  Possibly.  For example.

The main point of this process, as Keltham understands it, is that it's liable to produce relatively high-purity sulfuric acid at scale.

It will also - he suspects - be much easier to do, much much easier to do reliably, if they can learn to really use Prestidigitation.

Actual chemistry lecture this morning!  The newcomers should probably at least try to attend; it's not relying on math from Keltham's earlier lectures.

This is the Periodic Table of Elements.  It's ordered by protons, and organized by the arrangement of electrons as they fill up the least energetic orbital positions.

Keltham will spend much longer, this time, describing the electron orbitals; and how covalent bonds allow shared electrons to fill more of their shells; and how this lower-potential-energy state, where more electrons are closer to stronger attractive charges of protons, makes molecules tightly bound together in a way that takes heat to pry apart.

When you could-in-principle rearrange all the atoms in a reaction into a lower-energy state at the end, releasing heat along the way, it implies a potential more stable state the system might get into, and stay in afterwards.  To get there, you need a high temperature, so that the vibrations of heat will break apart some of the original molecules and let them randomly recombine, sometimes into the lower-energy state.  Ideally, you pick a reaction temperature that easily breaks apart the feed molecules, but doesn't so easily break apart the more tightly bound final forms.

The reaction they're trying for with sulfuric acid is:

Stage 1: burn sulfur to sulfur dioxide, this is exothermic but requires a high starting temperature

S + O2 => SO2

Stage 1.5: purify the sulfur dioxide?? Keltham doesn't actually remember how to do this, unfortunately

Stage 2: catalyze sulfur dioxide to sulfur trioxide, in the presence of oxygen, using Element-23 oxide as a catalyst

SO2 + V2O5 => SO3 + V2O4 (probably)
maybe SO2 + V2O4 => SO3 + V2O3 and then V2O3 + O2 = V2O5??

Stage 3: cool and dissolve the sulfur trioxide in sulfuric acid to form oleum

H2SO4 + SO3 => H2S2O7

Stage 4: react oleum with water to form concentrated sulfuric acid, twice as much as previously existed

H2S2O7 + H2O => 2 H2SO4

He's not really sure about the details, in the case of the sulfuric-acid synthesis pathway.  But a reason the vanadium catalyst might help in the first place - is if it's relatively hard to tear apart an O2 molecule, for example, and the exoergic reaction of a single SO2 to SO3 isn't enough pull to tear apart a single O2 molecule - it could happen with two SO2 molecules at once, cooperating to tear apart an O2 molecule that's already heated and vibrating, but then it's less likely for all those molecules to be in the right positions.  But V2O5 could resist the tug more weakly, giving up an O molecule to a demanding SO2 molecule; and then giving up another O molecule and turning into V2O3; and then swallowing down a whole O2 molecule to go back to V2O5 again.  Possibly.  For example.

The main point of this process, as Keltham understands it, is that it's liable to produce relatively high-purity sulfuric acid at scale.

It will also - he suspects - be much easier to do, much much easier to do reliably, if they can learn to really use Prestidigitation.