Biochemistry More on Thermodynamics & Free Energy

Hello and welcome back to Educator.com, and welcome back to Biochemistry.0000

In the last lesson, we started our discussion of thermodynamics; and today, we are going to finish off our discussion of thermodynamics, before we move on to another aspect of bioenergetics, which is going to be adenosine triphosphate.0003

OK, let's go ahead and continue on where we left off.0016

In the previous lesson, we ended with the following equation.0020

We said that the free energy change for a reaction is related to the equilibrium constant by this equation -RT, so ΔG = -RT ln K_eq.0023

Now, this is under standard conditions.0040

Let me go ahead and put that there.0045

This is under standard conditions, and when I say standard conditions at this point, I am talking about the transformed biological standard because now, we are no longer dealing with chemistry standards.0047

We are dealing with biochemistry standards, so everything is pretty much the same except the pH is now 7, instead of 0.0061

Excuse me.0073

However, but most - well, not most, all, I will just write most - most physiological conditions, most physiological circumstances, do not have standard reactant concentrations.0076

Well, you know what, I am not going to write that; let me just say, let me just start again here, so we erase this.0112

OK, but under physiological conditions, under physio conditions, things are not standard.0126

In other words, your concentrations are not all the same - 1 molarity or 1 atmosphere.0144

You have different concentrations of different things, so you might have 3M adenosine triphosphate, and you might have 0.1M adenosine diphosphate.0151

When we talk about a standard, again, this is just so we have some point of reference and frame of reference by which to measure something.0162

But again, physiological conditions, concentrations are not going to be 1M, and they are certainly not going to be the same for all species that are involved.0170

OK, the actual free energy change, the actual free energy ΔG is calculated as follows.0178

We have a way of finding the actual free energy under a given set of physiological circumstances based on what we have calculated under standard conditions, deviation from that.0197

So, it is calculated as follows, and this is the important equation that we want you to know.0211

ΔG rxn, reaction, but you know what, I will just put ΔG equals the ΔG standard + RT ln Q.0220

Now, notice this is RT ln Q, not K_eq because now, we are talking about a situation that is not at equilibrium.0234

The equation up here, this expresses a relationship between the equilibrium constant and the standard free energy change.0243

This says that these are just different ways of looking at the same thing, one is just a constant times the other.0250

What this equation tells us is that under a given set of circumstances, what is my temperature, what are my concentrations, given my base of a standard free energy change.0256

What is the free energy change under that circumstance; in other words, which way will the reaction actually go based on how things really are at that moment.0268

This is the equation that we want you to know; this is the one that is important.0277

Now, Q is exactly the same expression as K_eq, except it has concentrations not at equilibrium; but it has concentrations at any given moment - now, 3 hours from now, 30 seconds from now - at the particular moment.0281

Q, the expression is still the same as the K_eq, so it is going to be C raised to the C power, D raised to the D power, A raised to the A power and B raised to the B power, but it is at any given time, not at equilibrium.0298

The K_eq is a measure of the concentrations at equilibrium.0307

It is actually characteristic for that reaction under a given set of circumstances, at a given temperature, but here, Q, at any given moment.0324

That is how we calculate the free energy for any given set of circumstances.0333

OK, now, when ΔG ends up being less than 0, this means the reaction is spontaneous.0340

In other words, it will proceed to the right as written until it reaches equilibrium.0356

That is what is interesting about this.0362

Remember what we said, the free energy change is a measure of the potential to want to reach equilibrium.0378

Once a reaction actually starts, as the reaction moves forward and moves towards equilibrium, the ΔG is actually going to rise, so -50, -40, -30, -20, -10, -5, 4, 3, 2, 1, 0.0383

When the system reaches equilibrium, now, the system is exactly where it wants to be.0402

It does not have a tendency to go this way or that way.0406

In other words, it has no more free energy to release, so now, the ΔG is 0.0410

So, as a reaction moves forward, ΔG changes and moves towards 0 until it reaches equilibrium.0415

Now, at equilibrium, the ΔG actually equals 0.0425

This is very, very important.0435

OK, now, at equilibrium, ΔG = 0; this means all of the free energy available to do work has been used up- that is it.0440

It is done; there is no more, once the system reaches equilibrium.0471

OK, this is the take-home lesson.0481

The measure of whether a given reaction under given conditions will move forward spontaneously as written is ΔG, not ΔG standard.0489

Under a given set of conditions, it is ΔG that we calculate from this equation right here, that tells us whether a certain reaction will move forward.0538

The ΔG standard is a point of reference; things deviate from that.0549

Sometimes, it will be more spontaneous than the standard conditions; sometimes it will be less spontaneous than the standard conditions.0554

It might switch altogether depending on what the conditions are; that is what is important here.0560

OK, now, let's rewrite the equation, ΔG = ΔG standard + RT ln Q.0565

Well, we said that at equilibrium, the free energy is equal to 0.0580

At equilibrium, that means this left side is equal to 0, that means ΔG + RT ln Q.0586

Well, when I move this over, what I end up getting is my original equation that I had from the last lesson, equals -RT, except now, because we are at equilibrium, instead of Q, we write K_eq.0597

So, at equilibrium, you recover the relationship, and again, this relationship is a relationship between standard free energy change and the equilibrium constant; but at any given particular set of circumstances, it is ΔG that we are interested in.0613

That is what is going to tell us whether the reaction will move forward or will not move forward, and how much free energy we have at our disposal.0626

OK, now, ΔG is a measure of the maximum theoretical amount of free energy available.0635

The actual amount is much less.0675

ΔG is a measure of the maximum theoretical amount of free energy available.0679

So, let's say we have a particular reaction under a given set of conditions, and we calculated that the ΔG for this reaction is -50kJ/mol.0683

Well, that means that I have 50kJ/mol that I can do something with, whether it is build a macromolecule, or run some molecular motor or transport some solute across a membrane.0692

Whatever it is that the body needs to do it needs energy for, it has this 50kJ available to it, however, that is a theoretical number.0708

It is not going to use all of those 50kJ; the actual amount that ends up actually being used for useful work is actually a lot less than that, probably somewhere in the neighborhood of about 30 - 40% of that.0717

The rest of it is actually given off as heat.0727

OK, the actual amount is always less because some - actually most, I should say - free energy is always lost as heat entropy.0731

I have $50, but in order for me to actually go to a shop where I need to spend $50, I have to give $30 to the person at the door, so that leaves me with $20 to spend- that is what is going on.0772

I have to spend that $30 in order that I can do a little bit of shopping, if I want to do some shopping- that is all that is happening.0784

OK, let's do an example problem here; let's go ahead and do this in blue.0792

Oops, let's make sure we get it in blue here, so example problem no. 1.0797

OK, let's consider again, our glucose 1-phosphate reaction to glucose 6-phosphate and our ΔG standard for that, -7.3kJ/mol.0809

That is actually a little bit more than that; I must have gotten this number from a different...OK, so, -7.3kJ/mol.0828

Now, if the concentration of G1P equals 0.3M, and the concentration of G6P equals, let's say, 1.7M, OK, is this reaction spontaneous as written at 25°C.0835

OK, well, we have 0.3M; we have 1.7M, so these are not standard conditions.0875

The concentrations are different, but we have an equation that deals with this.0880

We have the ΔG under a given set of conditions equals the ΔG under standard conditions plus RT ln and Q.0885

Now, we just go ahead and put these in.0895

And, of course, Q is going to be the products, the G6P concentration over the G1P concentration, which is the reactants.0899

And again, we have to account for the units, so we have got, well, the ΔG standard is -7.3kJ/mol, so let's use -7300J/mol + 8.315 is our T is 298, and we have the logarithm of Q, which is going to be 1.7/0.3.0911

Concentration of G6-phosphate is 1.7; concentration of G1-phosphate is 0.3, and when we run this calculation, we end up with -7300 + 4298 for a total of -3002J/mol.0946

OK, now, the ΔG is -3002J/mol, which is -3.0kJ/mol.0978

Well, it is still spontaneous; it is negative, and it is -3kJ/mol.0991

But notice, under standard conditions, it is more than twice as spontaneous.0996

That is -7.3kJ/mol, so under these circumstances, I only have about 3000J of free energy available; and let's say we only end up using about 50% of that, so I am really going to only end up with about 1500J that is going to actually do useful work.1001

In this particular case, if the conditions were standard, I have 7.3kJ/mol, so I would have a lot more available.1021

I would have about 3.5, 3.6kJ actually available for real useful work.1029

Under these conditions, it is the ΔG that is going to tell us how much free energy we actually have available, if the reaction is still spontaneous.1037

Obviously, if these numbers change, we can actually change the spontaneity.1046

We can make it so that it actually does not want to move forward under standard conditions.1051

Perhaps, it wants to move backward depending on what these numbers are.1056

OK, now, let's see what we can do; now, let's leave it as blue.1062

Individual reactions that have species in common can be added, and common species can be cancelled on either side of the arrows to yield a net equation.1072

Do you remember back in general chemistry when you guys did Hess's law?1128

You wanted to find the enthalpy of a particular reaction.1132

Well, you knew that if you had the enthalpy of like 2 or 3 or 4 reactions, that when you added those reactions together, and manipulated them in such a way, so that when you actually added them straight down and species cancelled, if you ended up with your net reaction, what you can do is just add the enthalpies; and that will give you the enthalpy of the reaction that you want.1135

This is the same thing we are doing here; we can do the same for free energy, the same thing that we do for enthalpy.1154

We can add equations, cancel species, and the final equation that we get is perhaps the equation that we are actually interested in- that is what is going on here.1161

Now, the actual free energy of the net equation equals to the free energy of the first equation plus the free energy of the second equation, so the free energies are additive.1171

However, the equilibrium constant of the net equation is equal to the equilibrium constant times the second equilibrium constant.1185

So, the equilibrium constant of the first equation times the equilibrium constant of the second.1199

K_eq is multiplicative; the free energy is additive- very, very important.1203

Let's do an example here.1210

OK, now, the first step of glycolysis, which we will be getting to very soon, which is the breakdown of glucose, is glucose plus an inorganic phosphate.1216

Well, I will just write it as a single arrow - goes to glucose 6-phosphate + H₂O, and the ΔG for this is equal to +13.8kJ/mol.1241

This is positive; it is a positive ΔG.1259

What this tell me is that this reaction does not want to go this way under normal circumstances.1263

In fact, it wants to go the other way under normal circumstances because this is positive.1267

It is spontaneous in the reverse direction.1272

This reaction is not thermodynamically favorable; let's write it in red.1276

This reaction is not thermodynamically favorable, and if the reaction takes place all the time, so how does it proceed?1283

Very, very interesting, this is very, very profound- what we are about to do here.1313

Well, let's look at another reaction; I will do this in blue.1321