Thermodynamics of Membrane Diffusion: Uncharged Molecule
11. Biological Membranes and Transport
Thermodynamics of Membrane Diffusion: Uncharged Molecule - Video Tutorials & Practice Problems
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concept
Thermodynamics of Membrane Diffusion: Uncharged Molecule
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Video transcript
In this video, we're going to begin our lesson on the thermodynamics of membrane diffusion, but specifically for uncharged molecules. Now later in our course in a different video, we'll talk about the thermodynamics of membrane diffusion for charged ions. But for now, in this video, we're only going to focus on uncharged molecules. And so if we take a look at our image down below over here on the left hand side we have a little reminder that in this video we're only focusing on uncharged molecules. And so this little orange molecule that you see right here is our uncharged molecule. And again, we're going to be looking at the thermodynamics of this uncharged molecule as it diffuses across the membrane from its initial side over here on the left across the membrane to its final side over here on the right. And so it's also really important to note that in this video we're really going to be building off a lot of the knowledge from our older lesson videos, specifically where we talked about the change in Gibbs free energy equation. And so that was quite a long time ago for some of you guys. And so if you do not recognize this equation right here, if it looks completely foreign to you guys, then make sure to go back and check out those older lesson videos before you continue on here with this video. Because again this video is just building off of those older lesson videos. And so recall from those older lesson videos that the change in Gibbs free energy equation, specifically the Gibbs free energy under any conditions, is this equation that you see right here, where delta g is equal to delta g naught or the change in Gibbs free energy under standard conditions, plus r, which is the gas constant, times t, which is the temperature in Kelvin, times the natural log, ln, of the reaction quotient q. And recall that the reaction quotient q is really just equal to the ratio of the concentration of products over the concentration of reactants at any point in our reaction. And really, this equation that you see right here from our older lesson videos is really describing the change in Gibbs free energy that's due to chemical gradients or chemical concentrations, but it is not due to electrical gradients. And so the electrical gradients are going to come into play when we talk about charged ions later in our course. But again for now we're looking at uncharged molecules and so this equation only describes the delta g due to chemical gradients, not due to electrical gradients. Now this equation right here in our previous lesson videos, we really only applied it to calculate the Gibbs free energy under any conditions for specific reactions. However, this equation can also be applied to membrane diffusion. And so when this equation is applied to membrane diffusion, the Delta G here is actually referred to as the Delta G transport. And so this transport notation right here is again to remind us that it's referring to membrane diffusion. And so delta G transport is really just defined as the Gibbs free energy change associated with membrane diffusion. And so what's important to note here is that when the reaction in quotes is simply just membrane transport or membrane diffusion, then the delta G naught or the change in Gibbs free energy under standard conditions is actually equal to a value of 0. And so the reason for this is because membrane transport actually isn't really a real reaction. It's and that's why we have reaction here in quotes. And the reason that membrane transport is not a real reaction is because really no bonds are created or formed, and the molecule is exactly the same when it's on the initial side, as it is on the final side. All it's really doing is just moving across the membrane. And so, if delta g naught is equal to a value of 0 whenever we're applying this equation to membrane diffusion, what this means is that if we're applying this equation to membrane diffusion, we can pretty much take the delta G naught right here and just scratch it out, just eliminate it completely since it's equal to a value of 0. And so, if you take a look at our equation right here for the delta G transport, that would be equal to this delta G up above right here. So delta G transport is this delta G up here on the left. And notice that without the delta g naught, since we've eliminated that part, it's just going to be equal to rtlnq. And so you can see our r t right here, and notice that the ln and, q, this part right here is replacing the q, And recall that q is equal to the ratio of concentration of products over concentration of reactants. But again, in just, a reaction that is just membrane transport really there are no products or reactants because again the molecule is exactly the same as it is on the left hand side as it is on the right. So the product really is the same exact molecule as the reactant. And so instead, instead the q here is actually just going to be the concentration of this molecule, the ratio of the concentration of this molecule, on the initial side, which would be like the reactant for instance. The initial is like the reactant. It's what you start with. And then, it's gonna be the concentration of the molecule on the final side, which would be symbolic of the product. And so when we take the q and we say, okay, well concentration of product is gonna be on the top and the concentration of reactant is gonna be on bottom. Well, in terms of membrane diffusion, it's gonna be the concentration of that substance on the final side which again would be symbolic of the product, over the concentration of the molecule on its initial side. And again the initial side is symbolic of the reactant, which we know is in the denominator, and that's why it's in the denominator here. And again, this portion of the equation that you see right here is again going to explain or describe chemical gradients, and so notice that it has this blue background right here to be linked to chemical gradients. And so if we want to calculate the thermodynamics of membrane diffusion for any uncharged molecule, all we need to do is apply this equation that you see right here. And, plug in all of the values and you'll get the delta g transport for that particular uncharged molecule. And so, over here notice that what we have is a reminder of what this gas constant r, the value is equal to, which is equal to 8.315 joules, over moles times Kelvin. And then also a reminder that the temperature here is going to be the temperature in degrees Kelvin, and so you can't forget to convert the units to units of Kelvin. And so in our next video, we're going to show you guys an example of how to utilize this equation in order to calculate the thermodynamics of membrane diffusion for an uncharged molecule. But for now this here concludes our introduction to the thermodynamics of membrane diffusion for uncharged molecules, and again we'll be able to see an example of how to apply this equation in our next video. So I'll see you guys there.
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example
Thermodynamics of Membrane Diffusion: Uncharged Molecule Example 1
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11m
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Video transcript
Alright. So here we have an example problem that wants us to calculate the Delta G transport for glucose diffusion across a cell membrane to the inside of a cell if the glucose concentration on the outside of the cell is equal to 10 millimolar, the glucose concentration in the cytosol or on the inside of the cell is equal to 0.1 millimolar, and the temperature is equal to 20 degrees Celsius. Now in order to solve a problem like this one up above, really it's just a 4 step process that's going to allow us to get our answer. And so you can see that we're going to walk you through each of these 4 steps 1 by 1 down below. And so step number 1 here is just to determine the net charge of the diffusing molecule. And so if we take a look back at our problem, we can see that the diffusing molecule is glucose. And of course we know from our previous lesson videos that glucose is an uncharged molecule, and so we can go ahead and check off this checkbox here for uncharged. Now again, later in our course in a different video, we're going to talk about the thermodynamics of membrane diffusion for charged molecules. And if this were a charged molecule, then we would have to check off this box and we would have to use a slightly different equation than the equation that we introduced in our last lesson video. But thankfully here, again, glucose is the diffusing molecule and glucose is an uncharged molecule. And of course, what this means is that we can use the equation that we introduced in our last lesson video in order to solve this problem. And so really that is it for step number 1. We have determined that the diffusing molecule is glucose, We know that its net charge is uncharged, and so now we know that we can use the equation from our last lesson video to solve this problem. So now we can move on to step number 2. And in step number 2, what we need to do is determine the direction of diffusion across the membrane. And really what this means is that we need to establish the initial side of the membrane and the final side of the membrane with respect to the outside of the cell and the inside of the cell. And really in a lot of cases, drawing a sketch may be really, really helpful. And so notice down below, we have a little sketch here. You can see that we have the cell's plasma membrane, and we know that glucose is the diffusing molecule, and so notice that we have a glucose transporter here embedded in the membrane. And so if we take a look back at our problem, notice that it wants us to calculate the delta g transport for glucose diffusion across a cell membrane to the inside of a cell. And so this means towards the inside of the cell. And so if we label the left side of our membrane over here as the outside of the cell and the right side of our membrane over here as the inside of the cell, of course again to the inside means it's going to go towards the inside of the cell. And so this is the direction of the glucose diffusion across the membrane. And so, of course what this means is that the initial side is over here on the left, which is the outside of the cell. So the outside of the cell is the initial side of the membrane, and of course this means that the inside of the cell on this end of the arrow is going to be the final side of the membrane. And so really that is it for step number 2. We have determined the direction of diffusion, established an initial and final side of the membrane. So now we can move on to step number 3, and in step number 3, really all we need to do is check all of the units on all of the numbers, and if necessary, we need to convert the units to different units in order to ensure compatibility with the equation that we introduced in our last lesson video. And so when we take a look at the numbers that are given to us notice that we have the concentration of glucose outside of the cell as 10 millimolar, then we have the glucose concentration ins inside of the cell in the cytosol as 0.1 millimolar, so the millimolar here match each other. And then we're given the temperature in degrees Celsius. And so, notice that in our equation, the temperature must be in degrees Kelvin, not in degrees Celsius. And so that is definitely something that we're going to need to convert before we move on. And so if we go ahead and do that, recall that in order to convert Celsius into Kelvin, the equation that we need to use is, take the degrees Celsius and add 273.15, and then we'll get the temperature in Kelvin. So, the temperature that we're given in degrees Celsius, again, if we go up to our problem, is 20 degrees Celsius. So if we plug that in here, we'll say 20 degrees Celsius plus 273.15 is equal to our temperature in Kelvin. And of course, this comes out to an answer of 293.15 Kelvin. And so really this is the temperature that we want to plug into our equation. Now because the glucose concentration here and here are matching each other, then we don't need to convert them into units of molarity. And that's because these two concentrations are actually going to cancel each other out when we plug them into this ratio here. So, it doesn't matter if we were to convert if we were to convert this to molarity, if we were to convert both of these to molarity, we would still end up getting the same exact answer. So it's okay to actually not convert these to molarity again because they're going into this ratio and the units end up canceling each other out. So now that we've got all of the numbers converted to the correct units, we can move on to step number 4. And in step number 4, this is the most straightforward part, and all we need to do is plug in all of the given values that we are given in the problem, and of course with the appropriate units, into the correct equation, which we've already established, and then algebraically solve for the missing variable. And so again, when we're plugging in the given values into the correct equation, the equation that we're using is the one from our last lesson video right here. And so we're just going to plug in our values. So we can go ahead and start over here on the left hand side. So delta g transport is going to be equal to r and recall that r is the gas constant which is equal to a value of 8 point 315 joules per mole time Kelvin. So we can plug that in here. We'll put this in parentheses, 8.315 joules over moles times Kelvin. So there's our r and then it's going to be multiplied by the temperature, of course, up above by the temperature. But, of course, the temperature needs to be in Kelvin, so we've already converted it to Kelvin. So here is our temperature in units of Kelvin, 293.15 Kelvin. And then, of course, this is all going to be multiplied by the natural log, we'll do this in red here, the natural log of the ratio of the concentration on the final side of the membrane over the concentration, of the initial side of the membrane. And so recall that the, final side of the membrane is actually the concentration on the inside of the cell. And so, notice that the concentration on the inside of the cell or in the cytosol is 0.1 millimolar. And so, we know that that's gonna be the final side so that needs to be on the top number because remember the final side goes on top. So now that we have that, we can put in our number 0.1 millimolar on top here. So 0.1 millimolar goes here and this is going to be over, again, the concentration on the initial side of our membrane. And recall that the initial side is the outside of the membrane, and, notice that the concentration on the outside is 10 millimolar. So that's what's going to go on bottom, is 10 millimolar. And so now all we need to do is take our calculators and, really just algebraically solve here. And so, we can go ahead and rewrite this. So d delta g transport is going to be equal to 8.315 times 293.15 is equal to a value of 2,437.54 approximately, and the units are going to be joules over moles, since the units of Kelvin are going to cancel out. And then the natural log of this answer right here, which we'll put in parentheses, is equal to a value of about negative 4 point 605. And again, the units of millimolar cancel out here, so we don't need to worry about the units here. And so now all we need to do is take these two numbers and multiply them. So 2,437.54 times negative 4.605 is going to give us our answer. And so delta g transport is going to be equal to, our final answer, which is negative 11,225 point 3. And again, the only units that are left are joules per mole. And so, this here is our final answer, negative 11,225.3 joules per mole is the delta g transport in the final answer for this example problem that we had up above. And so a couple things to note here is that because the value is negative, we know that it's going to be an exergonic process. And so, exergonic, of course, means that it's going to be spontaneous and it's not going to require an input of energy. And so the glucose will be able to get transported from the outside of the cell towards the inside of the cell under these particular conditions spontaneously without an input of energy. And so really all of the problems here in this topic are going to follow these same exact four steps that we have listed right here, And so, be sure to practice this a couple of times and then you guys will get this on your exams no problem. And so this here concludes our example problem. Again, this is our final answer down here, and I'll see you guys in our next video.
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Problem
Problem
Calculate ΔGtransport for the diffusion of glucose from outside to inside a cell if extracellular [Glucose] = 5 mM, intacellular [Glucose] = 0.5 mM, & the temperature = 20˚C. Would this be an exergonic or endergonic process?
A
–5.61 KJ/mol.
B
5.61 KJ/mol.
C
–0.383 KJ/mol.
D
0.383 KJ/mol.
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Problem
Problem
Calculate ΔGtransport for the diffusion of glucose from inside to outside a cell if extracellular [Glucose] = 1 M, intacellular [Glucose] = 2.0 mM, & the temperature = 20˚C. Would this be a spontaneous or nonspontaneous process?
A
–15.15 KJ/mol.
B
15.15 KJ/mol.
C
–1.03 KJ/mol.
D
1.03 KJ/mol.
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Problem
Problem
Calculate the energy cost (ΔGtransport) of pumping an uncharged solute across a cell's plasma membrane, against a 1.0 x 104-fold concentration gradient at 25˚C. Would this be an exergonic or endergonic process?