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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8m

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Yeah, 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 were on Lee 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 Onley 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 molecules 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 Jeannot, or the change in Gibbs Free Energy under standard conditions Plus our, which is the gas constant Times T, which is the temperature and Calvin Times, the natural log Ellen 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 reactant 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 Grady INTs or chemical concentrations. But it is not due to electrical Grady INTs And so the electrical Grady INTs 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 on Lee describes the Delta G due to chemical ingredients not due to electrical gravy INTs. 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 Jeannot or the change in Gibbs Free Energy understand erred conditions is actually equal to a value of zero. And so the reason for this is because membrane transport actually isn't really a riel 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. Um, as it is on the final side, all it's really doing is just moving across the membrane. And so if Delta G not is equal to a value of zero 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 not right here and just scratch it out. Just eliminate it completely, since it's equal to a value of zero. And so, if you take a look at our equation right here for the Delta G transport, uh, that would be equal to this Delta G up above right here. So Delta G transport is this Delta G appear on the left and notice that without the Delta G, not since we've eliminated that part, it's just going to be equal to r T l n Q. And so you can see our our tea right here and noticed that the l N and Q this part right here is replacing the queue and recall that Q is equal to the ratio of concentration of products over concentration of reactive. But again, and just a reaction that is just membrane transport. Really, there are no products or reactions, 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 molecules, the reacting. And so instead, the cue 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 reacting. 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 the Delta G due to chemical Grady INTs. And so notice that it has this blue background right here to be linked to chemical Grady INTs. And so, if we want to calculate the thermodynamics of membrane diffusion for any uncharged molecules, all we need to do is apply this equation that you see right here and, uh, 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 are. The value is equal to which is equal to 8.315 jewels over moles times Calvin, and then also a reminder that the temperature here is going to be the temperature and degrees Calvin. And so you can't forget to convert the unit to unit of Calvin. 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 year concludes our introduction to the thermodynamics of memory 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.

2

example

Thermodynamics of Membrane Diffusion: Uncharged Molecule Example 1

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11m

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all right. So here we have an example problem that wants us to calculate the Delta G transport for glucose diffusion across the cell membrane to the inside of a cell. If the glucose concentration on the outside of the cell is equal to 10 million Mueller, the glucose concentration and the side assault or on the inside of the cell is equal to 0.1 million Euller, 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 four 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 four steps, one by one, down below. And so step number one here is just to determine the net charge of the defusing molecule. And so if we take a look back at our problem, we can see that the defusing molecule is a 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 check box 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 then the equation that we introduced in our last lesson video. But thankfully, here again, glucose is the defusing 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 one, we have determined that the defusing molecule is glucose. We know that it's net charges 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 two and in step number two. 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 cells plasma membrane, and we know that glucose is the defusing 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, noticed that it wants is to calculate the Delta G transport for glucose diffusion across the 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 two, 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 three and step number three. 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 is 10 million Mueller. Then we have the glucose concentration in inside of the cell on the side of Saul as 0.1 million Mueller. So the million Mueller 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 Calvin, the equation that we need to use is take the degrees Celsius and add 273.15 and then we'll get the temperature in Calvin. So the temperature that were 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. Calvin. 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 mole arat e. 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 tomb. Olara T. If we were to convert both of these similarity, we would still end up getting the same exact answer. So it's OK to actually not convert these two more clarity 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 four and and step number four. This is the most straightforward part. And all we need to do is plug in all of the given values that we, um, are given in the problem, of course, with the appropriate units, uh, into the correct equation which we've already established, and then algebraic Lee solved 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 could go ahead and start over here on the left hand side. So Delta G Transport is going to be equal to are and recall that are is the gas constant, which is equal to a value of 8.315 jewels per mole time. Calvin. So we can plug that in here. We'll put this in parentheses. 8.315 jewels over moles times Calvin. So there's our our and then it's gonna 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 and units of Calvin 293.15 Calvin. And then, of course, this is all going to be multiplied by the natural log Will 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 side of Saul is 0.1 million Mueller. 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 eso Now that we have that we can put in our number 0.1 million Mueller on top here. So 0.1 million Mueller 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 million Mueller. So that's what's going to go on. Bottom is 10. Milly Mueller. And so now all we need to do is take our calculators and really just algebraic we solve here and so we could go ahead and rewrite this So d Delta G transport is going to be equal to 8.315 times 293. is equal to a value of 2437. approximately, and the units are going to be jewels over moles since the units of Calvin are going to cancel out. And then the natural log of this answer right here, which will put in parentheses, is equal to a value of about negative 4.605 and again, the units of million Mueller, 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 2437.54 times negative 4.605 is going to give us our answer. And so, Delta G Transport, it is going to be equal to our final answer, which is negative. 11, 0.3. And again, the Onley units that are left are jewels per mole. And so this here is our final answer. Negative. 11,225 0.3 jewels per mole is the Delta G transport and 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 ex organic process and so X Organic, 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 is a couple of times. And then you guys will get this on your exams. No problem. And so this year 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 Î”G_{transport }for the diffusion of glucose from outside to inside a cell if extracellular [Glucose] = 5 mM, intacellular [Glucose] = 0.5 mM, & the temperature = 20ËšC. Would this be an exergonic or endergonic process?

A

â€“5.61 KJ/mol.

B

5.61 KJ/mol.

C

â€“0.383 KJ/mol.

D

0.383 KJ/mol.

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Problem

Problem

Calculate Î”G_{transport} for the diffusion of glucose from inside to outside a cell if extracellular [Glucose] = 1 M, intacellular [Glucose] = 2.0 mM, & the temperature = 20ËšC. Would this be a spontaneous or nonspontaneous process?

A

â€“15.15 KJ/mol.

B

15.15 KJ/mol.

C

â€“1.03 KJ/mol.

D

1.03 KJ/mol.

5

Problem

Problem

Calculate the energy cost (Î”G_{transport}) of pumping an uncharged solute across a cell's plasma membrane, against a 1.0 x 104-fold concentration gradient at 25ËšC. Would this be an exergonic or endergonic process?