Thermodynamics of Membrane Diffusion: Uncharged Molecule - Online Tutor, Practice Problems & Exam Prep

In studying the thermodynamics of membrane diffusion for uncharged molecules, the change in Gibbs free energy (ΔG) is crucial. The equation $\mathrm{\Delta G}=\mathrm{\Delta G}°+\mathrm{RT}ln\left(Q\right)$ applies, where Q represents the concentration ratio of the molecule across the membrane. For membrane transport, ΔG° equals zero, simplifying the equation to $\mathrm{\Delta G}transport=\mathrm{RT}ln(\mathrm{[C]}/\mathrm{[A]})$, highlighting the significance of chemical gradients in diffusion processes.

Thermodynamics of Membrane Diffusion: Uncharged Molecule

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.

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. If you do not recognize this equation right here, if it looks completely foreign to you, 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.

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 δG is equal to δG₀ 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. 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. 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, 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 δG due to chemical gradients, not due to electrical gradients.

Δ G = Δ G₀ + R T ln Q

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.

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 G₀ 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 isn't really a real reaction, 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 G₀ 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 G₀ 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 G₀, since we've eliminated that part, it's just going to be equal to RTlnQ. And so you can see our RT right here, and notice that the lnQ, this part right here is replacing the Q. And recall that Q is equal to the ratio of the concentration of products over the concentration of reactants. But again, in a reaction that is just membrane transport, really there are no products or reactants because 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, 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 going to 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 going to be on the top, and the concentration of reactant is going to be on the bottom. Well, in terms of membrane diffusion, it's going to 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.

Thermodynamics of Membrane Diffusion: Uncharged Molecule Example 1

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 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 one by one 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 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. 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. Notice that it wants us to calculate the delta g transport for glucose diffusion across a cell membrane to the inside of a cell. 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 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. The outside of the cell is the initial side of the membrane, which means that the inside of the cell on this end of the arrow is going to be the 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 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. 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 8.315 jou... Δ G ∗ transport = R T ln [glucose] inside [glucose] outside This equation gives us: Δ G ∗ transport = − 11225.3 J ∙ mol − 1 So, this here is our final answer, negative 11,225.3 Joules per mole is the Delta G transport for this problem. As the value is negative, it shows us that the diffusion process is exergonic and thus will be spontaneous, not requiring any external input of energy.

Calculate ΔG_transport 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.

Calculate ΔG_transport 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.

Calculate the energy cost (ΔG_transport) 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?

• A. –95 KJ/mol.
• B. 2 KJ/mol.
• C. 23 KJ/mol.
• D. –10 KJ/mol.

Here’s what students ask on this topic:

What is the thermodynamics of membrane diffusion for uncharged molecules?

The thermodynamics of membrane diffusion for uncharged molecules involves understanding the change in Gibbs free energy (ΔG) as the molecule moves across the membrane. The relevant equation is:

$\mathrm{\Delta G}=\mathrm{\Delta G}°+\mathrm{RT}ln\left(Q\right)$

where Q is the concentration ratio of the molecule across the membrane. For membrane transport, ΔG° equals zero, simplifying the equation to:

$\mathrm{\Delta G}{\mathrm{transport}}_{}=\mathrm{RT}ln\left(\frac{\mathrm{[C]}}{\mathrm{[A]}}\right)$

This highlights the significance of chemical gradients in diffusion processes.

How do you calculate the Gibbs free energy change for membrane diffusion of uncharged molecules?

To calculate the Gibbs free energy change (ΔG) for membrane diffusion of uncharged molecules, use the simplified equation:

$\mathrm{\Delta G}{\mathrm{transport}}_{}=\mathrm{RT}ln\left(\frac{\mathrm{[C]}}{\mathrm{[A]}}\right)$

Here, R is the gas constant (8.315 J/mol·K), T is the temperature in Kelvin, [C] is the concentration of the molecule on the final side of the membrane, and [A] is the concentration on the initial side. Plug in the values for R, T, [C], and [A] to find ΔG.

Why is ΔG° equal to zero for membrane diffusion of uncharged molecules?

ΔG° is equal to zero for membrane diffusion of uncharged molecules because membrane transport is not a real chemical reaction. No bonds are created or broken, and the molecule remains the same on both sides of the membrane. Therefore, the standard Gibbs free energy change (ΔG°) is zero, simplifying the equation to:

$\mathrm{\Delta G}{\mathrm{transport}}_{}=\mathrm{RT}ln\left(\frac{\mathrm{[C]}}{\mathrm{[A]}}\right)$

What is the significance of the reaction quotient (Q) in membrane diffusion?

The reaction quotient (Q) in membrane diffusion represents the ratio of the concentration of the molecule on the final side of the membrane ([C]) to the concentration on the initial side ([A]). It is used in the equation:

$\mathrm{\Delta G}{\mathrm{transport}}_{}=\mathrm{RT}ln\left(\frac{\mathrm{[C]}}{\mathrm{[A]}}\right)$

This ratio helps determine the direction and magnitude of the Gibbs free energy change, indicating whether the diffusion process is spontaneous or requires energy input.

How does temperature affect the Gibbs free energy change in membrane diffusion?

Temperature (T) directly affects the Gibbs free energy change (ΔG) in membrane diffusion. The equation:

$\mathrm{\Delta G}{\mathrm{transport}}_{}=\mathrm{RT}ln\left(\frac{\mathrm{[C]}}{\mathrm{[A]}}\right)$

shows that ΔG is proportional to the product of the gas constant (R) and the temperature in Kelvin (T). As temperature increases, the value of RT increases, which can either increase or decrease ΔG depending on the concentration ratio (Q). Higher temperatures generally increase the rate of diffusion.