Gibbs Free Energy - Video Tutorials & Practice Problems
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1
concept
Gibbs Free Energy Equation
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and this lesson, we're gonna cover Mawr Details about Gibbs free energy. Now, before we get to the other forms of the Gibbs Free Energy equations, let's first recall the standard Gibbs Free Energy equation and that relates the changes in free energy to the changes in entropy, temperature and the changes in entropy and recall that Gibbs Free Energy is the energy that's available to perform. Work and work is done in a reaction when the concentrations of that reaction or system change, and so recall that at equilibrium the concentrations of reactant and products do not change their constant. And so there's no work done at equilibrium, and the Delta G is going to be equal to zero at equilibrium. So this goes to show that the concentrations within a system influence the direction of a reaction. And so we're going to talk about in our next video, the reaction direction. So I'll see you guys in that video
2
concept
Reaction Direction
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So in our previous video, we talked about how the concentrations of a reaction impact the direction of that reaction. And so what's interesting to note is that cellular reactions are almost never at equilibrium, and this is due to several different factors, such as the conditions of the reactions changing and products being siphoned away into different reactions and reactant being constantly added. And so when a reaction is not at equilibrium, we need to make the appropriate adjustments. And one of those adjustments is to replace the equilibrium constant with the reaction quotient, which can be symbolized by the letter Q and notice that the reaction quotient expression is the same as the expression of the equilibrium constant. So it's the concentrations of products over the concentrations of react. It's and so the difference between the reaction potion and the equilibrium constant is that the equilibrium constant is specifically at equilibrium, whereas the reaction quotient is not at equilibrium. And so recall from your previous chemistry courses that last shot liaise principles so lush. Ali's principles state that when an equilibrium is disturbed, or when a reaction is not at equilibrium, such as cellular reactions, the reaction direction is going to proceed towards a direction to restore equilibrium so it will proceed to restore equilibrium. And so, in our example below, we're going to consider the reaction and then we're also going to complete the chart. And so in this reaction we have carbon monoxide gas interacting with hydrogen gas to produce methanol. And so again, recall that the reaction quotient expression is the concentration of products over the concentration of reactant. And so we only have one product here and that's methanol. So up here we can put methanol ch 30 h. Now, for the reactant, we have to react INTs the carbon monoxide which we can put over here. And we also have the hydrogen gas or H two. And again the coefficients, which is the two here, are expressed as, uh, exponents. And so we can go ahead and put this to Azan exponents here. And what you'll notice is that the reaction quotient expression is pretty much exactly the same as the equilibrium, constant expression. And the only difference is that the equilibrium, constants expression specifically has the concentrations of these substances at equilibrium, whereas the reaction question again is not at equilibrium. And so over here in this chart, where we're going to do is compare the reaction question with the equilibrium constant. Because if we know the equilibrium constant and we also know the concentrations of the substances at any particular moment in the reaction, we can predict the direction of that reaction. And so that's what we're going to do here. And so when the reaction quotient is smaller than the equilibrium constant, what that means is that the products are gonna be very small in their amounts, and the reactant are gonna be very large in their amounts. And so we'll have a whole bunch of reactant that's going to react and produce more product. So the reaction's gonna proceed in a forward direction. And so when the reaction question is smaller than the equilibrium constant, the reaction proceeds as written from left to right in a forward direction. Now, when the reaction question is larger than the equilibrium constant. What that means is that there there's a whole bunch of product and on Lee a little bit of reacting. And so we have a whole bunch of the product, and that's goingto go backwards and, uh, react to produce mawr of the reactant. And so when the reaction question is larger than the equilibrium constant, the reaction proceeds from right toe left in a reverse direction or a backwards direction. And, of course, when the reaction quote quotient is equal to the equilibrium constant, that means that we are at equilibrium and again at equilibrium. The rate of the Ford reaction is equal to the rate of the reverse reaction. So here we can put two arrows here showing that at equilibrium, the reaction rates are the same, going in both directions. And so what we're gonna see is what you're able to see is that, uh, the reactant and the concentrations of the reacting, some products, the concentrations impact the direction, and that is very, very apparent here. And so what's really important to note is that the concentrations are a part of the conditions of the reaction. And so in our next video, we're going to talk about how standard conditions impact the Gibbs Free Energy equation. And so I'll see you guys in that video
3
concept
Gibbs Free Energy (Standard Conditions)
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So from our previous video, we know that we can predict the direction of a reaction just by having the equilibrium, constant and the concentrations of reaction components. And so the concentrations are part of the conditions and scientists use standard conditions to allow them to compare different reactions under the same conditions. Since the conditions have a big impact on the reaction and this little symbol here, eyes used to represent standard conditions now recall from your previous chemistry courses that the equilibrium constant can be used to calculate the change in free energy under standard conditions. And so I recall that the little symbol here Delta, which is this triangle here, represents change. G represents the free energy. And again, this little symbol here not represents standard conditions. So this is the change in free energy under standard conditions. Now the Delta G without the not symbol represents the actual change in free energy under any conditions. And we'll talk about that one in our next video. So recall that standard conditions includes, uh, having a temperature at 25 degrees Celsius, which is equivalent to 298 kelvin. It also includes having a gas constant or are here, which is equal to 8.315 jewels per mole times, Kelvin. And so the gas constant, uh, magnitude. Here. The value of this number can actually change depending on the units. And so you might be familiar with a number 1.98 times 10 to the negative three. And this is when the units are in Q locales per mole times Calvin. And so it's good to be able to recognize that the gas constant can change depending on the units. And so the pressure, the atmospheric pressure under standard conditions is one atmosphere and the concentrations of reactant and products. The initial concentrations are one Moeller, and so you can see over here on the left that the Gibbs Free Energy understand Erred conditions is specifically shown as the following equation where the change in free energy under standard conditions is equal to negative are or negative gas constant times the temperature and units of Kelvin uh, times the natural log of the equilibrium constant understand erred conditions. And so we'll be able to apply this equation and some of our practice problems. Now, in our next video, we're going to talk about the Gibbs Free Energy under physiological conditions, which vary from standard conditions. So I'll see you guys in that video.
4
concept
Gibbs Free Energy (Physiological Conditions)
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8m
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So in our previous video, we talked about how scientists use standard conditions to be able to compare different reactions under the same conditions, and standard conditions occur in a very controlled environment within a test tube inside of a lab. However, physiological conditions within biological systems or living things very or can vary greatly from standard conditions. Now what's important to note is that by studying a particular reaction within a test tube in a lab, understand erred conditions were able to determine the change in free energy under standard conditions. And we can use the change in free energy under standard conditions to calculate the actual change in free energy under any condition or Delta G without the not and so you can see down here in this little blue box. Here we have the Gibbs Free Energy under any conditions. So this is used to calculate the change in free energy under physiological conditions, and so you can see that the change in free energy under any condition is equal to the change in free energy. Understand erred conditions plus the gas, constant times, temperature and units of Calvin times, the natural log of the reaction quotient And so we'll be able to use this, uh, exam this equation in a moment here when we talk about this example. And so in this example, what we're gonna do is calculate the change of free energy under standard conditions as well as the actual change in free energy. And what we're gonna do is consider a specific reaction that occurs in glide Collis ISS and recall from your previous bio courses. That Glen Collis is is the process of taking a glucose molecule and breaking it down into two Piru Bates. And so we'll talk more about Glen. Collis is later on in our course, But here's a specific reaction that occurs within Glen Collis is. And what this reaction does is that it takes a reactant known as di hydroxy acid tone phosphate or D H a P. And it converts this reactant into the product of Glessner outta hide three phosphate or G three p, and so you can see that the equilibrium expression of understand erred conditions is simply the concentrations of products at equilibrium over the concentrations of reactant at equilibrium, which we've already seen plenty of times in our previous videos and so uh, G three p. Here is our product. So we can put that here on the top and D h a p again is are reacting so we can put it on bottom. And so the equilibrium constant under standard conditions is given to us as 0.475 and so recall that Delta G, under standard conditions is equal to negative, are times the temperature times the natural log of the equilibrium constant under standard conditions. And so we can go ahead and plug in our values here to calculate the change in free energy under standard conditions. Now are here we know is equal to 8.315 equal to 8. The temperature under standard conditions is 298. Calvin recall. That's from our previous video. So we can put 298 for the temperature and then the natural log of the equilibrium constant under standard conditions is 0.475 So 0.475 So if you number crunches in your calculator, what you'll see is that this is equal to negative. I'm sorry. It's equal to 7550 jewels per mole. Jules Permal, and so notice that this is a positive value. And so the Delta G under standard conditions is an undergone IQ reactions because recall that positive delta Jeez, our undergone IQ or non spontaneous. However, we know that glide colossus is a process that happens all the time in ourselves. So if it's an organic, that means that it's got to be super energy intensive. But it's not. And so we have toe be able to produce this product over here. And so here's the thing. The spontaneity of a reaction within a cell is actually determined by the actual change in free energy of the system. So Delta G uh, without the not So we need to determine this Delta G in order Thio in order to determine the spontaneity off the reaction within a cell. So even though the spontaneity of the reaction under standard conditions in a test tube is, uh, non spontaneous, perhaps when we calculate the Delta G under physiological conditions, perhaps this will be a spontaneous reaction. And so we're gonna need some a little bit of additional information and what we're gonna need to know is that the concentration of D H AP under physiological conditions is actually two times 10 to the negative fourth Moeller. So this is the concentration of D. H ap under cellular physiology physiological conditions, whereas glycerol glycerol to hide three phosphate has a, uh a concentration, a physiological concentration of three times 10 to the negative six Mohler Fergie, three p. And so we're gonna need that for our equation here to calculate Delta G. And so now that we've calculated Delta G under, uh, standard conditions, we can go ahead and plug that value. And here 7550. So we've just taken this and we've plugged it in here for this part of the reaction of the equation. Now we're gonna need toe add our, which is 8.315 again. T is gonna be the temperature, um, the physiological temperature which recall from our previous videos, we're going to assume is 298 Kelvin. So we can keep it essentially the same as standard conditions and then the natural log of Q and recall that Q is the concentration of products, overreact INTs and so we have The concentration of our product is G three p is our product, so it's gonna be three times 10 to the negative six. So three times 10 to the negative six is our product over the concentration of reacting, which is D H a P. And that's two times 10 to the negative fourth Moeller. And so when we number crunch this, we do the natural log of this number here. Times 298 times 8.315 What we end up getting is, uh, let's keep this 7550. Here we end up getting plus or plus a negative 10, 410, 0.32 So 0. And so when we number crunch this, we end up getting that. The value is negative 2856. jewels per mole and so you can see that the Delta G, the actual Delta G under physiological conditions, is a negative value here. It's a negative value, which means that it is a spontaneous X organic reaction, which is what we anticipated. Since we know that Glen Collis occurs all the time and that this reaction should be spontaneous under physiological conditions. And so again, we'll be ableto use this reaction, Maurin, our practice videos and perhaps later in our course as well. But for now, this is a good summary of the, uh, Gibbs free Energy under physiological conditions, and I'll see you guys in our practice videos.
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Problem
Problem
Consider a reaction where Keq=1.6 but Q = 3.19. What direction will the reaction proceed?
A
Forward.
B
Reverse.
C
Forward & reverse reactions proceed equally.
D
Not enough information provided to make conclusions.
6
Problem
Problem
At equilibrium, the reaction A ⇌ B + C has the following reactant concentrations: [A] = 3 mM, [B] = 4 mM, and [C] = 10 mM. What is the standard free energy change for the reaction & is it endergonic or exergonic?
A
– 6418 J
B
6418 J
C
10,698 J
D
– 10,698 J
7
Problem
Problem
ΔG˚=141.7 kJ for the following reaction. Calculate ΔG: T=10˚C, [SO 3] = 25mM, [SO2] = 50mM, & [O2] = 75 mM.
