Gibbs Free Energy - Video Tutorials & Practice Problems
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Gibbs Free Energy Equation
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In this lesson, we're gonna cover more 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 enthalpy, 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 reactants and products do not change. They are constant. And so, there's no work done at equilibrium, and the delta G is going to be equal to 0 at equilibrium. So this goes to show that the concentrations within a system influence the direction of a reaction. And so, we're gonna 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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5m
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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 reactants 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 reactants. And so the difference between the reaction quotient 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 Le Chatelier's principles, so Le Chatelier'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 reactants. And so we only have one product here and that's methanol. So up here we can put methanol, CH3OH. Now for the reactants, we have 2 reactants, the carbon monoxide, which we can put over here, and we also have the hydrogen gas, or H2. And again, the coefficients, which is the 2 here, are expressed as, exponents. And so we can go ahead and put this 2, as an exponent 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 quotient again is not at equilibrium. And so, over here in this chart, what we're gonna do is compare the reaction quotient 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 going to be very small in their amounts, and the reactants are going to 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 is gonna proceed in a forward direction. And so when the reaction quotient is smaller than the equilibrium constant, the reaction proceeds as written from left to right in a forward direction. Now when the reaction quotient is larger than the equilibrium equilibrium constant, what that means is that there's a whole bunch of product and only a little bit of reactant. And so we have a whole bunch of the product and that's going to go backwards and, react to produce more of the reactants. And so when the reaction quotient is larger than the equilibrium constant, the reaction proceeds from right to left in a reverse direction or a backwards direction. And of course, when the reaction quotient is equal to the equilibrium constant, that means that we are at equilibrium. And again, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. So here, we can put 2 arrows here showing that at equilibrium, the reaction rates are the same going in both directions. And so, what we're going to see is, what you're able to see is that, the reactants and the concentrations of the reactants and 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 gonna 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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3m
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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, is 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, 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, 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, having a temperature at 25 degrees Celsius, which is equivalent to 298 Kelvin. It also includes having a gas constant or r here, which is equal to 8.315 joules per mole times Kelvin. And so the gas constant, 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 3, and this is when the units are in kilocals per mole times Kelvin. 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 1 atmosphere and the concentrations of reactants and, products, the initial concentrations, are 1 molar. And so you can see over here on the left that the Gibbs free energy under standard conditions is specifically shown as the following equation, where the change in free energy under standard conditions is equal to negative r or negative gas constant times the temperature in units of Kelvin, times the natural log of the equilibrium constant under standard conditions. And so, we'll be able to apply this equation in some of our practice problems. Now, in our next video, we're gonna 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 vary 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 under standard conditions, we're 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 knot. And so you can see down here in this little blue box here, we have the Gibbs free energy under any condition. 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 under standard conditions plus the gas constant times temperature in units of Kelvin, times the natural log of the reaction quotient. And so, we'll be able to use this, this equation in a moment here when we talk about this example. And so in this example, what we're going to do is calculate the change in free energy under standard conditions, as well as the actual change in free energy. And And what we're going to do is consider a specific reaction that occurs in glycolysis. And recall from your previous bio courses that glycolysis is the process of taking a glucose molecule and breaking it down into 2 pyruvates. And so we'll talk more about glycolysis later on in our course, but here's a specific reaction that occurs within glycolysis. And what this reaction does is it it takes a reactant known as dihydroxyacetone phosphate or DHAP, and it converts this reactant into the product of Glyceraldehyde 3 Phosphate or G3 P. And so you can see that the equilibrium expression of, under standard conditions is simply the concentrations of products at equilibrium over the concentrations of reactants at equilibrium, which we've already seen plenty of times in our previous videos. And so, G3P here is our product, so we can put that here on the top, And DHAP again is our reactant, so we can put it on bottom. And so the equilibrium constant under standard conditions is given to us as 0.0475. And so recall that delta g under standard conditions is equal to negative r 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, r here we know is equal to 8.315, so it's equal to 8.315. The temperature under standard conditions is 298 Kelvin. Recall that's from our previous video. So So we can put 298 for the temperature. And then the natural log of the equilibrium constant under standard conditions is 0.04 75, so 0.0475. So if you number crunch this in your calculator, what you'll, see is that this is equal to negative, I'm sorry, is equal to 7,550 joules per mole. Joules per mole. And so notice that this is a positive value, and so the delta G under standard conditions is an endergonic reaction, because recall that positive delta G's are endergonic or non spontaneous. However, we know that glycolysis is a process that happens all the time in our cells. So if it's endergonic, that means that it's gotta be super energy intensive, but it's not. And so we have to be able to produce cell is actually determined by the actual change in free energy of the system. So delta g, without the knot. So So we need to determine this delta g in order to to in order to determine the spontaneity of the reaction within a cell. So even though the spontaneity of the reaction under standard conditions in a test tube is, 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 DHAP under physiological conditions is actually 2 times 10 to the negative 4th molar. So this is the concentration of DHAP under cellular physiological conditions. Whereas, glyceral glyceraldehyde 3 phosphate has a, a concentration, a physiological concentration of 3 times 10 to the negative 6 molar for g 3p. And so, we're going to need that for our equation here to calculate delta g. And so now that we've calculated, delta g under, standard conditions, we can go ahead and plug that value in here, 7,550. 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 going to need to add R, which is 8.315 again. T is gonna be the temperature, 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 over reactants. And so we had the concentration of our product is, g 3p is our product, so it's going to be 3 times 10 to the negative 6, so 3 times 10 to the negative 6 is our product, over the concentration of reactant, which is DHAP, and that's 2 times 10 to the negative 4th molar. 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, let's keep this 7,005 150 here, we end up getting plus or plus a negative 10 1,004106.32. So 0.32. And so when we number crunch this, we end up getting that the value is negative 2,856.32 joules 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 exergonic reaction, which is what we anticipated since we know that glycolysis occurs all the time and that this reaction should be spontaneous under physiological But for now, this is a good summary of the, 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.