In this video, we're going to focus on problems associated with Gibbs free energy. So let's start with this question. Which of the following statements is false?
The change in free energy, delta G, is less than zero for a spontaneous process. Is that true or false? And what about B? Is it equal to zero at equilibrium? Or is D true?
Is it greater than zero for a spontaneous process? So what you need to know is that when delta G is basically positive, or you could say when it's greater than zero, this is true for a non-spontaneous process. Delta G is equal to zero during equilibrium, when you have a reversible process. When Delta G is negative or when it's less than zero this process is spontaneous Now, a reaction proceeds spontaneously when it could lower its energy.
In fact, a natural spontaneous process will occur in such a way to find the lowest possible energy state. And so we can see that when a change in free energy is negative, it's usually, it's always associated with a spontaneous process. And so that's something you want to keep in mind.
So which answers are true and which ones are false? So let's look at A. The change in free energy is less than zero for a spontaneous process.
That's true. Looking at B, the change in free energy is equal to zero at equilibrium. That's also true.
So we're looking for the false statement and see the maximum amount of work that can be obtained from a spontaneous process is equal to the change in free energy. That is also true. The maximum work that can be.
obtained is equal to Delta G for a spontaneous process. Now if you have a non-spontaneous process to make it work you got to put energy into it and so the minimum work required to drive a non-spontaneous reaction forward is also equal to delta G. Now we use the word minimum because you need to put in energy to overcome the energy losses that occur due to friction.
So if you're trying to drive a non-spontaneous process forward, you need more energy than delta G. So the minimum amount to get it started is equal to the change in free energy. But if you have a spontaneous reaction where you're trying to get energy from the reaction, then the maximum energy that you can get is less than the average.
or the maximum useful work that it can do, is equal to the change in free energy. So C is the true statement. Now D, the change in free energy is greater than zero for a spontaneous process.
It's greater than zero for a non-spontaneous process, so D is the false statement, which means D is the answer that we're looking for. Number two, the change in enthalpy and entropy for a certain reaction are those values. Calculate the change in free energy at 25 degrees Celsius.
So how can we do this? What formula do we need? The change in free energy, also known as Gibbs free energy, is equal to the enthalpy change minus the temperature multiplied by the entropy change. Now, you need to be careful with the units.
So notice that delta H, the enthalpy change, is in kilojoules per mole. And delta G is also in kilojoules per mole. But notice that the entropy change is in joules per mole per Kelvin. So we need to convert joules to kilojoules. And to do that, all we need to do is divide by 1,000.
So let me show you the conversion. So we have 212 joules per mole per Kelvin, and we know that there's 1,000 joules per 1 kilojoule. So we just got to divide this number by 1,000.
So the entropy change is going to be 0.212 kilojoules per mole per Kelvin. So just keep that in mind. So now let's work on this problem. So delta H is negative 46.5 kilojoules per mole.
Now, the temperature has to be in Kelvin. So 25 degrees Celsius plus 273, that's 298 Kelvin. And the entropy is 0.212 kilojoules per mole per Kelvin.
So notice how the unit Kelvin cancels. And so now this quantity is in kilojoules per mole, and this quantity also is in kilojoules per mole. So now let's get the answer.
So it's negative 46.5 minus 298 multiplied by positive 0.212. And so, delta G in this example is equal to negative 109.7 kilojoules per mole. Now, is the reaction spontaneous, non-spontaneous, or at equilibrium?
What would you say? Whenever delta G is negative, or when delta G is less than zero, what we have is a spontaneous process. And so that's the answer to the second part. Now this value corresponds to answer choice D, which is rounded to negative 110. And so that's it for this problem.
Number 3. Calculate the free energy change for the reaction shown below at 25 degrees Celsius. So what we need to do is we need to calculate delta G for that reaction. And we could use this formula. It's going to be delta H minus T times delta S. Now, using the information in this table, we could calculate delta H for the reaction and delta S.
And once we have that, we can plug it into this formula to get delta G. So let's go ahead and do that. Let's calculate the enthalpy change for the reaction.
These values that you see here, they represent the enthalpy change of formation for each of those subsets. And those values will always be zero for a pure element in its standard state. Now the formula that we're going to use is this formula. We're going to take the sum of the moles times the heat of formation for each reactant and product, but first this is going to be for the products, and then minus the sum of the moles times the heat of formation of all the reactants. So that's the formula we need to use.
So let's start with the products. All we have is a 2 in front of HCl. So it's going to be 2 moles of HCl times the heat of formation for HCl, which is negative 92 kilojoules per mole. And then for the reactants. So for H2, we have one mole of that.
The enthalpy change for H2 is zero. For Cl2, it's one mole. That's the coefficient. And then times zero as well.
2 times negative 92. That's going to be negative 184. And the unit moles will cancel. So what we have here is the delta H for the reaction is going to be negative 184 kilojoules. So we'll save that answer.
And now let's calculate the entropy change for the reaction. So we want to get delta S at this point. And we're going to use a similar process.
So we're going to take the sum of the product of the moles times the entropy values for each product. And then we're going to take the sum of the moles of the reactants times the entropy of each reactant. So we're going to have two moles of HDL, and the entropy for HDL is 187, the units being joules per mole per Kelvin. So once again, the unit moles will cancel, which will give us the entropy change in joules per Kelvin. Now let's focus on the entropy change in joules per Kelvin.
Focus on the reactants. So we have one mole of H2 and the entropy value for that is 131. And then one mole of Cl2 times the entropy value for that, which is 223 joules per mole per Kelvin. So in all cases, the unit moles will cancel.
So this is going to be 2 times 187 minus 131 minus 223. Thus, the entropy change for the reaction is going to be positive 20 joules per Kelvin. Now, let's go ahead and calculate delta G, since we now have everything that we need. Delta G is going to be equal to delta H minus T times delta S.
So delta H in this example is negative 184 kilojoules. The temperature is 25 degrees Celsius, and if we add 273 to that, that's going to be 298 Kelvin. and then we're going to multiply that by the entropy change of not joules but kilojoules. So we need to convert 20 joules into kilojoules. If we divide that by 1,000, that's going to be positive 0.02 kilojoules per Kelvin.
So the unit Kelvin will cancel and we'll get delta G for the reaction in kilojoules. And this is going to be standard delta G because we use standard delta H and standard delta S. So it's negative 184 minus 298 times 0.02.
So this is going to be negative 189.96 kilojoules. So we could say that is approximately negative 190 kilojoules. Therefore, answer choice A is the correct answer in this example. Number four, calculate the free energy change of the reaction shown below using the standard free energy change of formation values in the data table.
So this is going to be similar to the last problem, but we can calculate delta G directly using a table rather than the formula like we did in the last problem. So the free energy change for the reaction is going to be the sum. the moles times the Delta G values these are the standard delta G values of formation for all the products and then minus the sum of the moles of the reactants times the standard Delta G values of formation for the reactants So let's start with the products.
So we have two moles of iron oxide and the value for that is negative 740 kilojoules per mole. Now for the reactants we have four moles of iron oxide. We're going to multiply that by negative 255 kilojoules per mole. Now, oxygen is a pure element in its standard state.
So therefore, the delta G of formation for that is going to be zero. So we could just put plus one mole times zero. So once again, the unit moles will cancel. And we're going to get delta G in kilojoules. So we have 2 times negative 740 minus negative 255 times 4. So I got negative 460 kilojoules.
So be careful with this double negative here. There's a negative sign and another one. there.
So make sure you incorporate that. Thus the correct answer is going to be answer choice D. Now I want to take a minute to mention something. When dealing with thermal chemical equations, the energy change associated with a reaction, that is the amount of heat energy absorbed or released, it's typically in units of kilojoules. So for this first reaction, let's say it releases negative 20 kilojoules of thermal energy.
If we multiply the coefficients by 2, then we're going to get double the amount of energy released, because we've doubled the quantity of reactants that are converted into products. If we multiply the original equation by 3, the energy change will increase by a factor of 3. Now sometimes, instead of seeing kilojoules as the energy associated with the energy change that occurs for a reaction, you might see kilojoules per mole. And when you see that, that's basically the amount of energy that's absorbed or released per mole of reaction if we were to take negative 20 divided by 1 or 1 mole of a or you can think of it as just 1 mole of the reaction the energy change will be negative 20 kilojoules per mole now this ratio won't change Regardless of what the coefficients of the reaction will be.
So if you take negative 40 divided by 2, you're going to get negative 20 still. If you take negative 60 divided by 3, you're going to get negative 20. And when calculating the non-standard delta G value, it's important that we use this figure as opposed to these other numbers. And here's why.
Later in this video, we're going to use this equation. Delta G value is equal to the standard Delta G value plus RT ln Q where Q is the reaction quotient. R has the units joules per mole per Kelvin. Delta G we can have that in kilojoules per mole or joules per mole.
Temperature is in Kelvin. So in order for this equation to work, in order for us to algebraically add these quantities, they need to be in the same units. So if we were to report delta G naught as being in kilojoules and not kilojoules per mole, it won't work. well with this equation. So we need it to be in joules per mole and not just joules.
We can convert kilojoules per mole to joules per mole, but if we just have kilojoules and convert it to joules, we can't add these two quantities because we need them to be in joules per mole. So going forward, instead of just leaving the answer in kilojoules, let's say if I'm calculating delta H or delta G of the reaction, I'm going to be using kilojoules per mole so that it works well with this equation. So I want you to...
be aware of those units and their relationship to each other but for this equation to work just make sure that Delta G naught is in joules per mole and not just joules or kilojoules number five calculate the free energy change of the decomposition of dinitrogen pentoxide into dinitrogen tetroxide which is n2o4 and oxygen So how can we use the first two equations and the free energy associated with those equations to get the free energy of the third equation? What do we need to do? Well, we need to change these two equations in such a way that when we add them, we get this equation.
And so we can add the adjusted free energy change values to get this one. So how do we need to change equation 1 and 2 so that it can be added to this equation? Well, the first thing we need to realize is that, notice that in the second equation, N2O4 is in the right spot. And we have the right number of it, so we don't need to change the second equation.
Now the first equation we do need to change. Notice that N2O5 is found on the right side, but we need it on the left. And we need the coefficient to be 1. So therefore, the first reaction, we need to reverse it, and we need to multiply it by half.
So, half of 4nO2 is going to be 2nO2. And then, we'll first need to reverse it, so let me do that first. So, half of this will be 1. So, that's 1n205. And then, on the other side, half of 4nO2, that's going to be 2nO2. And then half of O2, we can simply write that as 1 half O2.
Now, because we reversed it, it's going to change from positive 60 to negative 60. And we multiplied it by half, so half of negative 60 is negative 30. So that's the new free energy change for this reaction. Now, this one we're not going to change, so I'm just going to rewrite it the way it is. And so that's going to stay negative 6. So now let's add these two reactions. So notice that NO2 will cancel.
It's the intermediate in this reaction. And so we're left with just one N2O5 molecule. it turns into one N2O4 molecule and half of an O2 molecule.
So now all we need to do is add up these two values. Negative 30 plus negative 6 is negative 30. So therefore, this is the final answer. So answer choice C is the correct answer in this example. Number 6. The standard free energy change for a reaction under certain conditions is negative 65 kilojoules per mole. At this instant, which of the following events will occur?
So will the free energy change increase or decrease? And will the reaction shift to the left or to the right? Or is the system at equilibrium?
So let's say if we have two species, A and B. Now, if delta G is negative, what's going to happen? Is the reaction spontaneous or non-spontaneous?
Anytime delta G is negative or less than zero, the reaction is spontaneous. So if we have a spontaneous process, what does that mean? What's happening in terms of the reaction?
Is the reaction going towards the right or is it going towards the left? When delta G is negative, you need to know that it's spontaneous in the forward direction. So that means that when delta G is negative, the reaction is going towards the right. It's shifting to the right. So it's not going to the left, so A is out and C is out.
And it's not at equilibrium. Delta G will be at equilibrium when delta G is zero. I mean, the reaction will be at equilibrium when delta G is zero.
That's what I meant to say. So we're between B and D. Now, as the reaction shifts towards the right, will the free energy change increase or decrease?
Right now, at this instant, the free energy change is negative 65 kilojoules per mole. As the reaction shifts towards the right, the products will increase and the reactants will decrease. And this will continue to happen until it reaches equilibrium.
And at equilibrium, the free energy change will be zero. So the reaction will proceed in such a way until delta G becomes zero, until the system reaches equilibrium. So to go from negative 65 to zero, will the free energy change? Will it increase or decrease?
Negative 65 is the lower number on the number line relative to 0. So, let's say if you draw a number line. Here's 0, here's 50, here's negative 65. The lower number is on the left side. So therefore, delta G has to increase from negative 65 to 0 as the reaction proceeds to the right going towards equilibrium. So therefore, we can eliminate answer choice D. It's not going to decrease.
It's going to increase. Number 7. Estimate the boiling point of bromine. Now, we're given the enthalpy of vaporization of bromine. It's 30.9 kilojoules per mole. And the standard entropy values for liquid and gaseous bromine are 152 and 245 joules per mole per Kelvin.
Now let's write a reaction. So we have liquid bromine turning into gaseous bromine. And so the enthalpy of vaporization is basically the enthalpy of this reaction. Because liquid bromine is being vaporized into gaseous bromine.
So the enthalpy of the reaction is positive 30.9 kilojoules per mole. Now how can we calculate the boiling point of bromine? What do we need? Well, we need to start with this equation.
Delta G is equal to delta H minus T delta S. And at the boiling point... These two physical states coexist.
So what that means is that this system is in equilibrium at the boiling point. The liquid state and the gaseous state, they coexist. They can go back and forth at that temperature.
Now, if this system is in equilibrium, what is delta G? At equilibrium, delta G is equal to zero. So now what we have is 0 is equal to delta H minus T delta S.
So let's take this term and move it to the other side. So T delta S is equal to delta H. And now we need to calculate T.
And T represents the boiling point temperature. And so to calculate it, it's simply the enthalpy of vaporization divided by the entropy change. So that's how you can calculate the boiling point of a liquid. Now, we already have the enthalpy change. What we're missing is the entropy change.
So to calculate the entropy change for the reaction, it's going to be the sum of the entropy values for the products minus the sum of the entropy values for the reactants. Now the only product that we have in this reaction is gaseous bromine, so I'm going to write G for gas and the reactant is liquid bromine and the entropy value for gaseous bromine is 245 and for liquid bromine, it's 152 So let's subtract those values. 245 minus 152. And so the entropy change is 93, and the units are joules per mole per Kelvin.
So now we have the entropy change, and we have the enthalpy change. So with that information, we can go ahead and calculate the temperature. Now we need to be careful with the units. As you can see, the enthalpy change is in kilojoules per mole.
The entropy change is in joules per mole. So I'm going to convert this to kilojoules by dividing it by 1000. So the temperature is going to be the enthalpy of vaporization, which is 30.9. kilojoules per mole divided by the entropy change 93 divided by thousand that's 0.093 and it's going to be kilojoules per mole per Kelvin so we can see that the unit kilojoules will cancel and the unit moles will cancel and so we're going to get the temperature in Kelvin so it's 30.9 divided by 0.093 and Now, this is equal to 332.26 Kelvin.
And if we subtract this by 273.15 Kelvin, we can convert it to Celsius. So the temperature is about 59.1 degrees Celsius. So we can round it and say the boiling point of bromine is approximately 59 degrees Celsius.
Therefore, C is the right answer in this problem. Number 8. Which of the following statements is false? Is it A, B, C, D, or E?
Well, let's go over some things that you need to know. Now, when delta G is less than zero, that means it's negative, we have a spontaneous reaction. And so that means that the reaction shifts to the right. If the standard delta G value is equal to zero, then that means that the system is at equilibrium.
And if the standard free energy change is greater than zero, then we have a non-spontaneous reaction. It's non-spontaneous in the forward direction, which means it's spontaneous in the reverse direction, so it shifts to the left. Now keep in mind when Delta G is less than zero it's equivalent to being negative And when delta G is greater than 0, it's equivalent to being positive. Now, when we have a spontaneous reaction, when delta G is less than 0, then K is greater than 1. And at equilibrium, K is equal to 1. And if we have a non-spontaneous reaction, K is less than 1, but it has to be greater than 0. K cannot be a negative number.
Now when K is greater than 1, it's product favored. Keep in mind, K is the ratio of the products to the reactants. So let's say if K is a significantly large number, like 10,000, that means that at equilibrium, there's a lot more products than reactants.
So anytime the reaction shifts to the right for a spontaneous process, it means that it's product favored. If delta G is positive, where K is small, like 1 times 10 to negative 8 or something, you have a non-spontaneous process. So it's spontaneous in the reverse direction. So it's reactant favored when K is small, but it's product favored when K is large.
So if we look at E, the reaction is reactant favored when K is significantly less than 1. That's true. And D, it's product favored when K is significantly larger than 1. That's also true. And Answer choice C, if the standard free energy change is 0, then k is equal to 1. That is a true statement. And for B, for a non-spontaneous process, K is between 0 and 1. And that's just something you need to know.
So make sure you write that down in your notes. Now, A has to be the false statement. For a spontaneous process, K is not less than 1. In fact, K is greater than 1 when spontaneous, because delta G is negative, it's less than 0. So A is the correct answer, it's the statement that is false. Number 9. The equilibrium partial pressure constant is 1.4 times 10 to negative 5 at 298 Kelvin for a certain reaction.
Calculate the standard free energy change at 298 Kelvin for that reaction. So we're given Kp, the equilibrium partial pressure constant, and it's equal to this value. So how can we use this number to calculate delta G? Now there is a formula.
The standard free energy change is equal to negative RT. times the natural log of the equilibrium constant so R in this example is the energy constant 8.3145 joules per mole per Kelvin and the temperature in Kelvin is 298 Kelvin and so we can see that the units Kelvin will cancel and K is unit list so we don't have to worry about that Now the units that we're going to have is joules per mole, and delta G has to be in kilojoules per mole, so we need to convert it to that. So let's take negative 8.3145, multiply it by 298, times the natural log of 1.4 times 10 to the negative 5. So the standard free energy change for this reaction is positive 27,700 joules per mole if you round it. Now let's convert it to kilojoules per mole.
So 1 kilo... joule is 1000 joules and now we can cancel this unit so 27,700 divided by a thousand is 27.7 kilojoules per mole And so that's delta G in this problem, which means that B is the right answer. Now let's think about this.
Notice that K is very small. It's 0.000014. So K...
is greater than 0 but is less than 1, which means that we have a reactant-favored reaction. And notice that delta G is positive. We said that if K is less than 1, delta G is positive, or delta G is greater than 0, which means that it's non-spontaneous in the forward direction, but it is spontaneous in the reverse direction, so it's reactant-favored.
It goes towards the left. Number 10, the standard free energy change for a certain reaction at 298 Kelvin is negative 215 kilojoules per mole. Calculate the value of the equilibrium constant for this reaction.
So how can we do this? What should we do? Now, notice that delta G is equal to a negative number.
And we know that when it's negative, the reaction is spontaneous. It's going to shift to the right, and it's a product-favored reaction. Now, when delta G is negative, is K very large or very small? If it's product-favored, K is going to be very large.
So k is going to be a positive number, and it's going to be 10 raised to some positive exponent. It's a large number. On the other hand, when delta G is positive, we have a non-spontaneous process. And so the reaction is reactant favored. And so k is going to be less than 1 but greater than 0. So because it's greater than 0, it's still a positive number.
But it's less than 1, so the exponent has to be negative. as opposed to being positive. But for this problem, delta G is clearly a negative value.
So we need a positive K value with a positive exponent. We can eliminate answer choice A because it has a negative exponent, which means that delta G is positive. This also... has a negative exponent.
Now, k can never be negative, so b is out. So, it's either between c or d. That's when k is positive and it has a positive exponent.
But let's get the exact answer. Now let's start with this equation. Delta G is equal to negative RT ln K.
So the first thing we need to do to isolate K is to divide both sides by negative RT. So lnk is equal to negative delta g over RT. Now, we need to get k by itself. When dealing with logarithms, here are some general things you need to keep in mind.
a raised to the c is equal to to B. So that's how you can convert a logarithmic expression into its exponential form. Now the base of a natural log is E, so E raised to everything on the right side is equal to K. Or we could say K is equal to E raised to the negative delta G over RT. So that's the formula that we're going to use to calculate K.
So I'm just going to rewrite it here. Now when using this formula, because R is in joules per mole, delta G has to be in joules per mole. So you've got to convert kilojoules to joules. So let's do that first.
So we have negative 215 kilojoules per mole. And we know that 1 kilojoule is equal to 1,000 joules. So this time we need to multiply by 1,000 as opposed to divide by 1,000.
So it's negative 215,000 joules per mole. so K is going to be E negative 215,000 now that's delta G and keep in mind there's another negative sign so it's negative delta G so we got two negative signs in this example and then R which is 8.3145 multiplied by the Kelvin temperature 298 don't plug in a Celsius temperature you won't get the right answer So this is equal to 4.8 times 10 to the 37 So that's the value of the equilibrium constant so C is the answer for this problem Number 11, which of the following statements is true concerning the physical reaction shown below? Is the reaction spontaneous at high temperatures? Low temperatures?
Is it always spontaneous and never spontaneous? Or is it spontaneous at the freezing point? So in this reaction, we have liquid water turning into ice.
So, which statement is true? If we take a step back and keep it simple, the best way to turn liquid water into ice is to put it in a freezer. If you put it in a freezer, heat energy is going to flow away from the liquid water, and so as water loses thermal energy, the temperature decreases.
And as the temperature decreases, it's going to freeze to ice. So therefore, we know that it's spontaneous at low temperatures. So just by intuition, you know that the answer has to be B.
But now let's talk about why that's the case. So first, this reaction, is it endothermic or exothermic? The freezing of water. In order to freeze water, in order to go from a liquid to solid, You need to remove energy.
So liquid water has to release heat energy into the surroundings. So let's say we have a beaker filled with liquid water. If we put it in a freezer, where let's say the temperature is negative 20 degrees Celsius, and the temperature of this water is 25, heat flows from hot to cold.
So heat is going to leave the system, and it's going to flow into the surroundings. Whenever the system releases heat energy, it's an X with a process, so delta H is negative. Now the next thing we need to analyze is the entropy change. Is it positive or is it negative? So which one has more entropy, a solid or a liquid?
A solid is very organized, so it doesn't have much disorder. So a solid has a very low entropy value. A liquid has a relatively higher entropy value than a solid. So we're going from a high entropy value to a low entropy value. So entropy is decreasing, therefore delta S is also negative.
Now you need to be familiar with this table. Delta H, Delta S, T, and Delta G. You may want to commit this to memory. So when Delta H and Delta S are both positive or both negative, The sign of delta G is temperature dependent.
So when they're both positive, and the temperature is high, delta G is going to be negative, which means that it's spontaneous under these conditions. It's spontaneous at high temperatures, and it's non-spontaneous at low temperatures. Now, when delta H and delta S are both negative, it's still going to be temperature dependent.
This time, delta G is negative at low temperatures. So it's going to be spontaneous at a low temperature, but not spontaneous at a high temperature. Now, when the signs of delta H and delta S are different, then it's no longer temperature dependent.
So when delta H is negative and delta S is positive, delta G is always spontaneous, regardless of the temperature. And when delta H is positive and delta S is negative, it's always going to be non-spontaneous. So in this example, delta H and delta S are both negative.
So we're dealing with this section here. So the reaction is going to be spontaneous only at low temperatures. And it makes sense.
Liquid water will only freeze at low temperatures. So that's why B is the answer. We can eliminate A, it's not going to be spontaneous at high temperatures, and C and D doesn't apply. Now what about E? At the freezing point, that is if the temperature is 0 degrees Celsius, then it's going to be...
be at equilibrium. Liquid water and ice can coexist at 0 degrees Celsius. So at this temperature, delta G will equal 0 and so it's going to be at equilibrium.
Now, if we decrease the temperature below zero, then it's going to shift in this direction. It's going to freeze. So at low temperatures, that is below the freezing point, it becomes spontaneous. Now, if we switch it, if we increase the temperature above zero, then it becomes non-spontaneous, which means that it's spontaneous in the reverse direction. So at high temperatures, water, or ice rather, is going to melt into liquid water at high temperatures.
temperatures. But at low temperatures, liquid water is going to freeze into ice. And so you can figure this out based on this table, or just using simple intuition.