2015 AP Chemistry free response 2c | Thermodynamics | Chemistry | Khan Academy

Because the dehydration reaction is not observed to occur at 298 Kelvin, the student claims that the reaction has an equilibrium constant less than 1.00 at 298 Kelvin. Do the thermodynamic data for the reaction support the student's claim? Justify your answer including a calculation of standard or change in standard Gibbs-free energy at 298 Kelvin for the reaction.

So let's first of all, let's review what the student's claim is. The student claims that since the reaction is not observed at 298 Kelvin, at that temperature the reaction has an equilibrium constant less than 1.0 at 298 Kelvin. That's their claim.

If these ideas of equilibrium constant or Gibbs free energy are completely foreign to you, I encourage you to review the videos on Khan Academy on Gibbs free energy and equilibrium constants. If they're familiar to you, you're like, okay, I don't know all the formulas that might connect the information that's given in the problem and how to calculate Delta G and maybe how do we go from Delta G to equilibrium constants and figuring out whether the equilibrium constant is going to be greater than, less than one, or equal to one.

The good thing is that they give you all the formulas that you need. It's on the first page of the free response section. You just have to understand which ones are applicable and what’s actually going on with those equations.

So let's first of all think about whether we can calculate Delta G with the information that they've given. If you go to the first page on one of the formulas they give you, you might even remember how you figure out whether a reaction is spontaneous or not and how you could calculate Delta G based on the temperature, the change in entropy, and the change in enthalpy.

You get this formula right over here, and I literally just copied and pasted it from what they give you when you take the test. They give us the change in enthalpy for the reaction, they give us the temperature, 298 Kelvin, they give us the change in entropy for the reaction. This is the change in standard entropy, change in standard enthalpy for the reaction. So using those, we can figure out what Delta G is going to be.

Then we can say, okay, is it consistent? Is it greater than zero, which is consistent with the reaction not being spontaneous, which seems to be what's observed?

Then how can we go from that Delta G to the equilibrium constant? Well, they also tell us that Delta G is equal to this second thing and this connects Delta G and the equilibrium constant. We know some things about R and T. In fact, we know exactly what R and T are if we need them.

So let's go ahead and apply these equations right over here. First, let me write down all the information we have. What is our change in standard enthalpy? We have to figure that out, and we have to figure out what our change in standard entropy is going to be. We know what the temperature is; we know that is 298 Kelvin.

Up here, when they gave us the reaction, they gave us our change in standard enthalpy and our change in standard entropy at 298 Kelvin. This is very convenient. Our change in enthalpy is 45.5 kJ per mole. Our change in standard entropy is 126 J per Kelvin mole.

This is very interesting; this is given in kJ, and this is given in J, so we want to make sure that we're being consistent. Let's make sure everything is in J. So I'm going to write this as the same thing as 45,500 J per mole for this reaction.

So let me write this down: this is 45,500 J per mole, and this is 126 J per Kelvin mole. Now we can figure out what Delta G is going to be. Delta G is going to be equal to our change in enthalpy, which is 45,500 J per mole, minus our temperature, 298 Kelvin, times our change in entropy, which is 126 J per Kelvin per mole.

Notice that the Kelvin cancels out. The units here are going to be J per mole, and then we just figure out the difference. So let's do this. Let me get my calculator out.

So this is going to be, let me multiply these two first. You're going to have 298 times 126. Let's see, I'm going to subtract this from 45,500. So let me just make it a negative and add it to 45,500. That is going to be equal to 7,952.

If we want to stay consistent with how many significant digits or significant figures we have, it looks like it's pretty consistently three significant figures. So we want three significant figures here. We could write 7,900, roughly 7,950.

So our Delta G is approximately 7,950 J per mole, and the fact that this is greater than zero tells us that this is not going to be spontaneous at that temperature. So it's already consistent with our observations.

That's always a good reality check: are the things that you're seeing consistent with what the question is describing? Delta G greater than zero is consistent with the reaction being not spontaneous at 298 Kelvin.

In the videos on Khan Academy on Gibbs free energy, we go into a lot more detail on what this is. One way to think about it is this is the change in energy that's available to do work. If your Gibbs free energy increases over the course of the reaction, that means the products have more energy to do work, and that means you have to put work into it in order for the reaction to actually proceed.

If your Delta G is negative, that means your products have less energy to do work than your reactants, which means that it can release that energy, and it can do work and be spontaneous. So this one, it’s greater than zero, so it’s not going to be spontaneous.

But that doesn't answer the question for us; we want to validate the claim that the reaction has an equilibrium constant less than one at 298 Kelvin. Luckily for us, they also give us the formula that ties our Delta G to our equilibrium constant.

We know the other things, we know R and T. We might not actually even have to think about them because they're not even telling us to calculate the equilibrium constant. They're just saying we'll validate that it's going to be less than one.

So if we take 7,950 J per mole is equal to -RT times the natural log of our equilibrium constant, we can solve for the equilibrium constant. If I divide both sides by -RT, I'm going to get the natural log.

Let me just write it this way; the natural log of my equilibrium constant is going to be equal to 7,950 J per mole over -RT. You could say e, this is just what power do I raise e to get k. So you could say e to the (7,950 J per mole over RT) is going to be equal to K.

We could actually calculate what this is. We know which R to use. We're dealing with J and moles, so it would be the first gas constant: 8.314 J per mole Kelvin. But we actually don't even have to do that because we just have to validate that K is going to be less than one.

What happens if you raise e to a negative exponent? This is going to be a negative exponent. This is positive; that is positive. We know it's 298 Kelvin. Since both are positive, my entire exponent is going to be negative.

So you could say that K equals e raised to a negative number. Remember, if the exponent is zero, e to the zero power is one. e to anything positive is greater than one, and e to anything negative is less than one.

Let me be careful: not less than zero; you actually can't get to less than zero. It's going to be less than one. So this by itself already validates the student's claim.

If you want to go further, you could calculate that you could just say K is equal to e to the (7,950 J per mole over R, which is what, 8.314 J per mole Kelvin, times 298 Kelvin). Those cancel out and J per mole divided by J per mole would cancel out.

So you would get your number, and let’s just calculate it just for kicks, just to really feel good about it. You can see this is going to be dimensionless, and equilibrium constants are dimensionless, so we are going to get this. This is kind of fun.

So let's see, let's do this denominator first. If you have 8.314 * 298, that's going to be equal to that. Now we're dividing by that, so let me take the reciprocal of that and multiply it by 7,950. That is equal to that. Now we want to raise e to the negative of that.

So let's make that negative and now let's raise e to that power. Let's see, we can raise e to that power. So we just press that, and there you have it. This is approximately 0.44, I guess you could say.

So this is, or 44. Let's just say approximately 0.404. Approximately, I already forgot the number. I have a bad memory. 0.044. The whole rules of significant figures get a little bit trickier when you're starting to deal with exponents like this.

We're not going to and they don't ask us to actually calculate the exact value, but hopefully this makes you at least appreciate that the equilibrium constant is for sure less than one.