# Video: Identifying the ΔH, ΔS, and ΔG Conditions Which Guarantee a Reaction Will Be Nonspontaneous in a Set of Conditions

Which of the following sets of conditions guarantees that a reaction will be nonspontaneous? [A] ΔH (+) and ΔS (−) [B] ΔH (−) and ΔS (+) [C] ΔH (+) and ΔS (+) at a low temperature [D] ΔH (−) and ΔS (−) at a high temperature [E] ΔG (−)

04:40

### Video Transcript

Which of the following sets of conditions guarantees that a reaction will be nonspontaneous? A) ΔH positive and ΔS negative. B) ΔH negative and ΔS positive. C) ΔH positive and ΔS positive at a low temperature. D) ΔH negative and ΔS negative at a high temperature. Or E) ΔG negative.

In chemistry, when we want to determine if a process or reaction is spontaneous or not, we generally use the Gibbs free energy, which is equal to the enthalpy minus the temperature times the entropy. If we calculate the change in Gibbs free energy for a reaction or process, and we find that it has a negative value, that reaction or process will be spontaneous. If the change in Gibbs free energy is positive, however, that process will be nonspontaneous.

With this in mind, we can immediately rule out answer choice E as we’re looking for conditions that will guarantee a reaction will be nonspontaneous. And by definition, a change in Gibbs free energy that is negative will be spontaneous. So, now, let’s take a look at the rest of our answer choices. We’ll have to play with the signs of the change in enthalpy and the change in entropy to figure out which one will give us a positive change in the Gibbs free energy. Because a positive change in Gibbs free energy corresponds to a nonspontaneous process.

Before we begin, we should remember that the temperature in the Gibbs free energy expression is the temperature in kelvin, which will always be positive. So, we don’t need to worry about that sign being positive or negative when we’re doing this problem.

For our first process, we have a positive change in enthalpy and a negative change in entropy. So, we have a positive number and then we’re subtracting a negative number from it. Subtracting a negative number, of course, makes a positive number. So, we would expect these conditions to give us a positive change in the Gibbs free energy. So, we’ve already found our correct answer since we’re looking for a nonspontaneous process. And nonspontaneous processes correspond to a positive change in Gibbs free energy. But still let’s walk through our other answer choices just so we make sure we understand the logic behind this kind of problem.

In our second answer choice, we have a negative change in enthalpy and a positive change in entropy. So, we have a negative number and we’re subtracting a positive number from it. Subtracting a positive number from a negative number would, of course, just give us a larger negative number. So, under these conditions, we would expect that the change in Gibbs free energy would be negative. So, these conditions would most likely give us a spontaneous process.

For our next answer choice, we have a positive change in enthalpy and a positive change in entropy as well. But we’re given the additional information that this process is occurring at a low temperature. Since the entropy is being multiplied by the temperature, we would expect that this means our second term might be small. So, we have a positive number. And from it, we’re subtracting a relatively small positive number. So, we might expect that we’ll have a positive change in free energy under these conditions.

However, we don’t know the exact numbers that we’re working with. It could be that the value for the enthalpy is relatively small, so we end up with a negative number for the value of the Gibbs free energy. We just don’t know. We don’t have enough information. So, this set of conditions does not guarantee that the reaction will be nonspontaneous.

In our last answer choice, we have a negative change in enthalpy and a negative change in entropy. We’re also told that this process occurs at a high temperature. Again, since entropy is being multiplied by the temperature, we expect that the second term in this expression will be large. So, we have a negative number. And from it we’re subtracting a large negative number. We might expect that the change in the Gibbs free energy under these conditions will be positive, so we’ll have a nonspontaneous reaction. But just like the last answer choice, we just don’t have enough information to guarantee that that’s the case.

So, like we’ve discussed earlier, answer choice A is the only set of conditions where we can guarantee that we’ll have a positive change in the Gibbs free energy. So, we can guarantee that the reaction will be nonspontaneous.