5.05 Entropy of the Universe

Loading...

En provenance du cours de University of Kentucky

Advanced Chemistry

414 notes

University of Kentucky

Advanced Chemistry

414 notes

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

À partir de la leçon

Thermodynamics

The overarching theme of thermodynamics is the prediction of whether a reaction will occur spontaneously under a certain set of conditions. Entropy and Free Energy are defined and utilized for this purpose.

Rencontrer les enseignants

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

In this module we're going to look at how we find the entropy of the universe

to determine whether or not a process is spontaneous.

Our objective is to understand how temperature affects the entropy of the

surroundings,

and in turn, how that affects the entropy of the universe.

When we look at entropy of the universe we know that we must have a positive value

in order to have a spontaneous process.

We also know that the change in entropy of the universe is equal to the change in

entropy of the system

plus the change in entropy of the surroundings. these are not necessarily

negatives of one another. What we have to look at

are the individual components to determine whether or not a process is

spontaneous.

when we look at the values at the delta S of the system and delta S of the

surroundings

we see that we can have both values as positive numbers, both fighters negative numbers,

or have one value that's positive and one value that's negative.

It's the balance and the signs of these numbers that determine whether or not a

process is spontaneous. Let's take an example where

this process is not always spontaneous. Looking at the freezing of water we know

that at lower temperatures this happen spontaneously. At higher temperatures

it is a non spontaneous process, so what we have to look at is

where's the energy going? We're dispersing energy

so we have to look at the temperature of the surroundings to see

which one is going to favor

the spontaneous process. For both of these we see that the Delta S of the system

is the same, but we see that the Delta S of the surroundings

is different and therefore the Delta S of the universe ends up being

different. It's the Delta S of the surroundings that depends on the temperature

because that's going to determine how that energy is dispersed.

So remembered that energy tells us about the dispersal of energy,

and the qualitative value tells us that entropy is a measure of energy dispersed

per unit of temperature.

As the temperature increases the amount of entropy for a given amount of

energy dispersed decreases, therefore, something at a higher temperature

has a lower impact and will not change the Delta S of the system as much.

As the temperature decreases or drops the amount of entropy for a given amount of

energy dispersed increases.

This means that at a lower temperature

we have a higher impact, so looking at the relative values of the Delta S

of the system

versus the delta S of the surroundings the temperature will determine

how much the Delta S of the surroundings will impact

the value of the delta S of the system and therefore how it will impact the

delta S of the universe.

So, how do we find the delta S of the surroundings? We know it's dependent on

temperature but we have to find the actual value of it.

For endothermic reactions, so that will be a reaction

where Delta H is less than zero,

so a negative value, we see that the delta S of surroundings is greater than zero

because an exothermic reaction is releasing energy. We're dispersing that energy

and so the Delta S of the surroundings is increasing.

An endothermic reaction when Delta H

is greater than zero, or a positive value,

shows we're absorbing energy from surroundings. We're

decreasing the dispersal of energy, and therefore delta S of the surroundings

will be

less then zero. But this doesn't give us the exact value of delta S of the

surroundings

to do that we have to combine both the enthalpy and the temperature

to understand how they are related.

So if I want to look at the relationship between

enthalpy and delta S of the surroundings I have to take the negative

of the delta H value, the negative enthalpy of the reaction of the system,

and divide it by the temperature. Remembered that the temperature must be

in units of Kelvin.

So now

we can find delta S of the universe if we know delta S of

the system and delta S of the surroundings. We have our way to find

delta S of the surroundings.

We looked earlier at how to find delta S of the system

by looking at products minus reactants, and from that information we could then

find delta S

of the universe, and then we can finally determined is a reaction

spontaneous or not

based on the Delta S of the universe.

so as we said before a small T, or a low temperature,

we have a large delta S of surroundings. For a large T

we have a small Delta S of surroundings.

Let's look at an example. Under which of the following conditions will Delta S

of the universe

always be positive?

So for delta S of the universe to be positive we know that our equation for that is

delta S of the universe

equals delta S of

the system plus delta S

of the surroundings. So, if delta S of the system is positive

and delta S of the surroundings is positive then Delta S of the universe

will also be positive

making B the correct answer. If we look at when both values are negative, if delta S

of the system is negative and delta S of the surrounding is negative

then our delta S of the universe will also be negative.

however, if we look at the options were one value is positive

and one value is negative we can't know

whether Delta S of the universe will be positive or negative

because it depends on the relative values of

those delta S of the system and delta S of the surroundings.

So, when will delta S of the surroundings be positive?

Delta S of the surroundings will be positive when we have an exothermic reaction

Remember, that for an exothermic reaction our delta H value

is less than zero. It's negative. so if I put a negative value

into my equation I'm going to end up with

a positive delta S value, or a value that is greater

than zero.

Let's look at a specific example of finding delta S of the surroundings.

We know that delta S of the surroundings

equals minus delta H

over T, remember that has to be in Kelvin.

So I'm going to take delta S of the surroundings

equals negative 34 kilojoules per mole,

and I'm actually going to go ahead and multiply that by a thousand because I

want to get this in the unit to joules because that's typically how

entropy is reported, in joules, and then I'm going to divide

by my temperature, which I have to convert to Kelvin,

If I add 273. Now, I could take delta S of surroundings

equal to -34,000 joules per mole

over 298 Kelvin

and what I find is that delta S of the surroundings

is equal to negative 114

joules per mole Kelvin.

now this doesn't necessarily mean that the process will be non spontaneous.

What's going to matter is what the change in entropy is of the system

so we can see how that compares to our negative 114 joules

for the Delta S of the surroundings.

Now we've talked about a lot of entropy terms here and we talked about a lot of

greater than zero and less than zero and what they exactly mean so we've got a little summary here

to go through.

Remember, that in order for reaction to be spontaneous delta S of the universe

must be greater than 0.

That's the second law of thermodynamics.

When I look at delta D of the system I see that it's greater than zero,

when I have more disorder, more dispersal of energy.

It's less than zero when I have less disorder and less dispersal of energy.

for delta S of the surroundings if I have an exothermic or negative value for

delta H

I'm going to end up with a positive Delta S of surroundings,

and for an endothermic process that has a positive delta H

I'm gonna end up with a negative Delta S of surroundings.

Next we're gonna look at Gibbs free energy.

Explorer notre catalogue

Rejoignez-nous gratuitement et obtenez des recommendations, des mises à jour et des offres personnalisées.

Coursera propose un accès universel à la meilleure formation au monde, en partenariat avec des universités et des organisations du plus haut niveau, pour proposer des cours en ligne.

© 2019 Coursera Inc. Tous droits réservés.

Télécharger dans l'App Store

Disponible sur Google Play