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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?
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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.
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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.
Dr. Allison SoultLecturer
Dr. Kim WoodrumSr. Lecturer
