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Another thermodynamic value we look at is Gibbs

free energy. The objective of this unit is to learn how to use the enthalpy of a system

and the entropy of a system to calculate the Gibbs free energy

and determine the spontaneity with respect to Delta G. Let's start where we

ended the last module.

Delta S of the universe equals delta S of the system plus delta S of the surroundings

We know that for a spontaneous reaction this value must be greater than 0.

However, we know that the delta S of the surroundings is equal to the negative

H of the system

divided by the temperature, so we can substitute that into our expression.

so now we have delta S of the universe equals delta S of the system

minus delta H of the system over T.

To be spontaneous this process still needs to be greater than 0.

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define this portion of my expression, delta H of the system minus T delta S

of the system as delta G, so I can use this to calculate delta G of the system

and what I see is that if Delta G of the system is less than 0

then I will have a spontaneous process because this corresponds to Delta S

of the universe

being greater than 0. If I make constant temperature

I see that I get delta G equals delta H minus T Delta S

because I see the difference in products and reactants for the Gibbs free energy

for Delta H and for Delta S, but my T value remains constant.

So, what is free energy? We talk about Gibbs free energy.

Well, it's the energy actually available or free

to do work. When we talk about the enthalpy of a reaction we know that energy is

given off for energies gained, but it doesn't mean that all that energy is usable for doing

work, only part of it is because some of it goes to the dispersal of

energy so what we want to worry about is what is the Gibbs free energy

that's the amount we actually have available to do work.

If there's no subscript as there aren't here we can assume that we're always

talking about the system.

So, delta H equals delta H of the system. The the same goes for delta

G and for Delta S. These are all in terms of the system.

So what does Delta G tell us?

Well, if delta G is less than 0 we have a reaction that is spontaneous in the forward direction.

If its greater than 0 it's spontaneous in the reverse direction or non spontaneous

in the forward direction.

If we have a delta G value equal to 0 then we know we have a system at equilibrium.

because it's neither spontaneous are non spontaneous in either direction.

So let's look at what happens when we have various values of Delta S and Delta H.

We could have both of these values as less than 0 or greater than 0,

so what we want to look at is the different combinations to determine the

relative value of delta G.

When will the process be spontaneous and when will it be non-spontaneous.

Let's first look at delta H is less than 0 and Delta S is a greater than 0.

so if delta H is less than zero then what I see is that my delta H value is negative

and my Delta S value is positive because it's great in 0

but the term, the second term, will be negative

so delta G will always be less than 0 in that case.

If I look at

Delta H greater than 0 and Delta S less than zero

I can't say delta H is greater than zero so it's positive

Delta S is less than 0, so it's negative,

but when I take the negative of a negative I get a positive value

and in that case I see that delta G is going to be positive.

Now we're going to look at an example to figure out what happens

when Delta S is less than 0 and Delta H is less than 0.

We have our equation delta G

equals Delta H minus T delta S. If Delta H is negative

and Delta S is negative which statement is true?

We know the sign of Delta G will depend on the temperature for Delta H and

delta S both less than zero.

The question becomes when will it be positive and when will it be

negative? We want to look at the difference in temperatures. When we're at a

high temperature

versus when we're at a low temperature. What I can do is

say I know I have a minus sign in front and this term,

delta S is negative, therefore minus T times a negative delta S will be greater

than 0 or have a positive value.

As T gets higher, a higher temperature, the T Delta S term becomes

more positive. It is increasing

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Now we need to see what happens at a low temperature.

If I have minus T delta S as greater than zero which I'll have because delta S itself is negative,

so that means that term will be positive,

at low temperature T delta S becomes less positive,

it's decreasing in value,

and so when I compare that to the negative delta H value

it's not going to be enough to compensate for that and as a result the

Delta H

overrules the Delta S term and because of that we end up with a Delta G that's

less than zero,

because the value of the delta H was more negative

than the T delta S term was positive.

At low T, delta G is less than zero and therefore it will be spontaneous

in the forward direction.

When we look at the last option of delta S greater than zero and Delta H greater

than zero

we can go through many of the same arguments, all be it in the opposite way as

we did for delta S

and delta H both being negative values.

We're not going to go to the explanation here, but what we find is that low T

means delta G will be greater than zero, or non spontaneous,

at high T delta G will be less than zero and therefore would be a spontaneous process.

Now let's look at an example where we actually find the value of delta G.

We know that Delta G equals

delta H minus T delta S.

I know that my temperature has to be units of Kelvin. It's given in

Celsius, so I'm going to say 25

plus 273 equals 298 Kelvin.

Then I'm going to notice

that my enthalpy value is in units of kilojoules

and my entropy value is in terms of joules, so I need to make sure I put those

in the same units.

Now I can can substitute in the values I know. I have

delta G equals Delta H, so

-92.22 kilojoules

minus my temperature in Kelvin, which is 298,

times my delta S value, which I'm going to go ahead and convert to

units of kilojoules per Calvin by dividing by thousand, so I have 0.1

0.19875 kilojoules per Kelvin,