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Another thermodynamic value we look at is Gibbs

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

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

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

Â ended the last module.

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

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

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

Â H of the system

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

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

Â minus delta H of the system over T.

Â 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

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

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

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

Â of the universe

Â being greater than 0. If I make constant temperature

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

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

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

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

Â Well, it's the energy actually available or free

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

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

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

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

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

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

Â talking about the system.

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

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

Â So what does Delta G tell us?

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

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

Â in the forward direction.

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

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

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

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

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

Â relative value of delta G.

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

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

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

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

Â but the term, the second term, will be negative

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

Â If I look at

Â Delta H greater than 0 and Delta S less than zero

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

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

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

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

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

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

Â We have our equation delta G

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

Â and Delta S is negative which statement is true?

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

Â delta S both less than zero.

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

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

Â high temperature

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

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

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

Â than 0 or have a positive value.

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

Â more positive. It is increasing

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

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

Â so that means that term will be positive,

Â at low temperature T delta S becomes less positive,

Â it's decreasing in value,

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

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

Â Delta H

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

Â less than zero,

Â because the value of the delta H was more negative

Â than the T delta S term was positive.

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

Â in the forward direction.

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

Â than zero

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

Â we did for delta S

Â and delta H both being negative values.

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

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

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

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

Â We know that Delta G equals

Â delta H minus T delta S.

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

Â Celsius, so I'm going to say 25

Â plus 273 equals 298 Kelvin.

Â Then I'm going to notice

Â that my enthalpy value is in units of kilojoules

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

Â in the same units.

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

Â delta G equals Delta H, so

Â -92.22 kilojoules

Â minus my temperature in Kelvin, which is 298,

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

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

Â 0.19875 kilojoules per Kelvin,

Â