>>In this module,

we’re going to talk about standard free energy,

equilibria and the relationship to free energy and work and it’s

relationship to free energy. So, delta g zero of a

reaction is the change in Gibb’s free energy

for a reaction under standard state conditions. What does that mean? Well, if it’s a gas, it

means that all the gasses are at one atmosphere of pressure. If it’s a liquid, means that the

liquids are all pure liquids. Solids are pure solids. Elements are in their most

stable form at one atmosphere and 25 degrees Celsius. And, all solutions

are one molar. That’s what we mean by delta

g zero in standard states. This is the thing

you can look up. These are tabulated. Now, if yet for a

chemical reaction, you can calculate delta g zero

for that reaction by looking up the delta g zeros

of the formation of the reactants and product. It ended up that change in, that Gibb’s free energy

works the same way that enthalpy does

and Hess’s law. Both just the same, only it’s

the delta g zero formation rather than delta

h zero formation. So, for our generic

chemical equation, aa plus bb in equilibrium

with cc and dd, we have this equation

describing the, that allows us to calculate the delta g zero

of the reaction if we can look up the delta g zero formations of all the reactants

and products. And, remember, the little

letters are the coefficients in the balanced equation. The big letters represent the

formulas for the elements. So, for example, in the

combustion of methane, this balanced equation

right here, the delta g zero of a reaction would be

the delta g zero formation of carbon dioxide times its

coefficient which is one. So, it’s just this. Plus delta g zero

formation of water liquid, and the state is

really important. That’s going to be different for

liquid than a gas, for instance. Times two because the

coefficient is two. Minus the delta g zero

formation of the reactants. Now, there’s methane, CH4. It’s coefficient is one. Notice I didn’t write

oxygen down here. That’s because the delta

g zero of formation of an element is zero

in its standard state. So, oxygen gas at

standard state. Delta g zero formation

of oxygen is zero. Make sure you remember

that because if you’re doing a

problem on a test and I didn’t give you the

delta g zero formation of some element, you’re not

going to know what to do if you don’t remember

that it’s zero. So, anyway, it ends up when

you look up these numbers here, delta g zero formation

of everything, there are these numbers here. Plug them in, and this

gives delta g zero of a reaction is

negative 1291 kilojoules. What does that tell us? Well, it tells us that

under standard conditions, this reaction is spontaneous

in the direction written. Remember, for a reaction to be

spontaneous, delta g zero has to be less than zero negative. Now, under most conditions, most conditions are not

standard conditions. And, that’s kind of

like a rare situation. It’s not delta g zero,

but rather delta g that determines whether a

process is spontaneous or not. And, they’re, most of the

time, those are different. So, now, these delta

g’s certainly depend upon temperature. And, when you calculate

delta g zero, if you get a large

negative value like we did for the combustion of

methane, that tells you that the products are favored

under standard conditions. And, if it’s a large

positive value, then the reactants are favored

under standard conditions. In other words, if you

get a large delta g zero for a positive delta

g zero for a reaction, that means that if you reverse

the reaction that direction, the opposite direction

will be spontaneous under the standard conditions. Now, it can be shown that, we

don’t show that in this class. It’s just little bit beyond

the scope of this class. But, it can be shown that the

relationship between delta g and delta g zero is

that delta g is equal to delta g zero plus RTln of q where r is 8.314

joules per Kelvin mole. T’s the temperature in Kelvin, and q is our friend the

reaction quotient from before. Now, remember it’s, well,

q looks just like k except that it’s not necessarily

at equilibrium. Memorize this equation, we’re going to see a

couple forms of it. It’s very useful. For example, for the combination

of hydrogen gas and iodine gas to form hydrogen iodide. It ends up, when you look it up

that delta g zero of formation, this is the, by the way, the

equation for the formation of this twice, two moles of it. Anyway, delta g zero for this

reaction, we looked it up. It’s 2.60 kilojoules per mole. Q is, you know, just

like k. It’s products over reactants raised to the

powers of their coefficients. In this case, it’s going

to be in atmospheres. So, the partial pressure

of hydrogen iodide squared over partial pressure of hydrogen times partial

pressure of iodine. And now, okay, so, because

delta g equals delta g zero plus RTlnq, delta g zero’s

positive, right? So, if we didn’t have this

term here, if delta g was equal to delta g zero, this would

not be spontaneous as written. But, because q can

be either negative or positive depending upon the

relative values of the numerator and the denominator, right? If this quotient is less

than one, q will be negative, and if it’s, if the

temperature’s high enough and it’s big enough,

negatively, then, it can overpower this positive

2.60 kilojoules per mole and make delta g’s negative which would make

this spontaneous. So, let’s see. So, at 25 Celsius, let’s

say the partial pressure of hydrogen iodide

was one atmosphere, but the partial pressure

of hydrogen and iodine were three

atmospheres. And, plugging into delta g

equals delta g zero plus RTlnq, there’s delta g zero. Notice we’re given delta g

zero in q joules per mole but because r has joules

per kelvin mole in it, I converted the kilojoules here

to joules just by changing it to times ten to the third. RTq is partial pressure

of hydrogen iodide squared over partial pressure

of hydrogen over partial pressure of iodine. Plug these numbers

in and check me. Make sure I did my

math correctly, but I believe we get

negative 2.84 times ten to the third joules per mole or a negative 2.4

kilojoules per mole. Now, that’s spontaneous under

those conditions at 25 Celsius and with these partial

pressures. Now, at equilibrium,

that’s when q is equal to k and equilibrium means

delta g is equal to zero. So, we can say zero equals

delta g zero plus RTlnk. Rearranging a little bit, we get delta g zero

equals negative RTlnk. Look right there. We have a relationship

between delta g zero and our equilibrium constant. Okay. Also, we can say because

we know delta g is delta h minus t delta s. So, delta g zero is

delta h zero minus t delta s. That’s still equal

to negative RTlnk. Rearranging a little bit, we

can get this equation here. This is nice. This equation here is in the

form of a straight line where ln of k would be y. The slope, m,

would be negative delta h zero over r. x would be one over

the temperature in Kelvin, and y would be delta

s zero over r. So, this means if you make a plot of

the natural log of k versus one over the temperature in Kelvin,

get the best fit straight line, the slope of that straight line

will be negative delta h zero over r. So, that means you

can get delta h zero just by multiplying the slope

by negative r. And, the y intercept is

delta s zero over r, so you can get the delta s zero

by multiplying the y intercept by r. Remember, r is 8.314

joules per Kelvin mole. All right. So, that’s real useful you

should memorize this stuff right here. You use it for sure. Now, finally, why is

it called free energy? Because it’s energy that’s

not because it’s energy that doesn’t cost money. No. It’s because it’s

energy that’s available, free to do work. As a matter of fact, the

maximum amount of work possible that can be done from a

process at constant temperature and pressure is equal to delta

g. But, that’s theoretical. In any real process, you will

never be able to get this amount of work out from that

process because some of that free energy

goes into not work but rather it’s lost as, you

know, dissipation as energy or friction, things like that. But, this gives you an upper

limit on the amount of work that you can do from the

free energy from a process. Subtitles by the Amara.org community