Chem102 Standard Free Energy Equilibria and Work

>>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 community

Leave a Reply

Your email address will not be published. Required fields are marked *