okay let's practice calculating free energy change
consider the following enthalpy and entropy data for water in both the
liquid and gaseous state so here is that data we have Delta H values for the
liquid and gas form of water and we have Delta S values for the liquid and gas
form of water so the question is what is the free energy change for the
vaporization of water at room temperature so we're looking at the
process of liquid water going to water vapor or water in the gas phase and we
want to know the free energy change associated with that process so if this
is confusing go ahead and check out my tutorial on thermodynamics and when
you're ready give this a try.
so let's put our data up top and then just to
remind ourselves we are looking at this process we are looking at water in the
liquid phase going to water in the gas phase that is a vaporization now we want
to calculate the Delta H and the Delta S associated with this process so for
Delta H since we have this enthalpy of formation data we know that we can take
the enthalpy of formation of products and subtract from that the enthalpy of
formation of reactants and that will give us the Delta H for that process so
we can say that the standard enthalpy is equal to negative 241.82, that's the
enthalpy of formation for water vapor, minus negative two hundred eighty six point
eight three that's the enthalpy of formation of liquid water and that will
be equal to 45 point zero one kilojoules per mole so that is the standard
enthalpy change for this vaporization now we can do the same thing for entropy
because we have this entropy data so standard entropy change will be 188.8
that is the data given for water vapor minus 70 point zero that is the
data given for liquid water and that will be equal to one hundred eighteen
point eight joules per mole Kelvin so now that we have enthalpy and entropy
data what are we going to do with that well we know that the standard free
energy change is going to be given by this expression so we have Delta G
equals Delta H minus T Delta S and so we now have all of this information so
let's plug some things in Delta G is going to be equal to Delta H which is
forty five point zero one kilojoules per mole minus T which is going to be 298
Kelvin remember we said this is done at room temperature that's 298 Kelvin and
then for Delta S we're gonna take the Delta S value that we had but this is
very important we need to make sure that the units on Delta H and Delta S agree
Delta H had kilojoules so the easiest thing to do here is to convert our Delta
S from joules per mole k to kilojoules per mole K which means we've to divide
by a thousand so that is going to be zero point one one eight eight
kilojoules per mole Kelvin a lot of students forget to do that and that will
give you an incorrect answer so make sure that your units agree and if we do
the arithmetic that gives us nine point six kilojoules per mole so that is our
Delta G value we want to understand that that is positive and so that corresponds
to a non spontaneous process so we have an endothermic process meaning it is
enthalpically unfavorable right from the standpoint of enthalpy this is an
unfavorable process and so even though the entropy change is positive meaning
it is entropically favorable because the water molecules are going from the
liquid phase to the gas phase there is a greater dispersal of matter so it is
entropically favorable the enthalpic unfavorability outweighs the entropic
favorability to give us a positive Delta G value making this a non spontaneous
process and that makes sense because if you have a glass
water sitting on the table in room temperature it does not spontaneously go
all into the gas phase it is going to pretty much stay just as
liquid water so that makes sense with our intuition and that is the answer for this problem.