Lecture 15. Getting to know the Gibbs Energy.
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Gibbs free energyUrocortinElectronic cigaretteTopicityLecture/Conference
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Gibbs free energyMill (grinding)Angular milHost (biology)CarbonLecture/Conference
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Transcript: English(auto-generated)
00:06
Ok, how are you guys doing? Got those mid-quarter doldrums, I know how you're feeling.
00:26
Ok, this is a Chapter 16 topic, we're really cranking away on Chapter 16 here, edging our way towards the end of thermodynamics which I hope will come probably in the middle
00:46
of next week. Now we posted the quiz scores this morning, this was the hardest quiz, might have been
01:01
too hard, first question was pretty easy, didn't you think? So these guys got A's, these guys got B's, there's a few C's here, we usually have hardly any C's.
01:22
Ok, quiz 5 is Friday, the first two are supposed to be easy, the first two problems, in this case the second one was actually not so easy. It was the body temperature one, ok I also posted the key in case you want to look at
01:41
that. So we won, and what's even sweeter is we beat SC, very very sweet. I'm always amused by the fighting anteater, the anteater is the most docile animal that
02:03
we know of in the animal kingdom, but to dress him up for the athletic department, here he is fierce, there's no such thing as a fierce anteater, in nature they're not known to be fierce.
02:20
Ok, so we're going to review a little from Friday, why does that say Wednesday? We're going to talk about how the Gibbs free energy varies with temperature and pressure, we'll do a couple of examples, just sort of easy ones to ease into this subject. Ok we'll do a bunch more examples on Wednesday. Alright, so on Friday, not Wednesday, we said look, there's three types of systems, ok,
02:55
and uniquely for this guy right here, it's an isolated system, there's no energy or
03:01
matter exchanged with the environment, the surroundings, we don't have to consider anything except the system when we think about the spontaneity of processes that occur within it. Ok, it's blocked off to the surroundings, it doesn't even know about them. And so we can say any process that has an entropy, positive entropy change is going
03:24
to be a spontaneous process for an isolated system. We don't even have to think about the surroundings, they're not part of our thought process. But we don't have isolated systems in chemistry too often, they're almost always in communication with the environment.
03:41
And so we have to consider open systems and closed systems as well and in those cases because there is communication with the surroundings, it's the total entropy, surroundings plus system that matters, ok, in terms of figuring out if this process is spontaneous.
04:00
Notice that the focus is totally on entropy, alright, we're not saying anything about the energy, the energy can do anything it wants. We're only focusing attention on the entropy to understand whether these processes are spontaneous or not. Ok, so we've got to have this term in here for the surroundings, we didn't need it for the isolated system.
04:23
Ok so now we're just going to do a little algebra, we're going to move the surroundings over to the right hand side, put a negative sign in front of it. And then we're going to remember that dS is minus dq over T and so we can make that substitution for the surroundings right here.
04:40
And then we're just going to remember that q is a conserved quantity, in other words if plus q is heat entering the system that heat had to come from the surroundings and so the surroundings has to be minus dq, there's conservation of q. And then we have to think back and remember that du is dw plus dq and so we can just
05:01
solve for dq in this expression, du minus dw. And if we plug that guy in for dq we get this guy right here. And if we consider only pressure volume work this dw is pdv.
05:23
Ok and so then we're going to multiply by T surroundings and move it over to the left hand side so we get rid of this T surroundings now, now it's over here and we just have du plus pdv. We're going to drop that cis subscript, if you don't see a subscript just assume we're
05:40
talking about the system. Ok so this is the pink equation, it only took us about four steps to get there from conservation of entropy, well not conservation of entropy, from entropy dictating which process is spontaneous or not.
06:01
We kept coming back to this pink equation on Friday and deriving different thermodynamic state functions from it. In fact we showed that if the process occurs under conditions of constant volume and constant entropy then it's the internal energy that tells us whether the process
06:22
is spontaneous or not. And if instead the process occurs under conditions of constant pressure and entropy, why it's the enthalpy that's going to tell us whether the process is spontaneous or not. But unfortunately we don't encounter these two sets of conditions very often.
06:42
It's virtually never the case that the entropy is constant, if you're a chemist you can ask, you know, how do you do that? How do you do an experiment with constant entropy? I don't know the answer. So when you're doing an experiment and you want to understand whether it's spontaneous or not, the chemistry that you're looking at, it's unlikely that you're going to be
07:01
paying attention to these two variables to figure that out. They're not going to help guide your decision making process in figuring out whether your chemistry is spontaneous or not. That's what we care about here. So we need some other state functions and we talked about one, the Helmholtz energy.
07:25
In chemistry temperature is frequently constant but not the entropy. Constant temperature, lots of constant temperature chemical processes that we can think about. So let's consider the case where DT is zero and the volume is zero.
07:41
We're going to need for two things to be zero, otherwise we're not going to end up with a state function. And so we'll also define a new state function A which is going to be called the Helmholtz energy. It's going to be defined as the internal energy minus TS. And so if we take the derivative now to get DA on the left hand side, we're going to
08:02
get DU and we're going to have DTS and so we can split that into two terms, DTS minus SDT. And then we can just solve for DU in this expression, DU is going to be equal to DA plus TDS plus SDT and of course the next thing that we do is we plug this thing
08:25
into the pink equation, put this expression for DU into the pink equation and then look for the terms that are going to cancel for us. So we've got DA, TDS, SDT and our expression for pressure volume work from the pink equation.
08:47
Okay and the first thing that we notice is that we got TDS here, we got TDS here. Now these two T's are different in principle. This is a T for the surroundings, this is a T for the system but as these, as we converge
09:00
on equilibrium these two temperatures will become very, very close. And under those conditions we can expect these two terms to cancel for us. And then under conditions where we said DT is zero and DV is zero, there's DV, we can cancel out that term and DT we can cancel out that term.
09:22
And we're just left with DA, there's nothing else left and so it's going to be DA is less than zero and so this Helmholtz energy is going to be a state function that we can use to tell whether the chemistry that we care about is spontaneous or not when temperature
09:42
and volume are held constant. And in the laboratory we can enforce those limits, we can do an experiment at constant temperature maintaining the volume constant, let the pressure do whatever it wants, okay we need one of these things, alright it's got a defined volume and it's built like
10:02
a tank and so even if the pressure changes a lot we're going to enforce constant pressure and in principle it's the Helmholtz energy that will tell us whether a reaction in this PAR bomb is going to be spontaneous or not. We would want to, if we do an experiment in here we would want to use the Helmholtz
10:22
energy to figure out whether it's spontaneous or not. Now if you do undergraduate research, how many people have done undergraduate research? How many people have seen a PAR bomb? Alright a few of you, who do you guys work for, you all work for the same person?
10:43
Oh, he's got a PAR bomb in his lab? Okay, so it's not completely impossible that you would use one of these things, alright they are in rather common use but I would say probably 99.9% of all the chemistry that
11:06
we're likely to do is not going to be in a PAR bomb, alright 99.9%. So we need a different thermodynamic function, the Helmholtz energy is fine but constant
11:21
volume is inconvenient for us to use because we needed a PAR bomb to do it in many cases. In chemistry it's even more useful to make predictions about processes occurring at constant pressure and temperature because that's dead easy. We live in an environment of quasi-constant pressure and so we can do chemistry that's
11:43
open to the environment and make predictions about whether it's spontaneous or not. To that we're going to use something called the Gibbs energy. Alright, we're going to define it as H minus TS, okay we're going to do the same kind of algebra we did before, DG is DH minus DTS and so we've got two terms here now and
12:06
then we're going to think back to Friday when we wrote an expression for DH, we said DH is DU plus PDV plus VDP, okay and so we can just plug that in for DH here.
12:22
Now we've got this long thing here that's equal to DG, okay and so once again we're just going to solve for DU, alright put all of this other stuff on the right hand side and then once we've got DU we're going to just plug it into the pink equation
12:41
again, there it is, put all of this stuff in for DU, we get this long thing here, okay and some of these terms are going to start cancelling for us as usual, PDV, PDV,
13:02
right remember these two pressures are not in principle identical, that's the system pressure, that's the surroundings pressure, alright but in the limit of equilibrium they will be the same, right TDS, TDS, same idea there and then because we're
13:21
talking about G we're going to make P constant and so we're going to lose that guy and DT so we're going to lose that guy so everything cancels out here except DG which is going to be less than or equal to zero and so that's going to be the state
13:42
function that we're going to want a key on most of the time as chemists, alright now if you're a physicist, you're some other kind of scientist these other state functions might be more important to you under other sets of conditions but for chemists it's all about the Gibbs free energy, the Gibbs energy we're not supposed to call it the
14:01
Gibbs free energy anymore, it's just the Gibbs energy, okay now I know that's tedious but this is important, this is actually one of the more important things in thermodynamics that we need to be able to understand, alright here's where we're doing chemistry and this guy right here we're open to the atmosphere and the Gibbs
14:24
function is going to tell us whether this blue stuff here is going to react spontaneously, alright we don't need the power bomb, okay so today and last
14:41
Friday we've taken the consideration, we've taken the condition for spontaneous change for non-isolated systems, we consider the total entropy change system plus surroundings that's got to be greater than or equal to zero and from that we derived all of these different conditions that apply for these different constraints, volume and entropy, temperature
15:05
and volume, pressure and entropy, temperature and pressure, we've got these four different and what I've told you today is look these two are not super useful to us as chemists these two are more useful and this one is way more useful than that, alright we derived
15:23
these all, we didn't have to assume anything, I'm very proud of that, it's hard. Okay now these conditions here also serve to tell us whether the system is proceeding towards equilibrium, it not only tells us whether the chemistry is spontaneous it'll
15:49
tell us whether the system is proceeding towards equilibrium or not, for example what I'm plotting here is the Gibbs energy for some chemical process and on this axis
16:05
I have the reaction coordinate so this represents reactants right here and this represents products this is 100% products this is 100% reactants but as you move along the axis in this direction we're converting reactants into products, alright that's what I mean
16:22
by a reaction coordinate, sometimes we'll call this reaction coordinate x or chi, okay reactants getting converted to products, very generic, what does dG that should be P and T less than zero, there it is, P and T, alright so first of all this difference
16:48
here between the reaction and product Gibbs functions that's the delta G reaction, makes sense, now let's consider a process that starts right here and ends right here, alright
17:06
we can ask is such a process going to be spontaneous or not, well we have a criterion here if we just change that V to a T, alright we know dG should be less than zero, okay
17:23
and so we can say G final minus G initial is that dG final minus initial is that going to be less than zero or greater than zero, what do you think, yeah it's a small number
17:41
minus a bigger number and so that difference is going to be negative isn't it, alright and so we would predict that's a spontaneous process, yes dG at constant T and P is less than zero, what about this guy, same conclusion, what about this guy, no, final minus initial
18:07
is going to be a positive number now, alright so that's not going to be a spontaneous process going from here to here, no, alright and what about that guy, yes, final minus
18:23
initial is going to be negative again and so that should be spontaneous, alright so basically what we're concluding is that if you're over here we're going to go spontaneously downhill in this direction and if you're over here you're going to go spontaneously downhill in this direction and that this minimum here in the Gibbs energy is going to indicate
18:46
the equilibrium position of this reaction, alright it's the point where dG over dx where x is now my reaction coordinate is equal to zero, at that point there's no more driving force for spontaneous change, we're at equilibrium, okay now amongst these four thermodynamic
19:12
potentials U, H, A and GG will be by far the most important to us, yes, yes, yes, how does G depend on temperature, alright how does the Gibbs energy depend on temperature,
19:28
well, that's a rather important thing for us to understand because as chemists if we want to accelerate a reaction G is going to tell us whether the reaction is spontaneous or not, alright we want to understand how temperature will influence that spontaneity,
19:44
G is equal to H minus TS we know that and so we can take the derivative immediately dG dT even I can take this derivative I get minus S, okay and so what this tells us is two things, first of all since we know that S is always a positive quantity there's no
20:03
such thing as negative entropy, S is always a positive quantity, right that tells us that G has to decrease with increasing temperature because that derivative is always going to be negative, right that's kind of surprising the Gibbs energy is going to go down as
20:26
the temperature goes up, that's counter intuitive, don't all energies go up when you increase the temperature, not this one, right the Gibbs energy goes down as you increase the temperature, not only that but the rate of change of G with temperature
20:46
is greatest for systems having high entropy, right the higher the entropy the greater the change in G is going to be with temperature, well what kind of systems have high entropy,
21:00
well gases, my laser pointer is dying, gases have the highest entropy and so their rate of change of the free energy with temperature is going to be the highest then liquids then solids, solids have the lowest entropy, okay so this plot is right out of your
21:23
chapter sixteen, gases biggest slope, right here's the Gibbs energy on this axis, here's these guys it's going down but it's going down at a rate that depends on the state,
21:41
gases show the largest decrease in Gibbs energy with temperature, liquids next, solids show the least, okay so a couple of things are surprising, I mean one thing that's surprising for sure is that the Gibbs energy goes down with temperature, alright it's an unusual energy isn't it that goes down with temperature, okay so we can evaluate this derivative
22:11
and then we can go back to this equation right here and we can just say we can solve for minus S, alright if I solve for minus S here I'm going to get G minus H over T
22:21
just solving for minus S in that equation right there, okay so I've got DG over T at constant P is G minus H over T and then we can rearrange that just split this into two terms move G over T to the left hand side, I don't know why we actually
22:51
did that because I don't think we need this result here, alright maybe we'll come back to this in a second but let's just look at this for a second, this is the derivative
23:07
of G over T, I don't know if that has anything to do with you, this is just the derivative of G over T, if I use the quotient rule to evaluate this derivative I've got 1 over T times the derivative of G with respect to T and I've got G times
23:23
the derivative of T with respect to T, right? Two terms in my quotient rule expansion, okay now the derivative of 1 over T is just minus 1 over T squared, right? Okay and so that's that derivative right there and this guy if we factor out 1 over
23:49
T so I'm going to pull the 1 over T out of both of these two terms and put it right there, alright now I've got this expression here and that is just the entropy,
24:01
right the derivative of the Gibbs energy with respect to T at constant P that's the entropy, okay and so I can plug that in to this expression right here, I still got G over T and then I can move, maybe that's why I did it, this alright is just
24:33
plugging in for G from that equation two slides ago, okay and so this is S over T,
24:45
this is S over T so we're going to get rid of the S over T, we're just going to be left with H over T squared, alright and this is your equation 15.62B, this is the Gibbs-Helmholtz equation which is important because it allows us to measure H
25:06
by looking at the temperature dependence of G and the temperature dependence of G is something we're going to be able to measure experimentally, alright and so we can get H directly from that using this Gibbs-Helmholtz equation. Now if this is a delta H and
25:29
that's a delta G this equation still is fine, okay so now let's ask some questions about the Gibbs function. We already asked a question about the temperature dependence,
25:43
we said the Gibbs function goes down with increasing temperature, surprising. The rate at which it goes down depends on the entropy, higher the entropy the faster the temperature rate of change of the Gibbs function. What about pressure? Alright we've got this expression here for DG and if we want to look at this at constant temperature we can
26:09
say DT is zero, alright and so that term is just going to go away, we've got DG is VDP and so to find out what the free energy, free energy, the Gibbs energy change
26:22
is there's the Gibbs energy at P final minus the Gibbs energy at P initial I can just integrate this VDP from initial to final pressure and if I know what that is, if these are molar quantities of course there's going to be an M there, an M there and an M there if these are all molar quantities and so this equation is not super useful to
26:45
us unless we know how this volume changes with pressure. But one thing that's obvious is for phases like solids and liquids that are essentially incompressible VM is virtually constant independent of pressure. There's not much compressibility of a solid phase
27:03
or a liquid phase right and so there isn't much pressure dependence of VM and so we can write a simpler expression. If we can pull VM into the front of this integral sign because it's constant then the integral just turns into this and we've got an extremely simple
27:22
expression that allows us to evaluate the pressure dependence on the Gibbs energy. It's just the molar volume times the change in pressure. So if you look once again this is the Gibbs
27:40
energy on this vertical axis here and this is the pressure on this horizontal axis and for liquids and solids you get virtually a horizontal line because they're incompressible. Their volume doesn't depend on pressure very strongly. Interestingly as the pressure
28:06
gets higher the Gibbs function goes up a little bit. With gases there's a much stronger effect. Gibbs energies of gas depend strongly on the pressure and you might expect them to
28:23
because gases are far from incompressible they're highly compressible. So their molar volume is highly dependent on pressure consequently their Gibbs energy is highly dependent on pressure. In fact their Gibbs energy goes up with increasing pressure. Now we can
28:43
actually figure out what this is for ideal gases very readily. We can just substitute for VM from the ideal gas equation, move the RT out front. So that's going to move out front. We've got 1 over P so we're just going to have log P final over P initial. That is the equation that describes the change in the Gibbs energy for an ideal
29:04
gas as a function of pressure. To change the pressure we have a very simple equation. Probably should go on your equations page for quiz 5. What is with this projector? What
29:24
is with my laser pointer? Too cheap to buy new batteries for it. Please, please. Okay so this is what the volume is doing as a function of pressure for an ideal gas.
29:44
It's following this purple line here. So if we want to evaluate this integral we're going to be integrating from some initial pressure to some final pressure. This is the area underneath this curve so this is the Gibbs energy. And it's obvious that as
30:02
we make PF higher and higher and higher this integral is going to get bigger right? And so it's obvious that the Gibbs energy is going to go up. Just based on that.
30:21
Now we can define a standard molar Gibbs free energy. We're constantly, there shouldn't be the word free in here. We're trying to get rid of the word free. It should be the standard molar Gibbs energy. Alright that's defined at a defined pressure which is one bar. Alright that's how we define the standard Gibbs energy. Or the standard entropy
30:45
or the standard anything. If it says standard it's one bar. Okay and so in this case we can write this expression it just follows directly from this guy right here except that we've now defined a particular Gibbs energy that applies when the pressure is one bar.
31:03
Okay so this initial guy is now that guy. Okay so all this plot shows, here's the molar Gibbs energy and here's the pressure and what we said earlier is that as I increase
31:20
PF this, I'm going to make this integral larger and so that tells us the Gibbs energy has got to go up with increasing pressure. Alright what I just showed on this slide right here is that we're going to make, we're going to define a special Gibbs energy at one bar and so that's what this becomes. This becomes one bar. Imagine this is one
31:44
bar and now we're integrating to higher pressure from that. Alright so the same intuitive picture applies. Alright if we move this to higher pressure the integral is going to go up and that's why this plot is going up, up, up, up, up but you can see that it's
32:02
got downward curvature. It's got downward curvature because here there's a big change in the Gibbs energy. Smaller, smaller, smaller. This has got upward curvature. So if we integrate this guy increases in pressure are going to have a smaller, progressively smaller and smaller effect on the Gibbs energy and that's what we're seeing here.
32:25
That's why there's downward curvature of this guy. Alright so the Gibbs energy goes down as we increase the pressure and up as we increase, sorry, goes down as we increase the temperature and up as we increase the pressure. Is all of this confusing? Absolutely.
32:49
I mean if you don't think so you're just not paying attention. Okay should we do some examples? The change in the Gibbs energy of 25 grams of methanol, mass density 0.791
33:12
grams per cubic centimeter when the pressure is increased isothermally from 100 kilopascals to 100 megapascals, oh I should say calculate the change in the Gibbs energy. There's
33:26
no verb. Calculate the Gibbs energy when we subject this methanol to a change in pressure. That's a change by a factor of 1000. It's a big change. So your first thought process
33:44
should be it's a liquid, it's incompressible, let's use the simplest equation we can. And the simplest equation for any liquid or solid is just Vm times delta P. Vm times delta P. Alright I immediately get the change in the Gibbs energy. Alright this is what
34:04
I did. Okay and so if I'm willing to give up these molar, forget molar I'm just going to calculate the difference. It doesn't ask me for the change in the molar Gibbs energy it just asks me for the change in the Gibbs energy. Alright so here I need the volume
34:22
of my 25 grams of methanol. Alright there's the density and so I can calculate the volume. That's 10 to the 5, that's 10 to the 8, so there's my factor of 1000 and so when I plug these numbers in I get 3.157 times 10 to the 9 cubic centimeters per cubic centimeter
34:42
Pascals. Alright that's the delta G that I get. I don't like these units but I can convert them later. Now so I converted them right here. Alright 3.157 times 10 to the
35:01
9 centimeters cubed Pascals, no I don't like that and so I can use the definition for here's the conversion factor for Pascals to ATMs, there's the conversion factor from cubic centimeters to liter, alright now I've got liter, atmospheres, alright and so then I can use the ratio between the two different Rs to do the unit conversion
35:23
to get joules, right? Tedious alright but when I do that I get 3.157 times 10 to the 3 joules you know or roughly 3 kilojoules. Alright that's what the change in the Gibbs energy is going to be. Now when I looked up the answer in the key it had something
35:43
far fancier alright it used the isothermal compressibility of methanol which you can look up in the back of your book that's the isothermal compressibility is 1.26 times 10 to the minus 9 per Pascal, well for goodness sakes, we can go through and do
36:02
it with the isothermal compressibility and see how different our answers will be. Right the isothermal compressibility is defined by equation 14.59 is 1 over V times the derivative of volume with pressure constant temperature and we can just linearize this differential
36:21
final minus initial because we know the change in volume is going to be small. Pressure difference is not small actually but we're going to linearize that also. Okay so we're going to make two approximations we're going to say the initial volume is just the final volume also it's not going to change much so that's going to become VI so I know
36:42
what that is alright and the final pressure is much, much larger than the initial pressure and so that difference right there I can just approximate as the final pressure alright the initial pressure is just a rounding error on the final pressure literally okay so that's the equation I'm going to use and so if I just do a little algebra split this
37:05
into two terms now solve for VF I get this expression right here which can be further simplified to give me this expression right here and then what I've got to do sadly is integrate that so I'm going to plug that in for VM and do the integral and so that's
37:24
what I did here I can calculate what this VI is that's the mass that's the density okay and here the two derivatives I'm going to do I'm going to actually pull this out front and then I'm going to take the derivative of dP well that's pretty easy to do and then
37:47
kappa T times the derivative of P alright and so I can run these two integrals alright this is just going to be 10 to the 8 minus 10 to the 5 this guy is one half
38:03
kappa T P squared and when I plug the integration limits into these two anti-derivatives alright this guy ends up being almost 10 to the 8 alright that's the rounding error that gets subtracted from him and these are the results for these other two terms
38:22
okay and so what I calculate is 2.96 times 10 to the 3 joules which is also about 3 kilojoules but if I do this carefully considering the isothermal compressibility of methanol I get a slightly different answer it's different by one part in 30 roughly alright that's
38:43
wrong that's right but pretty darn close to being right so it depends on the precision that you need on the quiz that might be good enough unless this is also an answer
39:04
that would be cruel wouldn't it one part in 30 difference in the answer no I would never do that everyone see how to do that isothermal compressibility makes it a little bit more complicated okay so the last thing to say today is that and this may be completely obvious to everybody
39:27
is that if you want to know the standard Gibbs free energy for a reaction like this reaction right here what you do is you look up in a table the Gibbs standard that free shouldn't be there again the Gibbs energy of this stuff the Gibbs energy
39:44
of this stuff and the Gibbs energy of this stuff and the standard energy difference is that minus that plus that alright I can look this up in tables now if I don't have a table of Gibbs energies well I've only got a table of entropies
40:03
and enthalpies which is not terribly uncommon you got to use this equation right here you can look up these enthalpies you can look up these entropies and if you know the temperature at which the reaction is happening you can figure out whether the reaction is spontaneous at that temperature or not you can look up the delta G this is the delta G reaction
40:26
so in the case of H we've got this delta sub R H what that notation means is that R is the reaction right this is the enthalpy change for the reaction
40:44
this is the delta H formation right that's what we're going to look up in the table alright what we want is we want to look up the delta H formation for all the reactants and all the products add up the reactant delta H formation
41:00
subtract sorry add up the product delta H formations subtract the reactant delta H formations to get the delta H for the reaction okay this zero means standard what does that mean pressure equals one bar temperature equals 298.16 K
41:26
okay that what is that new it's just a stoichiometric coefficient in front of these so that would be one one and one in this case okay we need to do the same thing for the entropy
41:42
here's the enthalpy here's the entropy we still get stoichiometric coefficients since we're going to look up these standard entropies products minus reactants so we can do that here's I've done it by golly that's for this guy right here which is what?
42:09
propionic acid and these two guys are for those two in no particular order okay and so the delta H is just that
42:23
the evaluation of that of all of those numbers okay and you can do the same thing for the entropy from a table of standard entropies you do exactly the same thing and the delta G then is just the delta H minus T times the delta S where we put plug in the T alright we get 72.6 kilojoules per mole
42:44
at the delta G we conclude that this reaction should be spontaneous at this temperature okay I think that's all I got
43:12
105 are there questions about this stuff? I know it's not riveting
43:24
alright we'll do hopefully it's going to get more interesting I think