Hey everyone this is Dave Laude your OnRamps Chemistry instructor and I'm here to um take you through practice exam two from unit four on thermodynamics this is um an exam that um is a step up in terms of the expectations for what you're able to do and that it actually wants you to do some of the kinds of table data thermodynamics calculations um involving things like Bond energies or Heats of formation um but the other thing is you're introduced to the second law of Thermodynamics uh where um entropy is evaluated in the context of whether or not a process um can happen or not and then it's all tied together by taking a look at free energy when it's done about a third of the questions on here you've seen pretty much identically on the first exam is so there's a lot of refresher that's going to take place especially early on so I'll work quickly through those um and then we'll sort of get bogged down working the the uh problems that have to do with the data in the tables and a bit on the entropy but then it wraps up with stuff which again you saw back on exam one so let's go ahead and do a share and take a look at these questions um the first question up is asking about um isolated systems and the flow of heat or energy and the answer is neither happens in an is isolated system the second question up wants to know about um um the the uh first law theory involving heat um and I'm sorry internal energy and enthalpy and so the first question is asking if internal energy is conserved and it is and the second question was to know if um um the internal energy of the system plus the surroundings is conserved yes because that's the internal energy and then it wants to know whether or not they have the same sign and they don't this was actually the practice question that was um on was a question that was on the first practice exam identically next question up is a work calculation in which You're simply calculating minus P Delta V and multiplying it times 100 so you have um in this particular case a pressure of 20 atmospheres and the volume um change is a 0.1 liter so you have to be careful about that and then that ends up being multiplied by 100 to converted into joules and when all of that is done this becomes uh 20 * 0.1 is is 2 * 100 is equal to 200 joules of work being done and in this particular case because the volume is increasing that means that work is done by the system on the surroundings and it's a negative the next question up is a very simple Q Plus W calculation um in this case here they tell you that the internal energy changes by 500 and that's going to be equal to Q + W and it says that the um um amount of energy that was involved here is a warming of your hands so that's work that's being done on the surroundings and so in this particular case if there was a change in internal energy where 500 uh KJ was used on the surroundings and that means that the amount of work done on the surroundings was minus 450 next up a calorimetry calculation you've done multiple of these and so this is going to be an MC delta T notice it's asking about a single gummy bear not multiple gummy bears and it does involve the calorimetry housing so you've got to take that into account so you just stick the numbers in here the mass is 250 for the water the heat capacity is four the change in temperature here is 20° the heat capacity of the bomb calorimeter is is uh 250 and the amount of heat change is also still going to be 20 and so when you add these numbers together uh let's see if I can do the math that's 4 time 250 is equal to a thousand so that's 20,000 right there and this over here is um If I multiply that by two that's 500 500 time 10 is equal to 5,000 so that's a total of 25,000 or 25 kJ of energy next question up is a mixed reaction enthalpy problem remember in the end I want to take these two values here and somehow rearrange them so that they are equivalent to what you see in terms of a reaction notice that c starts on the left side here and it starts on the left side there so I would leave that as it is B is on the right side so again you leave that how it is but I need to be able to get a over to the other side so to get a over to the other side I have to multiply it through times some number if I put a minus one in front of this this turns this into d goes to make 2 a I notice at that point that the uh D's cancel and then C becomes B plus 2 a so the only thing I had to do to change this was to multiply everything through by minus one so that means I've turned my minus 200 into plus2 200 and now when I add these up 200 + 200 is equal to 400 and so that's my answer for the Delta H of reaction here's a question asking about the definition of um enthalpies of formation which of the following is not true let's start at the bottom formation reactions produce a molar product in any state that's true um standard conditions are an atmosphere in 298k close enough yes reactants in a formation reaction are elements in their standard State yeah that's the definition of a formation reaction is from elements in their standard State and it says all reactants and products must be one mole that's not true the products must be one mole but the reactants can be any combination to make it stoichiometrically cancel out set equal to each other the next question up is asking a heat of formation reaction involving hydrogen peroxide so I need to carry that to the next page so I can still see it so I'm going to write it down up here for myself so I've got H2O2 I'm H2O plus a half a liter of water of oxygen excuse me goes to make H2O2 as a liquid so that's the thing that I'm making and um they give me different values here for these so for example the first thing up here when I look at it I notice that my H2O2 is minus 185 and my water is minus 285 and oxygen becoming oxygen is zero so this is the simplest of problem s it's how it ends minus 185 minus how it began which is a - 285 and the math when it's all done is plus 100 so there's a 100 KJ per mole change in the heat of formation next question up is a bond energy problem for these you have to be very careful and write everything out so that you're assigning the values for the bonds that you're forming in such a way that you don't overlook something so never take a relatively easy problem like this for granted so that's why you're seeing me painstakingly write down all of the bonds actually I'm writing down more than I needed to and over here I've got my four carbon hydrogen bonds all right so now it's time to plug them in change colors on this I have 400 kiloJoules in each of these four bonds so that's 1,600 kJ um oxygen double bonds are 500 so that's a thousand so I ended up having 2600 KJ of energy go into breaking the bonds that's always positive and then what came out the back end of it was 800 kJ that went into the carbon oxygen bonds and into the oxygen hydrogen bonds I had 450 and so adding them all up that's minus 1600 there and then this right here is 900 and 900 is another minus 1800 there and so when I compare these here I've got um um plus 2600 here but I've got minus 3400 here and so that's a difference of 800 kJ negative and that makes sense because this is a combustion reaction so I should expect to see a negative sign from it here's a theory question about bond energy which of the following is not true down here at the bottom it says the table of bond energies includes multiple values for the same bond energy and that's not true this is the average number of average amount of energy in a bond the next question up switches over to the second law for the last half of the test and asks which of the following compounds have the greatest absolute entropy um just remember in general as you go from liquids to solids I'm sorry from solids to liquids to gases the entropy goes up and as you go from smaller to larger it goes up and as you go from cyclo to um linear it goes up so in this case here um I want to find the ones that are in the gas phase so I'm going to rule out the liquid phase of immediately gas phas will have more absolute entropy more ability to move around and establish larger numbers of microstates and cyclohexane is constrained um because you're having the the carbon form a ring so it can't move freely but the linear hexane is able to flop around much more and establish greater numbers of microstates so it's the correct answer here next question up is asking you which shows the substance in decreasing order of their molar entropy same sort of argument here gases to liquids to solids are decreasing in entropy and with respect to size and molecules it works that way so uh it turns out that molecules have greater entropy than atoms do so as I look up here um I see lots of gases and only one liquid that's on the far end over here so I already know this is the right answer because liquids are going to end up having a lower um molar entropy than the gases do but as I look at the gases here they're ranking based upon molecular weight here so um and also on the amount of floppiness so there's a lot of vibrational modes sitting inside a CO2 um and then argon and neon don't have those because they don't have any bonds um and so they have fewer opportunities for movement so the order here is on the basis of size and then turning gases into liquids the next question up asks about predicting changes in entropy um in this one they want to know um in which of these will the in will the entropy um increase going from left to right so as I look at these I want to see which of these shows greater entropy um going from a solid to a couple of cats and anions dissolved in water is increasing in entropy absolutely is a solid becoming a solid and a gas um increasing in entropy yes because you're making gas molecules instead of solids do the liquid becoming a solid increasing entropy no it actually decreases it so the answers are one and two here next question up without performing any calculations toine determine whether the entropy of the system increases or decreases well as I look at this right here I notice I have a gas and a liquid becoming um aquous solution so in that case there I go for my chlorine being able to bounce around all over the place in the room you know 22 liters of volume at STP for a mole to where it's dissolved in solution so this is a decrease in the entropy next question up is a second law theory problem which of the following statements is not true concerning the second law and so the first question up here is the entropy of the surroundings is determined by the entropy of the system well that's not true um it turns out that Delta s of the universe is equal to Delta s of the system plus Delta s of the surroundings and Delta s of the surroundings is actually determined by Delta h of the system minus Delta H over the system over T is given me that value is it true that a perfect Crystal has an absolute entropy of zero yes because the log of one orientation is equal to zero in s equal K log W the number of microstates in the universe are always increasing and the entropy is always increasing these are both true statements here's a problem that's very similar to one that we've done before um only it was in the count context of Delta G not Delta H in this problem right here we're watching water as a solid become water as a liquid at minus 10° so the first thing you ask yourself is can water uh be turned into a liquid at minus10 degrees and the answer is no so Delta s of the universe must be zero must be negative because if it happens it's positive so right off the bat I am unable to choose the first or the second one and then the next question up is asking about uh Delta s of the system and Delta s of the surroundings if you look at this case here the Delta s of the system I'm going from a solid to a liquid so I am increasing the entropy and that makes out a positive value and so this one right off the bat has to be the only one that's correct because this is a process that doesn't happen but one in which Delta s of the system is increasing next one up is a calculation that you need to do um a hydrogen balloon explodes and releases a lot of energy what's Delta s of the the surroundings Delta s of the surroundings is minus Delta h of the system over T so it's just a plug-and chug problem here I've got 600,000 joules of energy and it's going to the surroundings so that's a negative sign divided by the um um temperature oh and I'm sorry it's room temperature I was looking for a number and so when you do the math on this right here um the ends up being um um canceling and canceling so that's 2,000 right there so those minus signs cancel so I've got a plus 2,000 joules per Kelvin change that's taking place and that makes sense as lots of energy leaves the system it creates chaos in the surroundings why is there a temperature at which the hydrogen balloon stops exploding well the answer is very simple here Delta a Delta G is equal to Delta H minus t Delta s and um since this is a reaction for which no um since this is a reaction for which uh Delta G is equal to Delta H minus t Delta s um uh you see here that it's an exothermic process and um you see here that uh you're going from through three moles of gas to two moles of gas so the entropy is also going to be negative when you have two values which are negative you um um um know then that the temperature is going to be dependent and it is at lower temperature that the process can happen at higher temperature it stops next up is relating reaction spontaneity uh to temperature for phase changes in chemical reactions what are the signs for water condensing at 105 again first thing you notice is that water doesn't condense at 105 degrees and so um when you're looking at this problem you know right off the bat that Delta G has to be a positive number so it's going to be either the first one or the second one and then if you look at the process and you look at Delta H you know that energy has to be leaving the system so that's going to be a negative sign and that means immediately then that this is the only correct answer here it's also the case that um Delta s has to be negative because going from a gas to a liquid and then the final question up is asking about spontaneity of chemical reactions and so yet another question like we just finished talking about this is asking what you would predict for the temperature dependence of the chemical reaction you're going to write down Delta G is equal to Delta H minus t Delta s so these values can be plus and plus plus and minus minus and plus and minus and minus and so as you look at this um this right here says that this reaction gets quite warm so I know that it's an exothermic process so it's going to be one of these two and it also says here as you look at this that I went from a solid and a gas to a solid so I am definitely watching Delta s become more negative and that means then that this is a reaction for which temperature um um plays a role and as it goes down the reaction becomes more likely and so that's why the answer is that it's more likely at lower temperature notice that about half of this test was exam one so make sure you study your exam one materials ahead of time then the second half of it is some relatively easy table data calculations and um then your ability to manipulate Delta G Delta H and Delta s over and over again in calculation so make sure you've got your stuff um down when it comes to being able to convert kiloJoules into joules and to be able to do those sorts of simple math calculations best of luck on this exam.