WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to talk about Thermal Chemistry.
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We just finished our discussion with several examples of Thermodynamics.
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Discussing energy and heat and work and things like that.
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I thought it would be best instead of discussing Thermal Chemistry theoretically and then doing problems, I think it is best just do the problems.
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And then within the problems itself, discuss anything that needs to be discussed theoretically which is actually very little.
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Much of this is stuff that you already know from General Chemistry.
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Just a couple of the problems are a little bit more sophisticated than what you are you are used to, except for the first which is going to be very basic.
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Let us jump right on in.
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This one was nice and easy, it says find δ H for the following reaction.
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We have titanium dioxide + chlorine gas goes to titanium chloride and oxygen gas.
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Notice we have solid gas, solid gas.
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Finding the δ H for the reaction is really simple.
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Again we know this from general chemistry.
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Let us see what color I’m going to use, I think I’m going to do blue.
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We know that the δ H is equal to what would you say the δ H of the reaction and
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we remember that this little degree sign up above there represents standard conditions.
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Standard conditions happen to be 25°C or 298 K and 1 atm pressure.
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It is just we need a particular standard so we choose that standard.
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It is equal to the sum, I have forgotten how to write this term here.
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The sum of the δ H is of formation for the products - the sum of the δ H is of formation for the reactants including coefficients.
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Again, this is something that we already know from general chemistry, it is products – reactants.
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Whenever you see a standard table of thermo chemical data or thermodynamic data, you are going to have the enthalpy a formation listed.
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You are going to have the standard free energies and you are going to have the entropy listed.
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And you can use those tables, you just take the enthalpy of the products, subtract the enthalpy of the reactants making sure to include the coefficients
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and that gives you the actual enthalpy change for the heat of reaction under constant pressure conditions,
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because under constant pressure conditions the enthalpy is equal to the heat.
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Let us just go ahead and calculate this.
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All we need is a table of thermodynamic data so we have δ H for elements and their standard state is 0.
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This is chlorine gas CO₂ so that is going to be 0 and O₂ gas that is going to be 0.
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We only have to worry about that one and that one right there.
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When we look up the enthalpy for the Ti Cl4, we have -803 kJ/ mol × 1 mol.
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That was canceled and we add 2 at the enthalpy for this which is 0 and we subtract the sum of the enthalpy of formation for the reactants.
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And we get, for the titanium dioxide, we look it up and -945 kJ/ mol.
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Again, the coefficient on that is 1 so it is × 1 mol + 0.
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It is going to be -803 --945 so we have a δ H for this reaction is going to equal 142 kJ or 142,000 J.
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Make sure you watch out for the units, it is really important.
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When you are looking at the table of thermodynamic values, the enthalpy of formation and the standard free energies of formation those are in kJ/ mol.
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The entropy is going to be in J/ mol/ K.
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Later on, when we actually get to the equation you are familiar with, δ G = δ H - T δ S, when you are mixing does up and
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when you are working with entropy, as well as enthalpy and free energy, we want to make sure the units match.
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We either need to convert to J or we need to convert the entropy to kJ.
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142 there you go, that is it.
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This is positive enthalpy which means this is an endothermic reaction.
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In other words, in order for this reaction to both go forward, it actually has to pull 142 kJ of heat from the surroundings, it has to go into the system.
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Let us go ahead, that was nice and easy.
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Let us go ahead and this one says calculate δ U for the reaction in problem 1, assuming the gases behave ideally.
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This is different, now we are not calculating enthalpy, we are calculating the change in energy.
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Let us go ahead and write our equation again just so we had on this page.
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We had a titanium dioxide which was solid + we had 2 Cl2 which was gaseous and it went 2 Ti Cl4 the titanium 4 chloride which was solid + 02 gas.
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Here is what is interesting.
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Notice that we have solid gas and a solid gas, when we run these reactions, when calculating heats of reaction,
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If any of the reactants or products are gaseous as they are in this case, the reaction has to be one of the bomb colorimeter.
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A bomb colorimeter, if you remember from general chemistry or if you do not I will explain it right now.
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It is basically is a colorimeter where the volume is held constant.
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We put everything in sight of this thing called a bomb and it is immersed in water and we run the reaction inside.
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We generally tend to keep the temperature and we tend to keep the volume.
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The volume is absolutely fixed because it is a bomb, it is a single volume, there is no expansion.
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This makes sense because if you are going to produce, for example this reaction produce oxygen gas, you have to contain that gas.
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You cannot just let it escape into the atm which is why we have to constrict and make sure it stays in one place.
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There are some things that we can control.
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In this case, we are controlling volume and we also make sure the temperature stays constant so the only pressure that ends up changing.
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Let us go ahead and see what is happening here.
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The transformation is this.
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Let us go ahead and see, the transformation is we have our reactants and it is going two products.
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It is going to be at a given pressure, the reactant is going to be at a given pressure P1, a certain temperature, and a certain volume.
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We are going to end up keeping temperature and volume the same experimentally but we are going to end up with a new pressure.
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This is what happens in terms of the state variables P1 TV to P2 TV.
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The relation between δ H and δ U, we are looking for δ U.
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We already calculated δ H from the previous problem.
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The relation is remember δ H = δ U+ δ PV.
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Remember the relation H = U + PV, in this case δ H = δ U+ δ PV that is the actual relation.
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But notice we are holding volume constant so it comes out.
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What you end up with is δ H = δ U+ δ P × V.
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Now δ H = δ U, δ P is just the final pressure - the initial pressure inside that bomb × the volume.
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Again, this says assuming the gases behave ideally.
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The ideal gas law is PV = nrt therefore P = nrt/ V.
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Therefore, pressure 2 = n2 RT/ T.
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Remember, T is constant and V is constant, R is a constant, it is the gas constant.
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The only thing that changes is the number of moles and P1 = n2 × RT/ V.
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Therefore, we will put these expressions into here and we end up with δ H = δ U + n2 RT/ V – n1 RT/ V × V.
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The V's cancel out and we are left with δ H = δ U + n2 – n1.
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I will pull out the RT δ n × RT.
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δ n is just the difference in the number of moles of the gaseous components of products and reactants.
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δ n is just the number of moles and product - the number of moles in the reactant of the gaseous products.
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The solids they do not count, they do not contribute to anything in terms of actual pressure of things like that.
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It is the gas that contributes to the pressure.
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This is our basic equation that we are going to need.
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What we are going to do is we are going to move this over the other side to get the δ U because δ U is what we want.
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Let us go ahead and calculate δ n.
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δ n = the number of moles of gaseous products - the number of moles gaseous reactants.
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Again that is just the coefficients δ n = 1 -2 = -1.
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δ H = δ U+ δ N × RT move this over and we are left with δ U = δ H – δ n × RT.
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We already calculated δ H, I’m going to do this in terms of, that was 142 kJ.
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We have 142,000 J.
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I’m just going to go ahead and convert them to J because R is here, I’m going to use the 8.314 J/ K mol - δ n is -1, R is 8.314 J/ mol K, and then RT temperature at 298, 25°.
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When I do this, I end up with δ U= 144000 J or 144 kJ.
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What is important was what we derive, this relation right here.
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It all comes from the definition of enthalpy H = U + PV.
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δ H = δ U + δ PV.
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V is held constant so I can pull that out so I just get δ H = δ U+ δ P × V.
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Pv = nrt so I put that in and I'm left with this final relationship here.
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Enthalpy energy is the relationship that we used.
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Let us go ahead and see what else we can do here.
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This is one of the problems that you probably did not see when you are in general chemistry and it is a very important problem
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because it is going to allow you to calculate enthalpy is a different temperatures.
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All your enthalpy is that we calculated in general chemistry where from the table of thermodynamic data and all that is done at 25°C 1 atm pressure.
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They told you this at the time but of course you did not do anything with it, that enthalpy reaction depends on the temperature which we run the reaction.
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The reaction was at 25°C vs. Reaction that is 225°C, the enthalpy is going to be different.
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We need to find a way to calculate the enthalpy at a new temperature.
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That is what I do with this problem.
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The following data holds at 1000 K.
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We have nitrogen gas + 3 mol of hydrogen gas goes to 2 mol of ammonia gas.
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Sorry about that, it should be right there, the δ H for this reaction at 1000° = -123.8 kJ/ mol.
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Here is some more data here, the substance nitrogen gas its molar heat capacity is 3.50 × R.
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Hydrogen has a molar heat capacity of 3.47 R and ammonia has a molar capacity of 4.22 R.
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We want to calculate the heat of formation of NH3 at 298 K.
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We want to find the heat formation NH3.
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In other words, we want to find this.
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We want to find N2 + 3H2 goes to 2 NH3.
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We want a find δ H of formation at 298.
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The δ H of formation is for formation of 1 mol.
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We actually have a something like this.
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This is actually what we want, the δ H of formation is the heat of a reaction for the formation of 1 mol of whatever it is that we are looking for.
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The reactants and their elements in their most basic state, nitrogen gas, hydrogen gas, goes to 1 mol of ammonia gas.
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This is what we are looking for.
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Let us get started.
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We will write this again so δ H values are generally temperature dependent.
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In other words, the enthalpy change for a reaction depends on the temperature at which a reaction is run.
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Let us begin with this one, let us begin with δ H = H of the products - the H of the reactants.
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That is what δ is, just products – reactants that is the definition of the δ.
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I will go ahead and differentiate this with respect to T.
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We have differential with respect to temperature not time.
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Differential with respect to temperature so we have DDT of this δ H = DH DT – DH sub R DT.
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Products and reactants, DH DT we know what that is.
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Anytime you have an enthalpy divided by a temperature that is the definition of heat capacity.
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DH DT that is the definition of constant pressure heat capacity.
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These are standard, let us just go ahead and put the standard, in other words, the heat capacity at 25°C and 1 atm pressure.
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Since that is the case, this D of the δ H with respect to temperature, this thing right here that = δ of the heat capacity,
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in other words, the heat capacity of all of the products - the heat capacity of the reactants.
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That is what we are saying here, that is what this is.
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Because DH DT = heat capacity is δ H is actually δ CP.
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We are just taking the heat constant pressure heat capacity of the products - the constant pressure heat capacity of the reactants.
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This is this and this is this.
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Since we have that, let us go ahead and move this over there and we have the following.
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We have the D of δ H standard = δ CP × DT.
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Let us go ahead integrate this.
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If we integrate this from temperature 1 to temperature 2, we integrate this from temperature 1 to temperature 2.
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Here, what this becomes is just δ H at temperature 2 - δ H at temperature 1.
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Write the integral of the D just go away and you are left with δ H and you do the T2 - T1 = the integral from T1 to T2, the difference in the heat capacity DT.
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Therefore, I just take this and move it over here and I get the change in enthalpy of the given reaction
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at a particular temperature T2 = the change in enthalpy at a given temperature.
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The way I keep this standard is actually the standard does not refer to temperature, the standards refers to pressure.
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When you see this thing right here, it actually means 1 atm pressure.
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It just happens that most of the work that we do in general chemistry happens at 25°C.
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We also include that 25°C and its degree sign but you often see the degree sign for different temperatures
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because that really just refers to the 1 atm pressure.
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It is when we change the pressure that this degree sign tends to disappear.
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I will move this over + the integral from T1 to T2 of δ CP DT.
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This is the equation that we wanted.
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This is the equation that we are going to use.
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Here is what is going on, it is telling me that if I want to calculate the change in enthalpy of the reaction at a given temperature like 1000°,
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I start by calculating the δ H at the temperature that I do know which is 25°.
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Then, I integrate the difference in the heat capacities of the products and the reactants from one temperature to the other.
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That is what this says.
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I have a way of calculating the enthalpy change for any reaction if I already know the change at 25°C
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and if I know the constant pressure heat capacities for the reactants and the products.
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Constant pressure heat capacities are tabulated for everything.
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These are just because I have to look them up in a table, the way you would in other bit of information in chemistry and physics.
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Let us go ahead and do this problem.
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This is the import the equation.
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Again δ CP, let me write that down, where this δ CP = the sum of the heat capacities of the products - the sum of the heat capacities of the reactants.
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That is all this is.
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Something very important to remember that is why I put several asterisk by it.
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When we calculate enthalpy for elements are going to be 0 like the O₂ and the Cl₂ it was 0.
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Heat capacity is never 0.
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Therefore, when you are dealing with elements you cannot ignore the heat capacity.
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The heat capacity for every product in every reacted must appear in this equation.
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The heat capacity for elements is not 0 like it is for the δ H of formation.
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The δ H of formation of elements is 0.
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All heat capacities must be accounted for and again multiplied by the respective coefficients.
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Here is what we have, we have that δ H at a given temperature = the δ H at some initial temperature + the integral from T0 to T of the change in the δ CP DT.
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We can set up this integral up in two ways.
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I will go ahead and show you the two different ways to set up and choose one to actually work with.
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We can set up this way.
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We are looking for the δ H in this particular problem, we are looking for the δ H of formation at 298°K,
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because the information they give us is 4000 K so we can set up this way.
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We can go δ H at 298 = δ H at 1000 + the integral from 1000 to 298.
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It is going to be from this temperature to that temperature.
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Generally, we tend to increase temperature.
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In this case, they gave us the δ H at a higher temperature.
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They want to know what it is for the normal or lower temperature.
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It is what you are given to where you are going, that is the integral.
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This is going to end up being negative of the δ CP DT or I can set up this way.
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I can set up as δ H 1000 = δ H of 298 + the integral of 298 to 1000 of δ CP DT.
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In this case, some people like to go, they like the lower limit integration to be a lower number, it is a personal choice.
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It does not really matter.
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In this case because this is the number you are looking for, you are given this one, you are going to calculate this one.
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When you do this arithmetic, you are just going to move this over to actually get that value.
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Let us go ahead and do it this way.
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The δ H at 298 is going to equal δ H at 1000 - the integral 298 to 1000 of δ CP DT.
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Let us go ahead and calculate δ CP.
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δ CP is equal to the capacities of the products - the capacities of the reactants multiplied by the coefficients.
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We will do this in red.
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Our equation was, I’m just going to write the equation right here.
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I had N2 + 3H2 = 2NH3.
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For NH3= 4.22 × R.
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Remember it was 4.22 R.
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So × 8.314 and then there are 2 mol × 2, that is our products.
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We will do the N2.
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The N2 was 3.50 R, 8.314, and there is 1 mol of it + for H2 it was 3.47 R 8.314.
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And of course, there are 3 mol of it, that is that.
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Our δ CP δ of the heat capacities of products and reactants = -45.48 J K.
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We can actually run the reaction.
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Our δ H of 298 = the δ H.
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That is what we are doing, we decided to use this version of it.
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Not this version but this version, δ H of 298 it was the δ H to 1000.
00:30:08.200 --> 00:30:13.900
That was given to us, the δ H 1,000 was -123.8 kJ.
00:30:13.900 --> 00:30:39.700
So -123,800 J - the integral from 298 to 1000 of the δ CP which we just calculated as - 45.48 DT.
00:30:39.700 --> 00:30:51.000
You can do this by hand or put a mathematical software, you end up with the following.
00:30:51.000 --> 00:31:10.000
When you actually do this calculation, what you are going to end up with is - 91,873 J but notice this is for the formation of 2 mol of the NH3.
00:31:10.000 --> 00:31:14.100
We were looking for the δ H of formation which by definition is the formation of 1 mol.
00:31:14.100 --> 00:31:17.600
We just divide that by 2.
00:31:17.600 --> 00:31:24.600
Our δ H of formation, sorry about that not the reaction.
00:31:24.600 --> 00:31:45.600
The δ H of formation is going to be at 298 going to be - 45.9 kJ/ mol, that is our final answer, that is what we wanted.
00:31:45.600 --> 00:31:52.600
And it is all based on this right there.
00:31:52.600 --> 00:32:05.100
This the δ H at 1 temperature is the δ H at any given temperature is that 298 - 25°C that we normally get from thermodynamic cables +
00:32:05.100 --> 00:32:10.400
the integral up the difference of the heat capacities of products and reactants.
00:32:10.400 --> 00:32:15.700
It is very important equation.
00:32:15.700 --> 00:32:17.400
Let us move on to the next one.
00:32:17.400 --> 00:32:19.400
This look like it is going to be long.
00:32:19.400 --> 00:32:22.400
Hopefully, it is not too bad.
00:32:22.400 --> 00:32:28.100
Part A, using a table standard thermodynamic data, calculate the heat of a precision of water at 25°C.
00:32:28.100 --> 00:32:30.900
This one is very easy, not a problem.
00:32:30.900 --> 00:32:37.900
Calculate the work done in vaporizing 2 mol of water at 25°C under constant pressure of 1 atm.
00:32:37.900 --> 00:32:40.900
It should not be too bad because we know what the definition of work is.
00:32:40.900 --> 00:32:48.200
We know the work is external pressure × change in volume.
00:32:48.200 --> 00:32:57.900
I will go ahead and write that down here so work = external pressure × change in volume, that is just the definition of work.
00:32:57.900 --> 00:33:04.200
Find a δ U for the vaporization of water.
00:33:04.200 --> 00:33:17.500
We are probably end up using that relationship δ H = δ U+ δ PV and we are going to probably work with that and manipulate that little bit.
00:33:17.500 --> 00:33:23.500
The CP for water vapor is this, the CP in other words the constant pressure heat capacity for liquid water is this.
00:33:23.500 --> 00:33:27.200
Find the enthalpy of vaporization of water at 100°C.
00:33:27.200 --> 00:33:29.000
We are going to do what we just did.
00:33:29.000 --> 00:33:36.300
We are going to find enthalpy in a given temperature and we are going to use it to find enthalpy at a higher temperature.
00:33:36.300 --> 00:33:40.300
Let us jump right on in and see what we can do.
00:33:40.300 --> 00:33:50.600
Part A, use a table standard thermodynamic data, calculate the heat of a precision of water at 25°C.
00:33:50.600 --> 00:34:02.700
Part A, our reaction is this, we are going H to a liquid to H2O gas and we want to find δ H of this reaction.
00:34:02.700 --> 00:34:10.000
This is the δ H of vaporization, in this reaction we are taking the liquid and we are converting it into a gas.
00:34:10.000 --> 00:34:21.700
Δ H of vaporization happens to equal the δ H of the reaction which is this reaction which is equal to.
00:34:21.700 --> 00:34:57.500
Let us look this one up, it is this – this, what you end up with is -241.814 - -285.830 so you end up with δ H of vaporization = 44.016 kJ/ mol.
00:34:57.500 --> 00:34:59.200
That is our answer.
00:34:59.200 --> 00:35:11.700
A 25°C 44.016 kJ of heat needs to be put into the liquid water in order to convert all of it to gaseous water, that is all the says.
00:35:11.700 --> 00:35:18.500
The δ H of vaporization is, how much heat is required to affect this transformation.
00:35:18.500 --> 00:35:25.500
If I'm at 25°C, in order to convert all that water I need to put in this much heat.
00:35:25.500 --> 00:35:32.700
Part B, what is the work required to affect this transformation?
00:35:32.700 --> 00:35:38.800
We said that the work = the external pressure × the change in volume.
00:35:38.800 --> 00:35:43.500
We know what the external pressures is 1 atm, this is happening under atmospheric conditions.
00:35:43.500 --> 00:35:46.800
You are just boiling water, it is what you are doing.
00:35:46.800 --> 00:35:54.800
Our δ V = the volume of the gas - the volume of liquid.
00:35:54.800 --> 00:36:00.000
In other words liquid has a certain volume, you are converting that all to water vapor gas.
00:36:00.000 --> 00:36:05.800
The change in volume is the difference between the volume that the gas occupies and the volume of liquid occupies.
00:36:05.800 --> 00:36:12.000
We have a little bit of calculation to do to find out what the volume of the liquid is.
00:36:12.000 --> 00:36:19.600
We only know what the volume of the gas is, we can just go ahead and use what we know from general chemistry which is 22.4 L/ mol.
00:36:19.600 --> 00:36:23.100
Let us see what we can do as far as the volume of liquid is concerned.
00:36:23.100 --> 00:36:30.400
The volume of liquid, volume of the liquid water, here is how you calculate it.
00:36:30.400 --> 00:36:40.400
25°C, the density of water is 0.9970 g/ ml.
00:36:40.400 --> 00:36:53.700
There are 1000 ml in 1 L and there are in 1 mol of water 18 g.
00:36:53.700 --> 00:36:58.700
G cancels g, ml cancels ml, I'm left with mol and L.
00:36:58.700 --> 00:37:07.700
The number is going to be 55.39 mol/ L.
00:37:07.700 --> 00:37:12.000
I need the molar volume so I need L/ ml.
00:37:12.000 --> 00:37:35.500
I’m going to this, when I reciprocated it I end up getting 1/ 55.39 mol/ L gives me 0.01805 L/ mol.
00:37:35.500 --> 00:37:41.700
1 mol liquid water at 25°C occupies 0.01805 L.
00:37:41.700 --> 00:37:47.000
This is just a simple conversion.
00:37:47.000 --> 00:38:02.500
I have 2 mol of water so I just multiply by 2 and I get 0.03611 L.
00:38:02.500 --> 00:38:08.200
2 mol of water at 25°C liquid water occupies 0.03611 L.
00:38:08.200 --> 00:38:12.700
That is the volume of liquid water.
00:38:12.700 --> 00:38:16.200
Let us go ahead and find the volume of the gas.
00:38:16.200 --> 00:38:19.000
The volume of the gas is very simple.
00:38:19.000 --> 00:38:26.200
We have 22.414 L/ mol.
00:38:26.200 --> 00:38:35.100
I have 2 mol of water so I have 44.828.
00:38:35.100 --> 00:38:38.300
Please confirm my arithmetic, notorious for arithmetic mistakes.
00:38:38.300 --> 00:38:48.300
The volume of our gas is that, δ V our change in volume is the volume of the gas - the volume of liquid.
00:38:48.300 --> 00:39:00.800
It is equal this - that so 44.828 -0.03611.
00:39:00.800 --> 00:39:06.800
I'm left with 44.792 L.
00:39:06.800 --> 00:39:12.300
Our δ V is that, now we can go to our problem.
00:39:12.300 --> 00:39:20.300
How our work = our external pressure × change in volume.
00:39:20.300 --> 00:39:30.100
Our external pressure is 1 atm, in this particular case our change in volume was 44.792 L.
00:39:30.100 --> 00:39:43.400
Therefore, our work = 44.792 L atm that is a unit of energy but we need to convert that J.
00:39:43.400 --> 00:40:00.600
Therefore, 44.792 L atm is × 8.314 J =.08206 L atm.
00:40:00.600 --> 00:40:07.600
Those are just the two different versions of R, the gas constant in J/ mol K L atm/ mol K.
00:40:07.600 --> 00:40:09.900
That is your conversion factor.
00:40:09.900 --> 00:40:22.600
When you do this, you end up with the work = 4.5 kJ/ 2 mol.
00:40:22.600 --> 00:40:29.700
Per mol = 2.25 kJ/ mol.
00:40:29.700 --> 00:40:31.700
There we go.
00:40:31.700 --> 00:40:43.700
What this says that if I have 1 mol water at 25°C, the amount of work that I do in converting that liquid water to gas,
00:40:43.700 --> 00:40:50.700
When I'm converting it to gas I’m pushing away the atm, the gas, the liquid, the water vapor is expanding.
00:40:50.700 --> 00:40:53.100
It is doing work, it is pushing against the atm.
00:40:53.100 --> 00:40:57.600
The amount of work it is doing per mol of water is 2.25 kJ.
00:40:57.600 --> 00:41:02.100
That is a lot of work.
00:41:02.100 --> 00:41:08.400
Part C, it wants you to find the change in energy for this transformation.
00:41:08.400 --> 00:41:13.100
We know that δ H = δ U.
00:41:13.100 --> 00:41:22.600
I will just start with the definition of enthalpy H = U + PV.
00:41:22.600 --> 00:41:33.100
Therefore, δ H = δ U+ δ PV.
00:41:33.100 --> 00:41:37.900
Pressure is constant, it is the atmospheric pressure so that P comes out.
00:41:37.900 --> 00:41:50.100
Therefore, what you have is now δ H = δ U+ P δ V.
00:41:50.100 --> 00:41:56.400
δ H = δ U + δ V that is just the work.
00:41:56.400 --> 00:41:59.400
We just did that, that is just work.
00:41:59.400 --> 00:42:05.700
Therefore, our δ U = δ H – W.
00:42:05.700 --> 00:42:14.500
We already calculated δ H that was 44.016 kJ/ mol.
00:42:14.500 --> 00:42:20.500
We calculate the work which was 2.25 kJ/ mol.
00:42:20.500 --> 00:42:35.000
Therefore, the change in energy is 41.766 kJ/ mol.
00:42:35.000 --> 00:42:47.700
In going from liquid to gas at 25°C the energy of the system increased by 41.76 kJ, that is a huge amount of energy, that is all this is saying.
00:42:47.700 --> 00:42:51.800
This is all based on just simple basic mathematical relationship that we already know,
00:42:51.800 --> 00:42:59.800
the handful of equations that we already know from our work with work and heat and energy in the first law of thermodynamics.
00:42:59.800 --> 00:43:04.800
We are just applying it to chemical situations.
00:43:04.800 --> 00:43:07.000
Let us see what we have for D.
00:43:07.000 --> 00:43:10.400
Let us see if I have another page actually available.
00:43:10.400 --> 00:43:13.100
I do, great.
00:43:13.100 --> 00:43:24.900
Part D, they want to know what the δ H is at 100°C instead of 25°C.
00:43:24.900 --> 00:43:32.400
The δ H that we actually calculated was this thing right here was the first that we get part A.
00:43:32.400 --> 00:43:38.900
This is a 25°C, that is the δ H in 100 ℃.
00:43:38.900 --> 00:43:57.600
First of all, let us go ahead and work in K so 25°C = is the same as 298 K and 100°C =373 K.
00:43:57.600 --> 00:44:20.200
Therefore, from the equation that we have from the previous problem our δ H at 373 K, 1 atm pressure its equal to the δ H at 298 K +
00:44:20.200 --> 00:44:36.200
the integral from 298 to 373 of the difference in the heat capacities of products and reactants, gas or water vapor, liquid water, DT.
00:44:36.200 --> 00:44:37.700
That is it, nice and simple.
00:44:37.700 --> 00:44:43.200
Let us go ahead and do this.
00:44:43.200 --> 00:44:53.200
Let us go ahead and write up the whole thing δ H 373.
00:44:53.200 --> 00:45:04.700
Let us go ahead and do that is going to equal δ H at 373 is going to equal δ H of,
00:45:04.700 --> 00:45:18.300
Let me write this again to keep it on one page + the integral from T0 to δ CP DT.
00:45:18.300 --> 00:45:35.500
Therefore, δ H 373 = δ H at 298 which we calculated which was part A 44.016 + the integral from 298 to 373.
00:45:35.500 --> 00:45:41.300
And now the difference in the heat capacities, they gave us those in the beginning part of a problem in part D.
00:45:41.300 --> 00:46:03.500
Gas, water, that was 33.577 and that is the molar heat capacity for water vapor - 75.291 molar heat capacity for the liquid water.
00:46:03.500 --> 00:46:20.100
We end up with 44.016 + -41.714 × 75.
00:46:20.100 --> 00:46:25.100
Just doing this by hand, this is this number and this is a constant so it pulls out.
00:46:25.100 --> 00:46:31.400
This integral of the DT is just δ T or δ T, between 290 - 373 is 75.
00:46:31.400 --> 00:46:45.900
What you end up with is δ H at 373 = 40.9 kJ/ mol.
00:46:45.900 --> 00:46:58.600
This tells me now at 25°C if I'm converting liquid water to water vapor, I have to put it that much heat 44.016 kJ/ mol of water.
00:46:58.600 --> 00:47:04.400
At 100°C, I only have to put in 40.9 kJ/ mol.
00:47:04.400 --> 00:47:06.600
Notice, this is less and that makes sense.
00:47:06.600 --> 00:47:09.000
I'm at a higher temperature so I do not have to work as hard.
00:47:09.000 --> 00:47:13.800
In order boil water, I have to get it to 100.
00:47:13.800 --> 00:47:22.800
A part of this energy here, about 4 kJ is used just to get my water from 25°C to 100°C and I can start boiling it off.
00:47:22.800 --> 00:47:28.500
If I’m already at 100℃, all that heat that I put in automatically goes towards just boiling it off.
00:47:28.500 --> 00:47:31.300
I do not have to raise it because it is boiling point temperature.
00:47:31.300 --> 00:47:34.500
This makes sense, empirically it makes sense.
00:47:34.500 --> 00:47:36.500
We know from experience.
00:47:36.500 --> 00:47:39.600
Let me write that down.
00:47:39.600 --> 00:48:07.300
Notice, it takes less energy to vaporize water at 100°C than it does at 25°C.
00:48:07.300 --> 00:48:13.000
And this makes sense.
00:48:13.000 --> 00:48:18.500
If you ended up with a higher number than 44 then something happened, something went wrong.
00:48:18.500 --> 00:48:22.600
A lot of times we get so wrapped up in the mathematics and I'm guilty of this just as much as anybody else,
00:48:22.600 --> 00:48:30.600
it is probably more so because to the mathematics that we do not pull back and stop and take a look if this actually make sense physically.
00:48:30.600 --> 00:48:37.300
It is very important that we want you to be able to develop mathematical skills that are important to solve these problems
00:48:37.300 --> 00:48:40.200
but we want you to understand what is happening conceptually.
00:48:40.200 --> 00:48:50.400
We want you to realize, wait a minute I have a 25°C of vaporizing water, a certain amount of heat is going to required for that, that is the δ H.
00:48:50.400 --> 00:48:55.900
If I’m at 100°C that is the boiling point of water already.
00:48:55.900 --> 00:49:02.200
I should stand back and say I should require less energy because I’m already at the boiling point.
00:49:02.200 --> 00:49:11.700
Therefore, the number that I get should be lower, sure enough it is lower 40.9 vs. 44.016.
00:49:11.700 --> 00:49:17.000
It makes sense physically.
00:49:17.000 --> 00:49:26.200
Let us round of this discussion of thermal chemistry with one more problem here and let us see what we can do.
00:49:26.200 --> 00:49:34.700
Liquid water has a molar volume of 18.0 cc/ mol, if the temperature is held constant.
00:49:34.700 --> 00:49:40.500
Temperature is constant and the pressure is increased by 15 atm.
00:49:40.500 --> 00:49:53.200
We take it from 1 atm to 15 atm, calculate the enthalpy change then calculate the enthalpy change by 15° increase in temperature when the pressure is held constant.
00:49:53.200 --> 00:50:04.500
Here what we are doing is we have a certain volume of water, we are asking if I hold the temperature constant in the first case and increase the pressure,
00:50:04.500 --> 00:50:11.000
what is the enthalpy change for that particular process?
00:50:11.000 --> 00:50:17.700
In the second process, if I hold the pressure constant the change in temperature what is the enthalpy change?
00:50:17.700 --> 00:50:29.400
That is what is happening here and the molar heat capacity 75.3 J/ mol K.
00:50:29.400 --> 00:50:32.600
Enthalpy change, remember one of the basic equations is the following.
00:50:32.600 --> 00:50:49.600
Let us go back to blue on this one , DH = the constant pressure heat capacity DT + DH DP.
00:50:49.600 --> 00:50:56.100
DP this is the general equation, remember that the change in enthalpy of the system, I can change the temperature,
00:50:56.100 --> 00:51:03.100
I can change the pressure, that is how I can change the enthalpy.
00:51:03.100 --> 00:51:16.100
The change is equal to the constant pressure heat capacity × the differential change in temperature + this pressure dependence in enthalpy.
00:51:16.100 --> 00:51:22.600
That is the rate at which the enthalpy changes per unit change in pressure × DP.
00:51:22.600 --> 00:51:26.000
If I hold the temperature constant this just goes to 0 only this matters.
00:51:26.000 --> 00:51:29.500
If I hold pressure constant this goes to 0 only this matters.
00:51:29.500 --> 00:51:32.800
We are going to do both.
00:51:32.800 --> 00:51:37.300
This equation is not valid for gas, it is valid for liquids and for solids.
00:51:37.300 --> 00:51:45.800
All of the equations that we developed so far they are valid for every single state of aggregation solid, liquid, gas, for every single phase.
00:51:45.800 --> 00:51:53.600
It does not really matter, we just need to have the particular values that are important to be able to solve the problem.
00:51:53.600 --> 00:52:17.100
The only thing that we need to recall is that for a liquid this DH DP at constant temperature it is actually equal to the volume of a particular sample.
00:52:17.100 --> 00:52:22.800
Let us go ahead and do the first part of temperatures held constant and the pressure is increased at 15 atm.
00:52:22.800 --> 00:52:41.800
In this particular case, the temperature is being held constant so that is going to go to 0.
00:52:41.800 --> 00:52:54.100
What I had is DH = DH DP T × DP.
00:52:54.100 --> 00:53:02.900
The DH DP = V so I have is DH = V DP.
00:53:02.900 --> 00:53:07.400
If I integrate this I end up with the following.
00:53:07.400 --> 00:53:19.400
I end up with δ H = V × the change in pressure so that is what I'm going to look at.
00:53:19.400 --> 00:53:23.600
I have the volume and I'm going to find the change in pressure.
00:53:23.600 --> 00:53:28.100
I already know what the change in pressure is, it is just 1 atm to 15 atm.
00:53:28.100 --> 00:53:32.400
I just need to know what the volume is.
00:53:32.400 --> 00:53:35.900
Let us go ahead and make some conversions here.
00:53:35.900 --> 00:54:00.500
I have 18.0 cm³ × 10⁻⁶ m³/ cm³ I need to convert this cm³ to m³ because when I multiply by a pressure
00:54:00.500 --> 00:54:06.700
which is going to be in a Pascal, the Pascal is defined in terms of the cubic meter not the cm³ .
00:54:06.700 --> 00:54:10.000
We have to watch for the units.
00:54:10.000 --> 00:54:32.900
When I go ahead and do that, I would actually change the 15 atm to Pascal, 15 atm × 1.01 × 10⁵ Pascal/ 1 atm.
00:54:32.900 --> 00:54:50.900
I’m just making some conversions here, = 15. 15 × 10⁵ Pascal and this is going to be 18 × 10⁻⁶ m³.
00:54:50.900 --> 00:54:54.300
Therefore, my δ H = V × δ P.
00:54:54.300 --> 00:55:00.600
My V is this, my δ P is this, I will just multiply them.
00:55:00.600 --> 00:55:37.300
So δ H = 18.0 × 10⁻⁶ m³/ mol and this is going to be 15.15 × 10⁵ Pascal, when I do this, my δ H going to be 27.27 J.
00:55:37.300 --> 00:55:44.100
I use my basic equation in this particular case temperature was held constant so this term is unimportant.
00:55:44.100 --> 00:55:52.600
This for a liquid happens to equal V so I get this and I’m left with δ H = V δ P.
00:55:52.600 --> 00:55:55.300
And what the volume is, I just have to make sure to be in appropriate unit.
00:55:55.300 --> 00:55:59.000
I know what the change in pressure is, I just need to make sure it is in the appropriate unit.
00:55:59.000 --> 00:56:00.700
I will go ahead and calculate the J.
00:56:00.700 --> 00:56:06.500
Notice, this is not very much 27.27 J.
00:56:06.500 --> 00:56:10.200
For constant pressure, let us do the same thing.
00:56:10.200 --> 00:56:20.500
We go ahead and move this back up here.
00:56:20.500 --> 00:56:23.000
For constant pressure, let us rewrite the equation.
00:56:23.000 --> 00:56:28.800
The basic equation it never hurts to actually write the equations that you are supposed to know and memorize over and over again.
00:56:28.800 --> 00:56:39.000
DH = CP DT + DH DP constant temperature DP.
00:56:39.000 --> 00:56:45.200
And this particular case, it is the pressure that is held constant so this goes to 0 that means there is no change in pressure.
00:56:45.200 --> 00:56:48.700
DP is 0 that means this is 0.
00:56:48.700 --> 00:56:54.000
What I have is DH=CP DT.
00:56:54.000 --> 00:57:06.700
When I integrate this, I end up with δ H = the constant pressure heat capacity × δ T the change in temperature.
00:57:06.700 --> 00:57:19.800
Therefore, the δ H I should go ahead and put P to let me know that this is happening under constant pressure = CP they gave it to us already.
00:57:19.800 --> 00:57:35.300
This is 75.3 J/ mol K and the difference in temperature was 50°.
00:57:35.300 --> 00:57:41.800
50°C the degree increment, the δ of Celsius is the same as the δ of K.
00:57:41.800 --> 00:57:46.500
Therefore, this is 50 K.
00:57:46.500 --> 00:57:54.300
What we end up with is 1130 J/ mol.
00:57:54.300 --> 00:58:00.800
That is a very big difference between 27 J/ mol and 1130 J/ mol.
00:58:00.800 --> 00:58:09.500
Basically, what this says is that if I have a liquid and if I put some pressure on it, it does not really change the enthalpy of the system.
00:58:09.500 --> 00:58:13.600
In other words, we are not really investing a lot.
00:58:13.600 --> 00:58:22.400
If I just change the temperature by 50° all of a sudden invested a lot of energy into this.
00:58:22.400 --> 00:58:27.100
The enthalpy has gone up by 1130 J.
00:58:27.100 --> 00:58:35.000
Clearly, for liquid and for a solid, temperature has a much greater affect on the enthalpy of the system than pressure.
00:58:35.000 --> 00:58:43.700
Obviously, if I put like 10,000 atm pressure on it, that is going to be different but normally with the levels of pressure
00:58:43.700 --> 00:58:49.500
that we work with the laboratory which is maybe 1 to 20, 30, 40, something like that.
00:58:49.500 --> 00:58:51.300
It really does not matter.
00:58:51.300 --> 00:59:00.800
The effect of pressure on enthalpy is for the most part negligible, that is really all that this is saying.
00:59:00.800 --> 00:59:03.400
That took care of the thermal chemistry via examples.
00:59:03.400 --> 00:59:05.600
Thank you so much for joining us here at www.educator.com.
00:59:05.600 --> 00:59:07.000
We will see you next time, bye.