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 the entropy of the universe and the surroundings.
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Let us just jump right on in.
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Let us start with the most fundamental equation that is important to chemists.
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Let me go to blue today.
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Let us start with δ G = δ H - T δ S.
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Every property of this equation, the δ G, the δ H, the δ S, and T, in fact every property that we have been dealing with
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in all the equations that we have dealt with, they are all properties of the system.
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Unless they specifically say otherwise, the assumption is that they are always property of the system not the assumption but they are.
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They are the properties of the system.
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Basically, this is δ G of the system equals δ H of the system - T δ S of the system.
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We will put it there because that is the general presumption.
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under conditions of constant temperature and pressure which accounts for 99.9% of work that is done in the laboratory,
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we know that if the δ G is less than 0 and that implies a spontaneous process.
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That is the very definition of the spontaneous process, the δ G has to be less than 0
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that means that the equation as written will move in the forward direction from left to right.
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δ G equal 0, this is the δ G of the system.
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Under constant pressure, we also know under constant pressure we know that the δ H happens to equal
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the heat of transferred in that process when the process is done under constant pressure.
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The heat that flows to the surroundings is just the negative of the heat that goes into the system.
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Heat has to come from somewhere.
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If it goes to the surroundings, it is coming from the system, it is coming from the surrounding and is going to the system.
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The heat that flows to the surroundings is just - Q that is equal to - δ H.
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Because Q is δ H so- Q is - δ H that is the heat.
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This is the δ H of the system, this is the heat that flows to the system.
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This is the heat that flows to the surroundings.
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If we suppose a reversible process the DS of the surroundings is equal to - DQ/ T.
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- DQ of the surroundings which is this thing right here = - δ H/ T.
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Let us stick with the differentials not the finite.
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- DH/ T or you can write it in terms of a finite difference, δ S of the surroundings is equal to - Q sub P surroundings.
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= -δ H/ T, if I want to know what the change in entropy of the surroundings is, not the change in entropy of the system, I just take the δ H of the system.
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Take the negative side of it and divide by the temperature at which this process is taking place.
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Since, we have δ S of the surroundings = - δ H/ T, we can solve for the δ H.
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We are just going to move some things around.
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We are going to get δ H = - T δ S.
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I’m going to take this δ H = - T δ S and I'm going to put back in for this δ H.
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This is what I get, I get δ G = - T δ S of the surroundings - T δ S.
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This δ S of the system.
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I get δ G = - T, I’m going to go ahead and factor out - T so I get δ S of the surroundings, the change in entropy of the surroundings + the change in entropy of the system.
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The change in entropy of the surroundings + the change in entropy of the system that is just a change in entropy of the entire universe,
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the universe that we are talking about, system + surroundings.
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What I have is a δ G = - T δ S of the universe.
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Let me go ahead and move forward here.
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The pages are a little sticky.
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Let me rearrange this and I have that the δ S of the universe = - δ G/ T.
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I just found the δ of the surroundings that was the - δ H of the system ÷ T.
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The δ of the universe that is just the - δ G of the system ÷ T.
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When δ G is less than 0, implying a spontaneous process that means δ G = - T δ S that means - T δ S of the universe is less than 0,
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that means δ S of the universe is greater than 0.
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In terms of entropy, in order for a process to be spontaneous it is the change in entropy of the entire universe that has to be greater than 0.
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We express the spontaneity condition using δ G, this is δ G of the system.
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This is very convenient because basically, we have only have to deal with some property of the system, this change in free energy.
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As it turns out, the change in free energy, the negative of the divided by the temperature actually gives us the change in entropy of the universe.
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This is the relationship, δ G is just another way of looking at δ S of the universe.
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That is what all these relationships say.
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That is what this says δ G and S of the universe they are actually the same thing that is related by the negative of the temperature.
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The spontaneity condition with a free energy change has to be less than 0
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is really the same as the change in entropy of the universe has to be greater than 0.
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We deal with δ G because it is a lot easier to deal with free energy and measure free energy changes than it is to measure entropy changes.
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But this is the relationship.
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Let me write that out.
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For a process to be spontaneous, it is the entropy of the universe that has to be greater than 0.
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Entropy, not the entropy of the system, it is the entropy of the universe that must be greater than 0.
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We elucidate it, it was just a very simple mathematics based on the fact that the δ G being less than 0 is the defining property for spontaneity.
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It turns out that that happens to be that the δ S of the universe has to be greater than 0.
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The entropy of the system in a spontaneous process can decrease as long as there
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is a more than compensating increase in the change in entropy of the universe.
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The entropy of the system can decrease in a spontaneous process as long as there is a more than compensating increase in entropy
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and the surroundings so that the overall entropy of the universe is greater than 0.
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Let us write this down.
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The entropy of the system may decrease during a spontaneous process as long as there is a more than compensating
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increase in the entropy of the surroundings such that the δ S of the universe which is equal to the δ S of the surroundings + the δ S of the system of this is greater than 0.
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In other words, δ S of the surroundings + δ S of the system greater than 0
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which means that the δ S of the surroundings is greater than that negative of the δ S of the system.
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If the system decreases, if the system’s entropy decreases by 10 units, in order for that process to be spontaneous,
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the change in entropy of the surroundings have to be greater than 10 units.
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10.1, 10.2, 11, 12,30, whatever, it has to be greater than the decrease in the change in entropy of the system.
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As long as that is satisfied, as long as this is satisfied, the process is spontaneous.
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Let us see here, the way I remember everything is the following.
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This is what makes this particular equation δ G = δ H - T δ S.
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It is a very important equation.
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The way that I remember everything is the following.
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Let me write that out.
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The way I remember everything is I take the δ G = δ H - T δ S our fundamental equation for that relationship,
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I go ahead and I divide everything by – T.
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I get - δ G/ T, I think my negative sign in the middle, I like to just put on the top for the numerator that way it is more consistent, = - δ H/ T + δ S.
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I have this is δ S of the universe, this is the δ S of the surroundings.
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Let me write that a little bit better, δ S of the surroundings and this is the δ S of the system.
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I have my equation and if I need my δ H.
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I'm sorry, if I need my δ S of the surroundings I just take the negative of the enthalpy of the system divided by the temperature which it take place.
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If I need the δ S of the universe, I can either add these.
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If I happen to know this or I can go ahead and take the negative of the free energy change for the equation
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and divide that by the temperature of which the process is taking place.
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That and that.
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Writing out formally, δ S of the universe = - δ G/ T.
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δ S of the surroundings = - δ H/ T and δ S of the system well that is just δ S system.
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One final thing regarding this,
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let us look at a non spontaneous process and verify that this is actually the case.
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Let us look at a known spontaneous process and you can do this with any spontaneous process.
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The best example, I did not choose the combustion example but it is probably the best example to choose.
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I chose the conversion of water vapor to water.
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Let us look at a non spontaneous process to verify what is it that we have just talked about.
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We would look at water vapor and 25°C condensing to liquid water.
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At 25°C, if I have some steam that is going to spontaneously turn into liquid water.
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Using a table of thermodynamic data, I get the following.
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I find that the δ H of the reaction was going to be products – reactants.
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It is going to be -286 and this is kJ/ mol but leave off the units.
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-242 so we end up getting - 44 kJ/ mol.
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I will go ahead and put it there.
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Table of thermodynamic data also includes free energy changes so I can do δ G as well from the table of thermodynamics data.
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We will only use all the resources of my disposal.
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The δ G of the reaction is -237 - -229 and that = -8 kJ/ mol.
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δ G = δ H - T δ S and this is of the system.
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Therefore, when we arrange this equation to solve for δ S I get the following.
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I get δ G - δ H ÷ - T = δ S.
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δ G is -8 - -44 ÷ 298.
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The table of thermodynamic data, everything is done at 25°C so 298°K.
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I get -0.121 as my change in entropy of the system.
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Let us go ahead and do the δ S of the surroundings.
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The δ S of the surroundings we said that = - δ H/ T.
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It is going to be - -44/ 298 and I end up with 0.148.
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Clearly, the change in the surroundings, the 0.148 is greater than the change the 0.121 of the system.
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The system decreases in entropy by 0.121 but the entropy of the surroundings changed by 0.148.
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If I take δ S of the universe = the positive 0.148 and I add to that the change in entropy of the system which is -0.121 and I end up with something.
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The number does not matter, it is positive, it is greater than 0, that is what it confirms.
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The spontaneous process is a change in entropy of the universe that is greater than 0 not the change in entropy of the system.
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The change in entropy of the system was negative but together combined with a change in entropy of the surroundings
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you end up with a change in entropy of the universe being greater than 0.
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+ 0.027 which is clearly greater than 0.
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Again, I have I do not know about my arithmetic, you can check it for me but you will find that this number is bigger than absolute value of that number.
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That is it, thank you for joining us here at www.educator.com.
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We will see you next time, bye.