 Hello guys, good evening all Am I audible? Yes, so tell me where were we in the last class? Yes, or what happened? I was sick. Yeah, I had fever that day on Monday So I thought of no taking rest So yeah, I was leaving that day fine now like it's what it was not No, go it and all so yeah, I was a bit worried but better now. Yeah Maybe some flu or something. Yes, so we Yeah, right. Yeah, so we had done of the second law of thermodynamics Correct, we have discussed about entropy and second law of thermodynamics. We finished Okay, so Next we are going to start Gives free energy, which is the last part of thermodynamics Correct and the second portion of this chapter is left then which is Thermochemistry so most probably today will finish this chapter Okay, second law of thermodynamics. We have none. Just a second We have done second law of thermodynamics, I can see the notes We have done till TDS is equals to du plus PDV, right? Yes, that is it second was done over there. Yes, that's it We are just statement of second law. That is it nothing much And then we have given the combined statement of first one second law, which is TDS is equals to du plus PDV So that is it. Okay. So next you see See just before taking this particular topic here Let's just Just a few things if you remember We had discussed a spontaneous process. Correct, right? Spontaneous process and there are two conditions for this What are those conditions? The conditions is Help me change negative and delta is the property change positive means randomness is increasing. Yes This is what we had discussed. If a process is taking place the process will continuous under two condition Either the enthalpy should decrease means the process would be exothermic or the entropy should increase Delta has written zero These are two conditions we had discussed, right? So first we'll talk about randomness and then we Okay to find out randomness or to measure randomness We have we got we've got a new thermodynamic term and that is entropy and the entire thing about entropy we discussed right Now here what we are going to understand because this is also Temperature dependent function. This is also temperature dependent function We know if temperature increases entropy increases. So both terms are temperature dependent So now what we thought Because suppose any process is given you need to first analyze enthalpy Or you need to analyze and copy right in order to understand the spontaneity of that particular process Right. So what we do instead of analyzing two different terms Now we thought of what that will combine the two term And we'll give a new thermodynamic term based on that we can say directly whether the process is spontaneous or not Okay, so in in this way like Like this was a thought process and what we got into this we got a new thermodynamic term That we call it as Gibbs free energy Gibbs free energy is represented by g And it is given to understand or analyze the spontaneity of any process like we Instead of analyzing two different terms delta h and entropy We will analyze You know Gibbs free energy We have a condition of Gibbs free energy like under this condition the process will be spontaneous So we will analyze only Gibbs free energy and if that condition is satisfied the process is said to be spontaneous Okay So what is Gibbs free energy? It is a thermodynamic quantity It is a thermodynamic quantity and this it is equals to the amount of Amount of useful work done the amount of useful work done by Work done by the system keep that in mind it is by the system amount of useful work done by the system is equal to is equal to Decrease in Decrease in the value of Gibbs free energy Gibbs free energy right So this is useful work done. We'll see there are two types of work done till now We were discussing about work done work done only. So when only work done is mentioned it means it is The pressure volume work done in general right if you have a piston cylinder system What useful work done is non expansion work done it is non expansion work done non expansion or We also call it as non pv. You can also understand this by this way Like for example, you have uh, you know, suppose you have a an objective Suppose I have this pen in my hand, right? If I place this pen from this point to another point from this room to another room, right? I have to do some work And this work done equals to decrease in my Gibbs free energy Right, so if you do any work You does work or sorry if you do any work that work will be done at the cost of Your Gibbs free energy Obviously, there is no pressure volume thing over here. There's no expansion or contraction we have here One second guys Yes, so useful work done is this it's non expansion work done Correct Where there's no pressure where there's no volume things in right That is useful work done so Gibbs free energy is equals to The useful work than magnitude wise if you see Okay, so mathematically mathematically you see It is defined as It is defined as d is equals to h minus ts So if you remember, I just said that in order to analyze this and this separately We'll combine these two With the help of temperature And we'll give a new thermodynamic term Right and based on that thermodynamic term. We can say the process is spontaneous or not So how do we combine the two? We combine h and s enthalpy and entropy like this Correct t is the temperature If you look at the change in Gibbs free energy Delta g is equals to Delta h minus t delta s This equation we call it as Gibbs m-holes equation Gibbs m-holes equation Right, we'll see what is the condition for delta g For Gibbs free energy for the process to be spontaneous. We'll see the condition Right, but the physical significance of delta g whatever I said it is the Decrease it is a useful work, right? What is the physical significance of every energy we have that you see physical significance? So like I said, we have two components of work over here and that is Expansion work and we have non expansion work expansion and non expansion Okay, non expansion nothing but like I said Is the useful work we have useful work? Okay, expansion the work in which Pressure volume term like p delta v that we calculate it is the expansion work Okay, suppose if you rub your hand, right? If you rub your hand, you will feel some heat That is non expansion work Okay, if you place the box from this point to another point That is an expansion work. So that kind of work is useful work. We have Okay, remember, I have given you a definition of a Gibbs free energy and there we have worked and by the system Correct. So what I am assuming here this condition The condition I am taking if q is the q is the heat absorbs Absorbed by the system by the system and W is the Work done W is the work done by the system Work done by the system Correct So if you see if I write down from first law of thermodynamics delta u Is equals to Is equals to q Minus w because we're working by the system. So minus w will write down this way. So basically q is equals to what? Delta u plus work done Now further if I write down because work done has two component Delta u plus we have worked done Expansion plus work done Non expansion Copy down this I'll go to the next slide Two component of work we have expansion and non expansion Okay, so Next time you see if I write down this Expansion work. So that would be q is equals to Delta u An expansion work we can write P delta v So worked on by the system. I have taken the sign convention earlier also if you see that's why I written q is equals to u plus w Right. So here I have taken p delta v Right minus w I have taken on the top Plus this I'll write down with this And this would be non expansion work non expansion right? What is this term? Could you tell me? This would be Then help me change Delta H Okay And this q we can write q is equals to T delta s Because entropy change formula is what it's q by t So q is t delta s isn't it So what we can write here you see T delta s equals to delta h Plus work done Non expansion So work done non expansion non expansion equals to minus Minus of delta h Minus t delta s So this is nothing but the change in gives free energy So further we can write this as The non expansion work done or useful work Is equals to minus delta g So if you write down this in this expression in words What you will write The useful work done by the system Or non expansion work done By the system is equals to the decrease in gifts free energy Of the system, isn't it? Okay So decrease in gifts free energy Is the measurement of useful or non expansion work done By the system write down this point This minus sign means gifts free energy is decreasing So we'll write down here decrease in Gifts free energy equals to equals to useful Work done By the system on this Clear? This is the physical significance Of gifts free energy Clear? Done Okay Next write down Standard free energy change Standard free energy change See standard free energy change is represented by delta g Not this dot Which is written over here This dot means standard It is condition is a standard standard a state we have Okay What is the standard state standard condition? Standard condition is We have one atmospheric pressure One atmospheric pressure and 298 Kelvin 298 Kelvin and one atmospheric pressure Write down it is the It is the free energy change change At standard condition Standard condition In which the reactants and The reactants and products And products are also in In Their standard state Standard state We'll talk about it What is the standard state of different No molecules atoms Okay later in thermochemistry Here it is not that useful just you need to know That the atoms must be there in Standard state the reactants or products that we have So at the standard state the free energy change will write like this Every term will be at The standard condition so delta g not Means free energy change at standard state standard free energy change Equals to the enthalpy change at standard state Minus the improperly change at standard state This is the relation we have at the standard state Copy this down What is this so this is No, it's not isothermal process. You are talking about this for isothermal because Temperature we have taken constant over there, right? Yeah, no, it's not like that. See the thing is any process generally we We're talking about in the reactions over here, right? Because in chemistry mostly we'll deal with all these things but with respect to a chemical reaction, correct? So in the chemical reaction what happens any chemical reaction takes place at constant pressure and temperature So whatever the you know Reaction we have that must we have at a given temperature Hence we'll take this temperature out over here condition We don't have such condition that it is for isothermal process Okay Yeah So this is a standard state we have now we have a relation here this relation you can you have to memorize But a little bit you should know about it You know the the you should know about equilibrium constant, you know, what is equilibrium constant? Tell me See, I'll tell you a little bit of about this equilibrium constant But we'll discuss this again in detail in chemical equilibrium Yeah, yeah, yeah, right. So a little bit we'll discuss about it Usually this chapter thermodynamics. I teach this chapter after finishing equilibrium because of this thing only, right? But since they have taken this chapter in the school early So because of the school thing I have taken this chapter also before equilibrium But usually we used to tease this chapter after finishing equilibrium Because of this thing only here you don't understand about equilibrium So let me just discuss few things about this equilibrium constant Just expression you should know and then we'll come back again to this point Okay, see first of all equilibrium constant Is you know represented by kc Is constant You can also it's simply k or kp also you can okay kc kp anything If you have a reversible reaction, it is defined for a reversible reaction If you have a reversal action a gives b Reversible means our backward both the reaction may proceed So we can always calculate the ratio of the product Concentration this square bracket means the concentration Concentration of what? Concentration of product by concentration of reactant We can always calculate this The moment reaction starts we can start calculating the concentration of Whatever it is So then the reaction is starts a starts converting into b So we'll get some amount of b So any You take the concentration of b and concentration of a and we can find out this Issue this ratio that has q That is reaction quotient Reaction quotient Right q is what is the reaction quotient Copy this down and go to the next slide Since a is converting into b obviously we can also write this as concentration of the product which is b Divided by the concentration of reactant which is at any point of time Now when you find out the ratio of product And reactant at time t is equals to t equilibrium When at equilibrium constant time you'll find out the ratio then this ratio is said to be kc which is the equilibrium constant Did you get my point? How many of you understood this tell me Ratio of product and reactant at equilibrium is called equilibrium constant If you find out the ratio of product and reactant at any point other than equilibrium So here this is t does not equal to t equilibrium But why means what? Ratio of the concentration of product and concentration of reactant Let's show you this yeah or just what's the one second I'm coming to your question She just got it So either we take concentration of reactant product or can also take pressure If the gaseous phase reaction is there So when you take the concentration of reactant divided by the concentration of product That that ratio is said reaction quotient understood She just understood So just to understand like this in detail we discuss this in chemical equilibrium, which is the chapter we have The equilibrium chapter is all about this equilibrium constant only So we'll discuss that but here just you need to keep in mind for any reversible reaction At any point of time take the concentration of Take the concentration of reactant Take the ratio of both you will get reaction quotient If the gaseous phase reaction is there Take the pressure of product Pressure of reactant Find out the ratio will be again the reaction quotient Okay, now equilibrium condition isn't a specific condition Right, it hasn't a specific properties also Right, it's not At any for any reaction equilibrium condition is very important Okay, so at equilibrium condition to define Whether the equilibrium condition is maintained or not For that we have defined a specific term of this ratio And that specific term is equilibrium constant Okay, so q is nothing but kc at equilibrium you can say If equilibrium is not maintained q does not equal to kc Right, so sometimes what happens if obviously we'll see this in A next chapter But why we have defined this term in order to understand the equilibrium condition is maintained or not So at equilibrium what we can write We can write q is equals to kc the reaction quotient Is nothing but equilibrium no k kc k is equals to at equilibrium It's not like that Anusha Value of k we are not bothering about it Okay, what should be the value of k it could be anything we don't know about it Right, but the value of k will be constant It depends only upon temperature that again we are not bothering about All these things we'll discuss in chemical equilibrium So k a value could be anything Okay, when equilibrium is not maintained If Equilibrium is not achieved Then q does not equal to kc We need to just keep this in mind nothing much The expression is what concentration of product by concentration of reacting This is the only thing which is required this chapter Other things we'll discuss in chemical equilibrium Okay So what happens you will see this condition or that some values will be given operating product To find out whether the equilibrium is maintained or not So what is the kind of equilibrium we'll find out this q And kc there will be data if this condition is coming to be it means equilibrium is maintained That's how specific term is given Yes correct Got it guys tell me why or n you can type yeah correct So what the relation of delta g and equilibrium constant or reaction quotient the relation is this again you have to memorize This relation is delta g Is equals to Delta g naught plus Rt ln q What is q you know now q is what q is reaction Concentration of product by concentration of React okay, so If you write down the equilibrium condition at equilibrium Delta g equals to zero. This is the condition of equilibrium. We'll see this also after some time and becomes kc At equilibrium so these two condition will substitute here. We'll get the expression Delta g naught is equals to minus rt ln kc This is the expression we have You know how to convert ln into log You can write down this as delta g naught is equals to minus 2.303 rt log kc Or in terms of Want to write down delta g is equals to minus rt ln kc from this we will find out kc equals to e to r minus delta naught by This is delta g naught Copy down this How did we get this expression? That is not the concern. You just keep this expression as it is in mind We'll discuss this later in other chapters. We'll see next chapter problem But this equation you must keep in mind at equilibrium condition Now if you draw the graph of this relation here, you see it is a logarithmic graph we have So obviously the graph will be like this So this one y axis is the Um, what is that reaction called q? And this side we have delta g naught x axis Is the logarithmic graph The graph goes like this Okay, logarithmic graph graph goes like this at some point. Definitely The value of delta g is zero that point is supposed this we have At this point supposed up is zero and If this is zero, you see here If this value is zero Then this means what the equilibrium is The corresponding value of kc is what delta is not is zero kc value is what this value is kc And this kc value is one if delta g naught is zero, isn't it? Right down this way So when delta g naught is so you substitute here log value is one here and since it is zero It is nothing the that's only one and i don't wait. Yeah, that's fine Now towards the left of this point We have negative Delta g naught and towards the right we have positive delta g So whenever delta g is positive, you see whenever delta g naught Is positive Is positive Means it is towards the right of it. The value of kc is what here also lied on k only The value of equilibrium constant k is what if it is positive the value k is what Is less than one If it is negative Less than zero, you see draw the graph the value of k is greater than one greater than one And when delta g is zero Is zero The value of k is equals to one which is The No doubt copied The purpose of the objective to define Gift-free energy is to find out the condition of spontaneity That was the whole purpose if you go back and see Discuss that enthalpy and property was the two condition So they combined the two term and they have given the new thermodynamic term that is gift-free energy in order to understand the spontaneity of any process So the whole purpose is to understand the spontaneity of the Spontaneity of the given process correct. So what is the condition of spontaneity? We have Okay, heading right down all of you condition of the spontaneity We know all we know all this fact that delta s total For spontaneous processes greater than zero This we know already, but this is the condition with respect to entropy only we have Right, this is the condition with respect to entropy What is the condition with respect to delta g that we are trying to understand? So you see here what I am assuming you see I am assuming Since any reaction takes place at constant temperature and pressure So we are first of all we are assuming constant Temperature And pressure so at this condition We are assuming that q amount of heat amount of heat given by the system to the surroundings to the surroundings and in this process Obviously some amount of work must be done. Okay, but heat is given by the system to the surrounding system is giving heat so its entropy is decreasing and we can say that the amount of heat is given mathematically we can write minus q of system isn't it? Right So minus q of sys equals to what we can write the heat Gained by the surroundings This is the relation we have magnitude wise Heat is given by the system Equals to heat is gained by the surroundings Okay, since we have constant pressure over here So this q of the system before the content of the system At constant pressure is nothing but 10th enthalpy so we can write this delta h Of the system negative we have here so what negative sign we have here Did you understand this relation? Yes, tell me Parvati you got it Parvati Anusha Aditi these three names I can see here Yes, okay Anusha is there Anjh Anjh Respondence, where is araya? Almost we are done more like 10-15 minutes more Yeah, okay So all of you understood this relation we have Now if I write down here delta s of Delta s of surroundings Is equals to what we can write Plus q of the surrounding Divided by t And this plus q of surrounding equals minus delta h of the system So can we write this equals to Minus of delta h Of the system divided by t Which further It becomes Delta s of surrounding equals to minus delta h system by t Now The total entropy change Expression we have delta s total Is equals to delta s Plus delta s Surrounding Delta s system you let it be as it is And delta s surrounding we can write Delta h of the system Divided by t With this expression this equals to this so I have written it over here Yeah Yeah Now just we need to solve this simplify it Left side we have delta s total Right hand side if you see it becomes Minus of Delta h of the system Minus t delta s Of the system Divided by So we can have t delta s Total It goes to what is this term could you tell me Delta h system minus t delta s system this is Delta g Of the system Right so delta g of system is what Minus of T delta s total And we know for the spontaneous process Delta s total is what Greater than zero And hence t delta s is negative so the point I went to understand here This is the relation we got In this we'll apply some condition and we'll understand the relation here So see so first of all For the spontaneous process For the spontaneous process Okay one second I'll go Okay So for spontaneous process we know the condition That is Delta s total Is greater than zero means this term is positive This is positive so delta h system is what Delta g System is always negative For the spontaneous process is the condition we have For any reaction Delta g if you find out to be negative it is a spontaneous process So rather than Analyzing enthalpy and entropy with two terms we'll just analyze Delta g If it is negative it's spontaneous If it is positive then non-spontaneous Right for non-spontaneous process The last one is at equilibrium We know delta s total Equals to zero When delta s total is zero Delta g of the system Also equals to zero at equilibrium Copy Okay So the last part is chapter we have That is the Role of temperature In spontaneity So we have the delta g Is equals to delta h My t delta s We have this relation Now We are having here two different conditions The first condition we have In the process could be Exothermic And it could endothermic also So I'm assuming exothermic process And we know for delta h is negative always First we are considering exothermic and then endothermic that is it Okay Okay So the expression The expression of gifts what we need to find out We need to find out the spontaneity Right For spontaneity what is the combination of delta g Delta g must be What For the spontaneous process delta g must be Negative Right So we need to keep this delta g negative here So what is the condition you see Delta h is already negative because we have assumed exothermic it is negative Right Depending upon t delta s If it is negative or positive we can easily understand This delta g is negative or positive So two possibilities also we have here If t delta s is positive If t delta s is positive Means this term is positive We already have a tip signed before Delta h is already negative Then delta g would be negative So no tell me Delta g is negative for all the pressure Remember the time you put It is negative Right negative temperature obviously it's not like minus 30 degree Celsius Right It is negative And hence the process is said to be Continuous at all temperature Spontaneous at all temperature For exothermic process If t delta if t is positive The process is spontaneous at all process Another condition is If t delta s is negative If t delta s is negative Then for delta g to be negative What is the condition The condition is the magnitude of delta h Must be greater than The nature of t delta s Isn't it Can we say this condition For delta g to be negative Are you getting it Tell me Yes Then only delta g would be negative This condition can be achieved by Because we need to decrease this term no Achieve by decreasing temperature So what we can say For exothermic process If t delta s is negative Then the process will be spontaneous at low temperature Right on this For exothermic process For exothermic process If t delta s is negative If t delta s is negative Then this will be at low temperature Yeah For exothermic process For exothermic process If t delta s is negative If t delta s is negative Then the process will be spontaneous at low temperature Okay Now The second condition we have If the process is endothermic Delta h is Is greater than zero Is greater than zero Is greater than zero So again you see the relation here The positive Delta s We need to see this right So If t delta s is negative What does this mean If t delta s is negative Then this is negative negative negative positive It means delta g is positive Always right It will never be negative And the process will be Non-spontaneous It will be spontaneous process in this condition If again the last one is If t delta s is positive Then the conditions quantity For delta g to be Delta g to be negative Right The condition is what The magnitude of t delta s Should be more than the magnitude of delta h And this condition can be achieved by Decreasing temperature Tell me any doubt in this Okay No doubt tell me Right Tell me Right So this is it for this particular thing Right thermodynamics we are done It's over Thermo chemistry the second part which is left We'll restart that also But before that we'll discuss few questions on this All three questions you try Done Okay 78 All the three questions you try Auro Then we'll start discussing Okay All the three questions you try And then we'll start discussing All of you Done Okay see first of all Question number 78 Combustion reaction it's a factual thing Combustion reaction is always exothermic So whenever a particle would substance Buns in presence of air There will be heat evolves Right So combustion is always exothermic process So delta h Is negative So wherever we have positive value Eliminate Okay Delta s is already positive given in both options So there's no point of you know A concluding what would be delta s for this So let's go directly to delta g Delta g what Delta g we know combustion reaction any example if you see It just requires a proper initiation And then it burns on its own Right So combustion reaction is spontaneous as well Of this two delta g is negative Right Delta h negative delta g negative We have only one option That is option z Okay Question number 79 For the reaction this delta u given Delta s is given We need to find out whether it's spontaneous or not So for spontaneous the condition is what for spontaneity Delta g Not must be less than zero This is what we need to find out If it is at zero then we'll be at equilibrium Delta g we need to find out Correct So you see here what is the formula of delta u Okay So delta h is equals to Delta u plus Delta mg rt Or that standard state so this Why we have to calculate delta h Because we know delta g we need to find out It is equals to delta h not Minus t delta s not This is already given in the question Right delta s not is given We need to find out delta h not So delta u not is equals to Minus 10 It's kilo joule here right It is joule Right You must have to Notice all these things Unit is joule Here it is kilo joule So what we'll write here We'll write 10 into 10 to the power 3 Joule delta u plus What is delta mg 2 minus 2 minus 1 that is minus 1 r value must be in joule So 8.314 Interpreter is 300 What is this expression could you tell me Tell me the value of delta h not here What is the value delta h not tell me Minus 12494.2 Is it So this is minus 12494.2 Okay Which further we can write Minus 12.49 kilo joule Approximately right Okay So now you have this delta h not Substitute this here In kilo joule Minus 12.49 Temperature is 300 And delta s over here Is given minus 45 So minus 45 into 10 to the power minus 3 Tell me this value Whether you are getting Negative or positive That is what you need to identify Because spontaneity we need to judge If it is negative then it's spontaneous If it is zero then Equilibrium No it is given in the question You see Anusha It's given 300 Kelvin Right Delta s is given at 300 Kelvin Yeah So what is this value Negative or positive So it will be 0.3 into 2 Sorry 0.3 into 45 Achha Are we getting this Minus 12.49 And this would be plus of 13. something is it So plus of 13. something It's all 5 and take Obviously this value is coming out to be greater than 0 So when delta g not Is greater than 0 we are getting So what is the meaning of this It is B Non-spontaneous Question number 80 Reversible reaction At temperature T Which is the following is correct Reversible reaction we have So we'll assume the equilibrium condition Right And that equilibrium Delta g would be zero And when delta g is zero T is equals to delta h by delta s So option C is correct here Clear 82, 83, 84, 85 Finish all four and then you can post your answer One second I'll go back Okay done Okay so The enthalpy of vaporization is given Question number 82 you see The entropy delta s Hello Can you hear me Yeah guys am I audible Yeah actually I sorry I lost connection Some power or interruption Yeah okay fine So this so you see here That the enthalpy is given in kilo joule And options are in joule So we'll just you know Unward this so 40.63 Into 10 to the power 3 Joule per oh divided by temperature must be in Kelvin So 373 What is this value you are getting All these kind of questions No units are very important must take care of that So answer here will be 108.9 approximately Option B is correct I guess Yeah so it is approximately 109 108.9 Joule per mole Kelvin Option B is correct for this one Okay now the next question you see 83 The spontaneous process at all condition Of temperature which of the following is correct Spontaneous process of delta s must be You know negative Right delta s must be negative So this A and B option we can eliminate easily And at all temperature It was it is the first case that we discussed Because How do you proceed in this particular question You write down the expression of delta G It's equals to delta H minus T delta s So at all temperature it is negative This is negative This is already negative exothermic So for this temperature here you If this enthalpy is negative over here Then for whatever the value of temperature you put Delta G is definitely negative Means delta H must be negative It is possible when the process is exothermic Answer is option D isn't it Yes so s is positive you know that's worked This is positive Question is about the spontaneous Okay I make my I thought it is delta G Yes yes yes that's right that's right My bad guys I thought delta G is written Okay no problem So spontaneous delta s must be positive So negative value option we can eliminate right This we can eliminate And delta H is negative so we'll have this option D Yes that's B I thought delta G my bad Okay so this is the answer we have The spontaneous process which is the following correct Delta s total is positive I've done this already Option C is right For reversible isothermal expansion Of an ideal gas tell me the answer 85 Reversible isothermal expansion Right Reversible isothermal expansion The temperature of system and surrounding must be same Right So the entropy of system if it decreases The entropy of surroundings will increase by the same value So their magnitude would be same But the sign is opposite right So if delta S system is positive Delta S surrounding is negative So option B is the answer we have here Tell me any doubt you have in this All these conditions we have discussed already in the theory part right That's what I said you know in the beginning That here in this chapter we have various conditions And that conditions you need to keep in mind And what conditions how do we approach Right We see all these things we have discussed already So you'll get questions like this in this chapter That you need to solve I'll share the assignment also on this Okay So we are moving next to the That's a one second Move like one thing also Okay these three questions you see A one second one second evaluation of when is this Or let this be this we haven't done yet Few questions we haven't done yet let it be So next we are going to start thermochemistry