 Welcome back to the final segment of this thermodynamics discussion. In the previous sections we have discussed about the important thermodynamic properties like energy, work and heat, how they are connected together to form the basics of the first law of thermodynamics. Then we derived the important parameters like enthalpy and entropy and redefined the second law of thermodynamics and from there we understand when a reaction becomes spontaneous from there we even go to the third law of thermodynamics. After that we look into the free energy parameters, the Helmholtz free energy and Gibbs free energy and how they are connected with respect to the work done by the system and from there we actually extrapolate that to different applications with respect to the electrochemical cell, with respect to acid base equilibria, with respect to solubility equilibria and at the end we have also discussed a little bit on corrosion which is also can be understood better with the extension of the thermodynamics. So, after we covered all these processes in this final part we will do a few problems and this problem set will help you to understand the underlying concept of thermodynamics better. So, let us start the problem set. So, over here I am going to write the different problems first explain it and solve it. So, the first problem says that there is an electric motor which produces 15 kilo joule of energy as mechanical work 2 kilo joule of heat to the surrounding. Now we need to calculate the change in internal energy of this motor. So, over here there is an electric motor is present there. So, say this is defining the motor it actually produces 15 kilo joule of mechanical work. So, it itself has done 15 kilo joule of work and at the same time it lose 2 kilo joule of heat to the surrounding. We have to measure what is the overall change in internal energy during this process. So, this is nothing, but we have to measure the delta E the internal energy change. We have given the work done we have given the heat exchange. So, these 3 parameters comes together and play well in the first law of thermodynamics which says delta E is equal to Q minus W. Now again the problem is easy, but over here the important factor is the sign of this parameters. Over here it is said that the heat is actually lost to the surrounding. So, that means the heat is coming from the system to the surrounding. So, that means heat is lost so that sign of heat will be negative when you are talking with respect to the system or the motor in this case. So, when you are talking about delta E this will be minus Q because heat is lost. Now the next is W the work the work when the system does work on its own then it is taken as negative and that is exactly what is happening. The electric motor produces 15 kilo joule of energy as mechanical work that means it is doing the work itself. So, it will remain as negative sign because the system itself is working. Now once the signs are actually figured out the rest is simple the heat is minus 2 kilo joule, work is minus 15 kilo joule altogether minus 17 kilo joule. So, that much of energy change is happening. So, it is negative which is saying that in the final state the motor is actually 17 kilo joules lower compared to when it actually started this process and why it is so? Because 15 kilo joule is lost for doing the work and 2 kilo joule is lost as a heat. So, again the important factor is the signature if heat is lost from that system to the surrounding it is negative and if work is done by the system it is negative. So, that is the first problem let us go to the next one. The next problem says there is a spring made out of steel is suppressed with 100 kilo joule of work. During this process 15 kilo joule of heat escapes the system again the quotient remains same what is the change in internal energy. So, that is the question. So, over here we have a spring in the beginning and then it is pressed. So, during this pressing a work is done on the system. So, over here the work is done on the system so that is why now the work will be having a positive sign because work is done on the system not by the system and at the same time some heat actually escapes from the system to the surrounding. So, it will be negative because heat is going out of the system. So, looking back into the first law of thermodynamics we can redefine over here for this particular system delta E is minus Q because heat is lost from the system and plus W because work is done on the system. So, if we combine this thing together then straight forward just put the values minus 15 kilo joule plus 100 kilo joule so altogether 85 kilo joule. So, it says now the system has more energy compared to where it start and that is quite obvious by looking into the picture. Now it is suppressed it has more capacity to do work which is defined as the energy. So, again for first law of thermodynamics be very careful while you are looking into the signatures heat if the heat is going out of the system it is negative if it is coming to the system is positive work work if it is done by the system it is negative work is done on the system it is positive. So, with respect to that we will go to the next set of the question. The next question is it is said carbon dioxide can be removed from the atmosphere by reaction with barium oxide. How? So, carbon dioxide and barium oxide can you can put together where carbon dioxide is a gas, barium oxide is a solid and you can produce barium carbonate carbon salt is formed and you can remove the carbon dioxide. The question is compute the enthalpy change in this reaction using the following data the following data are as follows barium oxide solid plus 2 HCl which is a gas producing Bacl2 solid plus H2O is a gas and delta H0 value that means the reaction is actually considered the standard state and standard enthalpy change for this reaction is minus 360 kilo joule. There is another reaction given over there barium carbonate solid plus 2 HCl in the gaseous form barium chloride in solid form H2O in gaseous and carbon dioxide in gaseous form. The standard enthalpy change for this system is minus 94 kilo joule these are the two equations given and now I have to find out the enthalpy change for this reaction. So, this is nothing but an example of Hesse's law. In Hesse's law we actually discussed that we can combine a particular unknown reaction to the known reaction such a way that we can just subtract or add the enthalpy values and we can get the enthalpy value for the reaction we are actually looking for. So, basically over here what you need to do reorganize these two reactions such a way that combining them I can get this the reaction given in this question. So, how to do that let us take a look. So, I have barium oxide solid plus 2 HCl gas it is giving me barium chloride solid plus H2O gas. So, that is fine it is given now at the end result what I need is barium oxide solid plus carbon dioxide gas to barium carbonate solid. So, that means the other reaction I have to balance such a way that I have a barium carbonate on the right hand side and carbon dioxide on the left hand side and if we look to this particular reaction over here it is such a way that the carbon dioxide is on right hand side and barium carbonate on the left hand side. Now I need to do the opposite so what I need to do I just need to shift the reaction other way. So, what we did is the following barium chloride solid plus H2O gas plus carbon dioxide gas is given into barium carbonate solid plus 2 HCl gas. Now we have turned the reaction on the opposite way. So, what will be the effect on the delta H value it will be exactly same value but opposite sign. So, now it will be plus 94 kilojoule why because look into this equation before it was minus 94 when the reaction is barium carbonate to barium chloride and carbon dioxide and now we put barium chloride and carbon dioxide together to form barium carbonate. So, it is actually opposite whereas the first reaction it is written in the same way as the values are given. So, it remains minus 360 kilojoule. Now if we combine these two reactions together you can see the barium chloride and barium chloride getting cancelled H2O H2O get cancelled HCl HCl get cancelled. So, it is a just simple addition which leave us with only barium oxide carbon dioxide and barium carbonate. So, the overall delta H for this particular reaction will be minus 360 kilojoule for the first reaction and plus 94 kilojoule because we have reversed the reaction the second one and all together if you put it all together it is going to be minus 266 kilojoule. So, you can see over there by using the Hess's law we actually recombined some known reaction such a way that we can go to an unknown reaction by simple addition and subtraction and find out what will be the enthalpy value of this reaction that is the beauty of Hess's law because the enthalpy change can be used in this particular manner and we can find out the enthalpy change even for some unknown reactions. Next we go to the next problem which says calculate the entropy of fusion of ice if its heat of fusion that is nothing but delta H fusion is 6.01 kilojoule per mole at 273 Kelvin. So, you have to calculate the entropy the heat of fusion is given temperature is given you have to find it out. So, as we know the entropy change for a phase transition can be expressed as delta S is equal to delta H fusion in this particular case and temperature of melting. So, over here delta H fusion is given 6.01 kilojoule per mole temperature is 273 Kelvin. So, be very careful even if it is given 0 degree centigrade you have to convert that to absolute temperature because this is absolute temperature. So, you have to put that in Kelvin scale. So, that is very important factor and then we have to do this calculation. So, it is kilojoule we can break it down to joule. So, it will be 6010 joule per mole inverse divided by 273 Kelvin and if you do calculation you will find it will be very close to 22.01 joule per Kelvin inverse into mole inverse. So, the two important factors of doing this problem is make sure you are going to use the absolute temperature over here otherwise it will not work do not divided that by 0 degree centigrade and 0 it will be undivided and you will get not the correct answer. Over here the other important factor is the unit. So, you can see the enthalpy values are generally in the kilojoule unit whereas, the delta S value are generally in the joule unit that is because it is multiplied by this particular temperature when we are considering the delta H which is T times delta S. So, delta S is divided by the temperature and this temperature generally has a huge value. So, it divides the factor by at least 3 orders of magnitude and that is why this joule and kilojoule factor is particularly present there. So, while you are doing this calculation make sure you want to do this calculation such a way that the kilojoule to joule transformation you multiply the 1000 in the proper way. Next we will do another problem of the same sort. So, the question is calculate the increase in entropy when one mole of water evaporates at 100 degree centigrade and what is given is the latent heat of vaporization and that is 540 calorie per gram at 100 degree centigrade. So, the question is very similar to the last time, but it is a little bit different because right now I have to ask to measure the entropy change for an amount of water. In previous case we just asked to add the entropy of fusion. So, we calculated that with respect to per mole unit over here it is given like how much water is there. So, sometime instead of water it can given like the amount of water in the volume or the weight. So, you have to do that perfectly. So, over here first I need to find out what is the overall heat of vaporization because it is given 540 calorie per gram not per mole. So, I have to calculate the mole of the water to the gram. So, transform it that and then calculate the overall heat exchange during that process. So, to do that what we will do first we will calculate what is one mole of water that is nothing but 18 gram of water. Now, the total latent heat of vaporization that is involved in this case is equal to 540 calorie per gram into 18 gram because over here 18 gram of water is present there. So, they actually cancel each other the unit and I get 540 into 18 calorie which you find it out close to 9720 calorie. So, this much of calorie is actually heat is actually exchanged. So, this is nothing but this Q reversible during this vaporization and as we know delta S is Q reversible divided by T. So, this is going to be 9720 calorie divided by temperature. Temperature is 100 degree centigrade but again the 100 degree centigrade should be converted to Kelvin scale first which goes to 373 Kelvin. So, I have to divided it with 373 Kelvin and we get the value close to 26.06 calorie per Kelvin and you can write it per mole unit right now over here this is gram per mole unit. So, I should have write that mole unit. So, that mole unit is coming over here. So, over here the important factor is 2 when it is given the amount of water in presence of volume or in terms of grams you have to multiply it or consider it properly. Then the temperature you have to consider it with respect to the absolute temperature and only then you can go to the right calculation. So, then we go to the next question where we go to some of the application part of this overall calculations. So, the question is calculate the free energy and entropy change per mole of liquid water when it boils at one atmospheric pressure and the other data given over there enthalpy of the vaporization and that is given in the unit of 2.0723 kilo joule per gram. So, that is the heat of vaporization of enthalpy change of vaporization is given. So, now enthalpy of vaporization if I want to carry because over here we have one mole of liquid water that means again 18 gram of water comes here. So, the overall enthalpy change for this process will be the following 2.0723 kilo joule per gram into 18 gram per mole because that is one mole of water ways and all together you get 37.30 kilo joule per mole that is overall enthalpy change. And the process what you are talking about it is happening at 100 degree centigrade that means 373 Kelvin because it is takes as this boiling condition. So, that means is the 100 degree centigrade temperature. So, the entropy change will be delta S delta H vaporization divided by T delta H vaporization of enthalpy we have already calculated over here. So, it is 37.30 kilo joule per mole divided by 373 Kelvin which comes out to be 0.1 kilo joule per Kelvin per mole inverse or if you want to convert it to the joule unit it is 100 joule per Kelvin per mole inverse. So, that is the overall entropy change. Now, we have to also calculate the free energy change. So, the free energy change for this system is delta G which is given by delta H minus T delta S. So, delta H we know it is 37.30 kilo joule per mole inverse T is 100 Kelvin sorry I am doing the same mistake 373 Kelvin into 0.1 kilo joule per mole inverse. So, over here you can see it will cancel out and give you 0 kilo joule per mole inverse is the overall Gibbs free energy change it is coming to this particular figure because it is in the equilibrium during the vaporization of the system of the water. So, that is how you calculate the energy and enthalpy changes accordingly using the free energy conversions. Now, the penultimate problem there is a zinc silver cell and we have to calculate the electromotive force of this zinc silver cell which is operating at 22.3 degree centigrade over here it is given zinc sulphate is present there its concentration is 0.191 mole silver nitrate is present there its concentration is 0.0289 mole to other standard reduction potential values are given zinc plus 22 zinc it is given minus 0.76 volt and silver to silver iron to silver is given as plus 0.80 volt. So, now I have to calculate the EMF for this whole system. So, now over here which side the reaction will first go that we have to calculate and over there you can see the E 0 value of silver plus to silver is plus 0.8 which is a positive. So, delta G is negative for this particular system whereas zinc plus 2 to zinc is negative value. So, delta G is positive for this particular direction. So, by looking into that we can say the first process will be spontaneous this process will be not the opposite process will be spontaneous because the opposite process will be E 0 value of zinc to zinc plus 2 the oxidation that will be plus 0.76 volt that will be spontaneous because the redox potential values are mostly given in the reduction side. So, when you are talking about oxidation the value remains same only the sign changes because it is exactly the opposite reaction. So, over here we can say silver plus 1 will go to silver that reaction will happen and over here this reaction becomes negative then. So, zinc will go to zinc plus 2. So, now we know the direction of the reaction in this electrochemical cell silver will getting reduced zinc is going to get oxidized. So, now we will create the cell. So, zinc is going to get oxidized to zinc sulfate and its concentration is given 0.191 molar. The other part of the cell this double line signifies the separation of the 2 half cells is silver plus ion in the present in the form of silver nitrate is going to silver 0 and this is present in 0.0289 molar. So, this is overall cell. So, now the overall reaction is the following zinc is going to zinc plus 2 plus 2 electron and now there are 2 silver ions has to act over here to balance the overall electrons is going to be 2 silver because there are 2 electrons are released by the zinc during oxidation and these 2 electron will reduce 2 silver ions because 1 silver ion goes to silver with taking with 1 electron. After balancing the reaction we find the reaction is zinc plus 2 silver plus going to zinc plus 2 plus 2 H. Now, we are going to use the E cell which is nothing but the EMF value we need to measure. It is going to be the E 0 cell at the standard condition and by using the Nernst equation we can write 0.059 by N log of the product side is zinc plus 2 into silver and there are 2 silver. So, it will be squared the reactant side is silver plus squared and zinc the square terms coming because they are stoichiometric coefficient is 2 for both the silver. E 0 cell is nothing but the difference between the E 0 Ag plus Ag the cathode minus E 0 of the anode zinc plus 2 zinc it is already given in minus. So, do not need to convert it to zinc to zinc plus 2 again. So, this value will be 0.80 then minus minus 0.76 volt. So, 1.56 volt. So, that is the E 0 cell value. So, we are going to write it over there 1.56 volt minus 0.059 N is the number of electrons involved over here there are 2 electrons involved. So, it will be 2 log scale of zinc plus 2 solution is given value over here 0.191 silver is in the solid state. So, activity is 1 zinc in solid state activity is 1. So, you do not need to consider silver plus concentration is 0.0289 then squared and if you do all the calculation this is also given in the value of volt all together it comes to 1.4908 volt and this is the EMF of the cell. So, that is how you use the standard deduction potential and the Nernst equation to find out the overall EMF of an electrochemical cell. Now, the last problem. So, over here is a solubility equilibria problem. The solubility of a sparingly soluble silver chromate salt AG2 CRO4 salt is 8.25 into 10 to the power minus 5 mole per liter. So, that is the solubility it is found experimentally. Now, our question is we need to find the solubility product this particular salt AG2 CRO4 assuming. So, there is assumption over there that the salt is completely dissociated. So, considering this assumption we have to find out the Ksp or the solubility product of this salt. So, let us took into the equation first. So, it is AG2 CRO4. So, when it splits out it is going to give me 2 silver plus ion and 1 CRO4 2 minus. So, now you can see the solubility is given over here. So, how much it is actually present as a salt and it is 100 percent dissociating. So, the concentration present in the solution is 8.25 into 10 to the power minus 5 mole per liter. So, when it dissociates it completely dissociates. So, 100 percent of it goes to silver ion and chromatin it is going to create 2 equivalent of silver and 1 equivalent of chromatin. So, it will be 2 into 8.25 into 10 to the power minus 5 mole per liter silver ion present and 8.25 into 10 to the power minus 5 mole per liter chromate ion present there. So, that is we know the ion concentration. Now, what is the solubility product as we know it is going to be the silver ion squared because it has 2 coefficient over there and CRO4 2 minus the coefficient is 1. So, there is no other thing coming into here and now what we need to do we have to just put the concentration we just figure it out over here. So, it will be 2 into 8.25 into 10 to the power minus 5 mole per liter square into 8.25 into 10 to the power minus 5 mole per liter. And if you all do the calculations around it comes out 2.246 into 10 to the power minus 12 mole cube by liter cube. So, that is the solubility product. So, over here if you find this kind of problems coming into your way you have to first find out what is the ion concentration and that is given by the solubility and over there very careful about how much equivalent of ions you are producing over here because that is going to be multiplied with respect to the solubility and that is also going to be coming over here as the power of the system is respect to the stoichiometric coefficient. So, this is all the application part we actually discussed with respect to the problem solving and that is the concluding section of the thermodynamics. I hope that is quite an enjoyable segment for you in the near future. Thank you very much.