Entropy

Summary

In this engaging chemistry lecture, Florence Joie F. Lacsa dives into the intricate topic of entropy, part of chemical thermodynamics. The discussion explores entropy’s definition, its relationship with microstates, and how it predicts the nature of chemical and physical processes. Drawing from historical scientific discoveries, the lecture illustrates entropy's fundamental role in understanding molecular disorder, spontaneity, and thermodynamic calculations. Rich in problem-solving, Joey uses various examples to depict entropy changes in ideal gases and chemical reactions, explaining both statistical and heat-based approaches for calculating entropy. The lecture serves as an essential guide to grasping the underlying principles of entropy and its effects on thermodynamic systems.

Highlights

• Joey explains entropy with reference to historical figures like Carnot and Clausius 🤓.
• The lecture describes entropy as a measure of molecular disorder, tracing its roots to Boltzmann’s statistical mechanics 📈.
• Problem-solving illustrates entropy change calculations for chemical reactions using statistical and heat-based methods ✍️.
• Examples show how entropy increases with volume, temperature, and complexity of molecular structures 🌡️.
• The discussion includes predicting entropy changes in various chemical scenarios, enhancing understanding 🌟.

Key Takeaways

• Entropy is a state function indicating the level of disorder or randomness in a system 🌪️.
• It’s closely tied to the number of microstates available to a system's particles 🧩.
• Entropy changes predict the spontaneity and feasibility of chemical reactions 🔄.
• Historically, entropy was developed to improve the understanding of heat engines and spontaneity 📚.
• Different calculation approaches for entropy lead to the same results, illustrating its robust nature 📊.

Overview

In this intriguing lecture, Florence Joie F. Lacsa takes us through the essential aspects of entropy within the broader topic of chemical thermodynamics. The discussion starts by defining entropy and linking it to system microstates, providing a historical context through the contributions of scientists like Carnot and Clausius.

Joey presents entropy changes in various states of matter, offering a deep dive into molecules moving through different phases. She also tackles complex conceptions of entropy using examples from chemical reactions, showcasing both statistical and classical thermodynamic approaches to calculating entropy, highlighting their validity and equivalence.

The lecture ends with practical problem-solving segments, illustrating how entropy predictions apply to real-world scenarios and chemical reactions. This makes the lecture not just an informative session but also an interactive guide for learners eager to understand how entropy influences thermodynamic behavior and predictions.

Chapters

• 00:00 - 00:30: Introduction In this chapter titled 'Introduction', Joey Laksa, a chemistry lecturer for engineers, introduces the third topic under chemical thermodynamics. The discussion is part of a lecture series in the fifth week of the semester. The specific focus of this lesson is on entropy, which is identified as the second part of a larger four-part series in the subject.
• 00:30 - 01:00: Understanding Entropy This chapter focuses on the concept of entropy within the realm of chemical thermodynamics. By the end of this chapter, you are expected to be able to: 1. Define what entropy is. 2. Explain the relationship between entropy and the number of microstates in a system. 3. Predict the sign of the entropy change for chemical and physical processes. 4. Calculate the standard entropy of a reaction.
• 01:00 - 01:30: Review of Past Discussions In this chapter, we reviewed our previous discussions on the concept of spontaneity in the study of the second law of thermodynamics. We revisited the idea that a process is considered spontaneous if it proceeds on its own without external assistance and is irreversible. Furthermore, the spontaneity of a process is influenced by the temperature at which it occurs.
• 01:30 - 02:00: The Steam Engine and Carnot's Research The steam engine's operations were initially not well understood, with their efficiency at a mere three percent. Sadie Carnot undertook research to enhance the efficiency of heat engines. Drawing from Joel's work, it was discovered that there is a quantitative relationship between heat and work.
• 03:30 - 04:00: Clausius' Inequality and Definition of Entropy The chapter explores Clausius' Inequality and its connection to the concept of entropy.
• 05:00 - 05:30: Boltzmann's Statistical Mechanics The chapter discusses the process of converting heat into mechanical work, particularly through the expansion and compression of gas. It explains how heat removal leads to a temperature drop and the subsequent rise of a piston that compresses the gas, causing the temperature to rise. The cycle is described in terms of its continual movement, influenced by interactions with a heat source.
• 06:30 - 07:00: Entropy Changes in Physical States The chapter discusses the concept of entropy changes in physical states, examining how physical work is expected to produce heat following Joule's experiment. The key insight from the experiment is that heat sources used to perform mechanical work have limitations. It is observed that as cycles continue, the temperature differential between reservoirs cannot be re-established, indicating a limitation in the heat source. This understanding underlines the thermodynamic principle that not all energy transformations can be perfectly efficient due to entropy changes.
• 08:00 - 08:30: Solving Problems of Entropy In this chapter, Rudolf Klosus works on understanding observations from the Cardinal Engine. He determines that the integral of a complete cyclic system is never positive, a concept known as Klosus's Inequality. This can be denoted as the cyclic integral of differential heat transfer amounts, divided by temperature, always being less than or equal to zero. He goes on to define this understanding as 'delta q'.
• 09:30 - 10:00: Standard Entropy and Calculating Reactions The chapter titled 'Standard Entropy and Calculating Reactions' delves into the concept of entropy, emphasizing its nature as a state function that is not dependent on the path taken. The chapter credits Clausus with defining entropy as the measure of thermal energy that can no longer do work.
• 17:00 - 17:30: References and Resources This chapter discusses the concept of entropy as a thermodynamic property, alongside internal energy and enthalpy. It explains how entropy is an extensive property, meaning its value is dependent on the system's thermodynamic parameters. The chapter also presents an alternative perspective on entropy through Ludwig Boltzmann's viewpoint, which suggests that the individual motions of tiny particles within a system define its thermodynamic behavior.

Entropy Transcription

• 00:00 - 00:30 so hello everyone this is joey laksa once again your lecturer in chemistry for engineers so we are now on our third topic under chemical thermodynamics so this is still part of our chemistry for engineers lecture series and we are now on the fifth week of our semester and in this particular lesson we are to discuss about entropy which is the second of the four parts of
• 00:30 - 01:00 our discussion under chemical thermodynamics so at the end of this particular lesson you are expected to be able to do the following define what entropy is explain the relationship between entropy and the number of microstates in a system predict the sign of the entropy change for chemical and physical processes and calculate standard entropy of a reaction
• 01:00 - 01:30 so last meeting we discussed about one of the two important terms in the study of the second law of thermodynamics which is spontaneity wherein we learned that a process is said to be spontaneous when it proceeds on its own without any outside assistance and that it is irreversible we also learned that the process spontaneity is dependent on the temperature by which the process is taking place
• 01:30 - 02:00 even when the world started to use the steam engine the fundamental scientific principles behind their operations remain to be a mystery sadie carneau made a research to improve the efficiency of heat engines which during that time was only three percent from the work of joel we learned that there is a quantitative relationship between heat and work cardinal
• 02:00 - 02:30 wanted to know if the heat source that is used to perform mechanical work is unlimited or not in cardinal's model a heat engine is placed between reservoirs of different temperatures to allow heat flow as the piston moves downward the engine absorbs heat from the source which results to the gastric spine and as the piston moves
• 02:30 - 03:00 further downward the heat is removed from the system and is converted to mechanical work but this results in temperature drop the piston starts to rise the gas is compressed but the temperature is driven off as the piston continues to move upward the cold gas is compressed and the temperature will rise the cycle moves on the heat source is
• 03:00 - 03:30 expected to do work and it is also expected that the work can produce hit following the concepts of joel's experiment however this did not occur because there was a depletion on the temperature differential the difference on the temperature of the reservoirs were not re-established as the cycle went on which led to the conclusion that the heat source that is used to perform mechanical work is limited
• 03:30 - 04:00 rudolf klosus tried to quantify the observations from the cardinal engine he found that the integral of the whole cyclic system is never positive this is what we call as the closest inequality it can be expressed as the cyclic integral of all the differential amounts of heat transfer divided by the temperature is always or equal to zero he defined this delta q
• 04:00 - 04:30 over t as entropy and further discovered that it is not path dependent hence it is a state function it was clausus who described entropy as the thermal energy that is considered that is considered as the measure of thermal energy that is no longer available for doing
• 04:30 - 05:00 useful work entropy became one of the thermodynamic properties along with internal energy and enthalpy in addition entropy is an extensive property and its values is dependent on the thermodynamic parameters of the system it describes entropy on the other hand was viewed in other means ludwig boltzmann they believe that the individual motions of tiny particles of a system define its thermodynamic behavior that
• 05:00 - 05:30 is the more spread the energy is the less useful it is this led to more popular definition of entropy which is a measure of the molecular disorder of a system he came up with a statistical means of calculating the change in entropy of the system based on the final and initial microstates of the system so it was him who is considered as the
• 05:30 - 06:00 father of the statistical mechanics okay a microstate is the number of possible arrangements of the elements or substances that comprise a system in the given illustration we are seeing four microstates so one two three four so meaning there are four types of arrangements okay possible arrangements so when there is an increase in the randomness of a system
• 06:00 - 06:30 there is also an increase in entropy which is denoted by a positive delta s on the other hand if there is a decrease in entropy a negative sign for delta s is applied okay in general the enthalpy of the system increases when the number of microstates possible for a system also increases which may be due to an increase in volume
• 06:30 - 07:00 increase in temperature and an increase in the number of molecules that usually comprise the system this is because any of these changes increase the possible positions and kinetic energies of the molecules making up the system now let us examine the increasing entropy of water from its solid to its vapor form we can see that in a solid phase the ice molecules have rigid and crystalline structure which
• 07:00 - 07:30 restricts the motions of the molecules into vibration only among the three this has the smallest number of microstates hence the lowest entropy in the liquid water we can observe that the molecules comprising the system are free to vibrate and rotate there is also an increase of freedom with respect to molecular translations in the vapor phase the molecules are spread out and there is a complete
• 07:30 - 08:00 freedom for the molecules to translate vibrate and rotate with this the vapor phase of the water is considered to have the largest number of microstates among the three okay in the photo is a dissociation of an ionic substance in a solvent notice that the ions here that compose a compound cause an increase in the microstate of the com of the system after its dissolution
• 08:00 - 08:30 hence there was an increase in entropy on the other hand the reaction between nitrogen oxide and oxygen gas results in a decrease in entropy this is because the formation of nitrogen dioxide resulted in lower number of microstates now let us answer this problem to illustrate the application of the concepts learned
• 08:30 - 09:00 for each of the following pairs choose the substance with the higher entropy per mole at a given temperature okay so it should be let me change it to blue for letter a we are to choose between the liquid argon and the gaseous good the one with the higher entropy is the gaseous are argon okay
• 09:00 - 09:30 because gaseous molecules can move freely and have more dispersed energy compared to the liquid molecules for letter b we have one mole of helium gas in 15.0 liters or one mole of helium gas in 1.5 liters it is the one mole of helium gas moving in a 1.5 0 liter
• 09:30 - 10:00 that has a higher entropy okay so let me just underline it gases why why is this so because gases feel in the container where they are put in hence they also assume the volume of the container higher volume means bigger space for them to move around which results in higher number of microstates this is due to the higher translational energy levels
• 10:00 - 10:30 as the number of particle locations is increased okay for number three we are to choose between one mole of hydrogen gas at five atmosphere or one mole of hydrogen gas at 1.5 atmosphere pressure okay the higher the pressure the lower the volume of a gas okay so which means this one has a lower volume this gives gives us the answer for a higher entropy for the one mole of the hydrogen gas
• 10:30 - 11:00 at 1.5 atmosphere pressure okay following the previous explanation the answer is this one okay because we have a higher volume compared to this one okay for letter d we are to choose between sodium and rubidium both of them are solid however rubidium is heavier than sodium which means that its energy levels
• 11:00 - 11:30 are higher this leads to a higher number of microstates so the answer for letter d is rubidium okay for letter e we are to choose between a copper metal at 273 kelvin and the same copper metal up to 98 kelvin the one with a higher entropy is the copper metal at 298 kelvin right this is because when temperature
• 11:30 - 12:00 increases the average kinetic energy of the particles also increases hence they will be more dispersed this results in an increase in the number of microstates then we have solid sodium chloride and aqueous sodium chloride okay between the two it is the queue use sodium chloride that has a higher entropy because dissolving a molecular solid leads to an increase in entropy this is because when ionic solid
• 12:00 - 12:30 dissolves in water the crystals break down and the ions will have an increase in freedom of motion as they separate with this the number of microstates is also increase and for the energy we are to choose between nitrogen oxide and dinitrogen tetraoxide both are gashes the one with higher entropy is a dinitrogen tetraoxide okay um because it has more bonds that
• 12:30 - 13:00 vibrate compared to that nitrogen oxide okay the the higher the bonds the number of bonds that vibrate it what is what makes um its entropy higher then we have a gaseous methane and a liquid ethanol okay it is the gaseous methane that has a higher entropy because even the liquid ethanol has a more complex
• 13:00 - 13:30 structure the freedom of molecules of the methane in its gaseous state dominates the entropy that results from the molecular complexity of the liquid ethanol there are two approaches of quantifying entropy changes the one made by boltzmann which is based on the number of microstates in the system and that of clausus which is based on the heat involved in reactions or in the system this two approaches give the same result
• 13:30 - 14:00 as can be illustrated on the succeeding problem so here's the problem calculate for the change in entropy for one mole of an ideal gas undergoing an isothermal reversible expansion from 1 liter to 2 liter at 298 kelvin show that either the statistical approach or the heat based equation get will give the same answer
• 14:00 - 14:30 okay so i think i'm using a blue pointer so let me change this one to y so let's first start with the statistical approach okay so let us have two identical liter one liter flasks connected to each other to illustrate the expansion process
• 14:30 - 15:00 assume that these two flaws are identical and they are of one liter volume each okay let us assume that the volume of the pipe connecting the gases are negligible compared to the volume of the flasks okay so pardon me with my drawing it isn't good so they're connecting just imagine them to be connecting
• 15:00 - 15:30 okay so i'm having trouble with my pen top i'm not comfortable with it all right so there um and then let's see before the expansion process the valve that connects the two so there is a valve between these two is closed okay and then imagine all the particles of one mole of the ideal gas to occupy this
• 15:30 - 16:00 one liter flask and then the bulb will be opened so there was an opening of the ball so paradise it's still one liter so with the opening of the bulb the gas molecules from here will simultaneously spread out to both of the flasks
• 16:00 - 16:30 okay to occupy both of the flasks this changes the volume from only one liter to two liters and the particles comprising the system can move freely inside the container but to make matters simpler to illustrate the concept of the microstates let us take one particle from one mole of the gas please recall that one mole of the substance
• 16:30 - 17:00 is equivalent to your avogadros number [Music] 6.02 times 10 to the 23 particles [Music] all right so moving back [Music] so say we have one particle uh in this flasks
• 17:00 - 17:30 okay we can say that the particle may either occupy flask a or class b okay [Music] so we can say that there are
• 17:30 - 18:00 two number of microstates so this is number of micro states or possible microstate for one particle what if we take two particles in this um arrangement so let's see how many microstates is possible so can i increase the size allow me to increase the size of the flasks because we now have two particles but
• 18:00 - 18:30 this should be the same all the flux should be the same because they only mean the same thing okay so they're around four so let's see so can be that the flask has particle a on this side and b on the other one and then it can be exchanged
• 18:30 - 19:00 both are here and both are there so it's more like of studying your simple statistics because uh microstates the concept of microstates and entropy is actually a statistical in nature so let's measure so if you will count we have four possible arrangements meaning there are four microstates what if we have three particles so let's see how many
• 19:00 - 19:30 microstates are possible so i think you have around eight let me draw the flasks first pardon me with my drawing anyway this isn't a drawing class okay
• 19:30 - 20:00 two three six seven eight okay so let's see sub a b where's my red bed so let's say there are three of them it can be that this has a b and c or the three particles are here
• 20:00 - 20:30 it can also be that a b is here and c is there then and that can be that ac is here and b is alone on the second flask and it can be a is alone in the second class and you have b and c there okay so b and c and that is a so there are around
• 20:30 - 21:00 four microstates all and all so if you try to notice the number of microstates exactly two this one is uh let me change the color okay so this is actually two raised to 1
• 21:00 - 21:30 this is 2 raised to 2 and this is 2 raised to 3 so and so on and so forth so 2 is the possible arrangement is the 3 is the number of particles and notice that we have n a particles all in all for your one mole okay so this actually pertains to your number of microstate w
• 21:30 - 22:00 okay from the boltzmann equation we have the change of entropy in your system is equal to your kb times your loan of w okay but we also know from our discussion that your kb is the same as the universal gas constant r divided by n of a so you have here lon 2 raised to ends of a
• 22:00 - 22:30 okay from our trigonometry the identity we know for a fact that your long to ends of raised to n is a n of a can be written as n a ln of 2 that will cancel out the n a okay so let me cancel out the n a so you are left with r times lon of two okay so your
• 22:30 - 23:00 delta s of the system is equal to your r which is given as eight point 8.314 joel per kelvin times your loan of 2. so let's have our calculators and let's check it so lon of 2 is equivalent to 0.69 times
• 23:00 - 23:30 8.314 this will give us 5.76 tools so this is 5.76 joules only only have one significant figures there for one and two so may i write it here this is six chose okay so
• 23:30 - 24:00 let's solve the problem using the hit base equation the closures all right so says said that the delta s of the system is equivalent to the integral of your dq over t and this is for the
• 24:00 - 24:30 reverse reversible process because a reversible process is the basis of its derivation okay so we also know where will i write from the first of thermodynamics that your delta u is equal to your q plus w and your w is negative p this of v
• 24:30 - 25:00 however for an isothermal process your delta u is equal to zero so that will give you your q is equivalent to negative w so let me substitute this definition to the working equation i don't have any other colors but
• 25:00 - 25:30 minimal all right so you have where are we so the integral of your integral of your negative w over your t okay or d w over t is equal to your negative so negative p
• 25:30 - 26:00 dv all over t so this is the same as the integral of your pdv over t okay but since we are dealing with an ideal gas we can use the ideal gas equation pv and rt uh to solve for p therefore p is equal to your n r t all over your v
• 26:00 - 26:30 okay so this will give you the n r t over t the integral of your d v over v so we can cancel out the t [Music] i'm sorry for the train everything is closed so can i just rewrite this equation here
• 26:30 - 27:00 so you have your delta s of the system is equivalent to your n r okay so this is evaluated from v 1 to b f the lon of v evaluated from v i to v m so this is your nr loan of v2 over this of one so substituting it you have you're substituting the given you have
• 27:00 - 27:30 one mole of your ideal gas times your r which is 8.314 joule per mole kelvin okay is equal to oh i'm sorry time salon is ln
• 27:30 - 28:00 times ln of 0b2 2 liters over 1 milliliter so let's check on the units okay the mole cancels out liter cancels out so let's solve for the delta s of your system so your delta s of your system is equal to it's the same loan of 2
• 28:00 - 28:30 times 8.314 sold on 2 times 8.314 still 5.76 so this is 5 point seventy six joel per kelvin which is your six joules per kelvin let me check my unit to be joel for kelvin sorry for this
• 28:30 - 29:00 chopper kelvin all right okay for sample problem number two should be number two not number four predict the sign of the entropy change of the system for each of the following reactions let's see for letter a for letter a the predicted sign of the
• 29:00 - 29:30 reaction is the positive okay so for here delta s is positive y because there is an increase in entropy of the system so as observed there are the same number of molecules you have one and two for the product and two for the reactant sides however the entropy increases as a result of the separation and
• 29:30 - 30:00 mixing of the molecules and then we have letter b the combustion reaction of your propane for this one your delta s is negative because this process is uh gives us a decrease in entropy as you can see the reactant sites are both made of gaseous molecules which are freely moving they have a very uh gases have very high values of entropy and then the gaseous reactants the gaseous products rather are also
• 30:00 - 30:30 available but we have here the presence of a liquid system which means that there was any in entropy of course prediction are very difficult to do but you don't need to worry because it is not all the time that prediction of entropy sign of a chemical reaction comes in we do not need to worry because we can solve for the standard entropy of a given reaction using the standard entropy values of various
• 30:30 - 31:00 products following the same equation or yeah the same equation as what we had during the calculation of the standard heat of reaction only we are to change the standard um heat of reaction into standard entropy of the different substances that comprises the chemical reaction so let us solve for this one
• 31:00 - 31:30 using the standard entropy values calculate delta s for the reactions in the previous problem and check if the prediction previously made is correct so let us copy you have hydrogen gas and i2 so let's say let's see let me rewrite this one so for letter a
• 31:30 - 32:00 so for um letter a we have hydrogen gauss combining with i2 bus to give you two h i guesses so i already searched for the standard entropies for all the products and reactants so let me just write it here
• 32:00 - 32:30 you have 130.6 jolt per mole kelvin and then for i2 that is 260 joel per mole kelvin and for the hi you have 206.13
• 32:30 - 33:00 joel per mole kelvin so again this standard values can be found on in different texts in different books in different sites later on at the end of the discussion we'll try to look for one okay so these are but constants usually they can be found for the reliable values it is better for you to look for books they are found in the appendices the
• 33:00 - 33:30 general chemistry books but um just like our quiz and thermal chemistry i will be providing the values that you're going to need during the process so that you won't have any troubles okay or using internet connections but during face-to-face classes the students have photocopies of the appendices of the books and it is them who are going to look for one
• 33:30 - 34:00 so values okay so let me check this is 130.6 260.58 and 206. so we have the delta s of your reaction okay i hope you can still recall the meaning of this little dot from our thermo chemistry topic this means that
• 34:00 - 34:30 the entropy of this substances was taken at standard temperature standard of temperature and pressure which is one atmosphere and at 298 kelvin which is 25 degrees celsius so we have the summation of n s of your products minus the summation m s of your reactants okay so for the delta s of your reaction
• 34:30 - 35:00 we have so this is 2 mole of your h i times 206.13 joel per mole kelvin minus one mole of your h2 hmm
• 35:00 - 35:30 let me get it one mole of your h2 which has a standard entropy value of 100 30.6 joel per mole kelvin so this is a plus then one mole for your iodine times 260 point
• 35:30 - 36:00 58 joel per mole kelvin okay so let's check on our units the mole cancels out ball cancels out all cancels out your entropy is energy unit has a unit of energy per temperature so please uh hold on your calculators
• 36:00 - 36:30 and let's compute this uh delta s of reaction okay you have 2 times 206.33 minus 130.6 minus 260. 58. so let me check thirty another six one thirty point six plus two sixty
• 36:30 - 37:00 it is three ninety we have two times 206.33 minus answer for 22.06 just 22.06 joel kelvin are you getting the same 2 times 206.33 minus 130.6
• 37:00 - 37:30 plus 260 point 58. 206.33 minus 130.6 plus 260.58 21.48 sorry
• 37:30 - 38:00 so this is 21.48 [Music] 21.48 so the sign is positive so let us check but it has positive twenty one point forty eight function forty one point forty eight yeah you know what two times two oh six
• 38:00 - 38:30 point thirty three minutes so let's go to the combustion of propane so you have there c3 h8 gas plus five o2 gauss will give you three co2 nagashus
• 38:30 - 39:00 plus 4h2o liquid or glass liquid [Music] liquid okay so again i took the values of the standard entropy of each from the appendix of brownlee may and
• 39:00 - 39:30 burstin this chemistry the central science so this is 269.9 joel per mole kelvin and then you have 205 joel per mole kelvin and then 2 1 3 4.7 joel
• 39:30 - 40:00 per mole kelvin for the carbon dioxide and the last one near liquid water 69 point 94 joel per mole kelvin okay so so for your delta s of your reaction a standard entropy of reaction so the summation of m times s
• 40:00 - 40:30 of your products minus the summation of your n it just denotes the number of moles of each of the products and reactants so your delta s of reaction is three moles of carbon dioxide times two one three point seven joule
• 40:30 - 41:00 per mole kelvin plus 4 moles of your hydrogen 69.94 joule per mole kelvin okay minus you have one mole of your propane and the standard entropy is 269.9
• 41:00 - 41:30 joel per mole plus 5 moles of oxygen times 205 joule per mole kelvin so notice that you have now a value for oxygen even if it is a free element when you talk about this standard entropy okay at this one is at 298 kelvin unlike
• 41:30 - 42:00 in the standard hero formation wherein you have free element that your oxygen is zero why is this so because remember that entropy talks about the energy of the system so you know for a fact that oxygen is made up of two atoms okay oxygen molecule is made up of two atoms of um oxygen so there are rotations translations and
• 42:00 - 42:30 vibrations okay there are motions along with the bonds of the oxygen which okay gives the value for your s okay so please take note of that form of
• 42:30 - 43:00 carelessness okay all right so all units of the mole cancelled out so delta s of your reaction is so you have 69.9 plus 5 times 205
• 43:00 - 43:30 three two one three plus 1694 negative 374.04
• 43:30 - 44:00 so i don't think four three negative 374. 24 joel kelvin i hope i got it correctly and double check three times two ones three times two one three plus four times 69.94 minus 269 minus
• 44:00 - 44:30 five times two oh five negative twelve three times two one three point seven plus four pi 69.94 minus 269.9 minus five times 205. 374.04 habila
• 44:30 - 45:00 21.4 in one single future for sultan so for the second one we have a negative sign okay delta s is negative so let's see there was a decrease in entropy so we predict we predicted the chemical reactions correctly so okay these are the list of references thank you for listening so let us check some websites
• 45:00 - 45:30 wherein we can find the entropy values what happened see google so you have standard
• 45:30 - 46:00 entropy thermodynamic values so you have your dhdf so let's check that quantum iosa water h2o liquid value water is actually on the oxygen part oxidant some water
• 46:00 - 46:30 [Music] water water yeah liquid 69.91 okay so i use 69.94 it is
• 46:30 - 47:00 so that is the reason why i am giving it to you during exams okay let's check for propane propane propane 369.9 at least parents are propane and then oxygen oxygen [Music]
• 47:00 - 47:30 because it is a free element but when you have a standard of entropy you have a value because you have motion so it cannot be zero so that is 25.138 what i use is only 205 so again um don't worry because during exams increases i'll be again giving you the values for your thermodynamic uh tables okay for your hs and
• 47:30 - 48:00 gibbs free energy so next not epic nothing you gives free energy okay so there again the thermodynamic values are constants and they can they are because of that uh they are accessible anywhere okay so there um you know mean it's not so sorry 30 minutes okay all right so see you in our discussion of the second
• 48:00 - 48:30 law of thermodynamics it will come shortly next night