Free Energy (B)

Summary

In this engaging chemistry lecture by Florence Joie F. Lacsa, the intricacies of Gibbs Free Energy and its relationship with enthalpy and entropy are explored. The video dissects the dependencies of Gibbs Free Energy on various thermodynamic parameters and elucidates the concept of equilibrium in chemical reactions. The lecturer navigates through the detailed calculations involved in estimating boiling points and calculating Gibbs Free Energy under standard conditions. Practical examples include solving for the maximum useful work derivable from combustion reactions, highlighting the strong connection between theoretical formulas and real chemical processes. With step-by-step problem-solving, this video serves as a comprehensive guide to understanding free energy in chemical thermodynamics.

Highlights

• The lecture explores how Gibbs Free Energy combines enthalpy and entropy to form an extensive property and state function.
• Detailed explanation of temperature and pressure dependencies on Gibbs Free Energy is provided.
• Practicals covered include estimating boiling points, using thermodynamic data to solve real-world chemistry problems.
• Calculation of Gibbs Free Energy at standard conditions, with emphasis on importance of equilibrium in reactions.
• Coverage of maximum useful work obtainable from combustion, illustrating practical applications.

Key Takeaways

• Gibbs Free Energy combines enthalpy and entropy, reflecting both properties in equal measure. 🌡️
• Free Energy is dependent on the temperature, pressure, and state of reactants/products. ⚖️
• The video provides clear walkthroughs of calculating boiling points using thermodynamic values. 📈
• The lecture emphasizes the dependence of Gibbs Free Energy on standard conditions and equilibrium. 🌍
• Understand how spontaneous and non-spontaneous reactions differ through practical calculations. 🔄

Overview

This dynamic lecture by Florence Joie F. Lacsa takes students on a deep dive into the world of Gibbs Free Energy. You'll discover how this fundamental concept in thermodynamics is intimately tied to both enthalpy and entropy, two critical properties that define the behavior of chemical systems under different conditions.

Students are equipped with the skills to calculate Gibbs Free Energy in various scenarios, focusing on equilibrium conditions. Using clear examples, the lecture walks through the estimation of boiling points and highlights the importance of standard thermodynamic data in understanding chemical reactions.

From defining the parameters that determine reaction spontaneity to dissecting real-life applications such as combustion processes, this lecture offers an engaging and thorough exploration of Free Energy. The practical examples and step-by-step demonstrations make complex topics accessible and engaging for all learners.

Chapters

• 00:00 - 03:00: Introduction and Review of Gibbs Free Energy The chapter introduces the concept of Gibbs Free Energy, essential in chemistry for engineers. It builds on prior discussions, emphasizing the combination of enthalpy and entropy in the Gibbs Free Energy equation. The lecturer, Jovi Laksa, continues from the previous lecture to deepen the understanding of this fundamental thermodynamic quantity.
• 03:00 - 06:00: Thermodynamic Parameters and Standard States This chapter discusses the concept of Gibbs free energy as an extensive property and a state function. It highlights the dependence of Gibbs free energy, enthalpy, and entropy on temperature, pressure, and the state of reactants and products in a chemical reaction. The chapter emphasizes the importance of understanding these dependencies to fully grasp the thermodynamic parameters and states involved in chemical processes.
• 06:00 - 18:00: Problem Solving on Boiling Point Estimation The chapter discusses problem-solving strategies for estimating the boiling point of substances using thermodynamic principles. It begins by explaining the concept of standard states, emphasizing that they relate to conditions of one atmospheric pressure and temperature of 298 Kelvin, with all reactants and products in their pure states. The importance of Gibbs free energy in these calculations is discussed, reinforcing how its standard state is defined by the enthalpy and entropy of the system.
• 18:00 - 31:00: Delta H, Delta S, and Delta G Calculation The chapter discusses the concepts of standard energy of formation and standard free energy of formation for reactants and products. It explains that these are important thermodynamic quantities that can be accessed from various sources. The lecturer also mentions a correction about quiz number two regarding the values previously stated.
• 31:00 - 37:00: Spontaneous vs Non-Spontaneous Processes In this chapter, the focus is on distinguishing between spontaneous and non-spontaneous processes in the context of thermodynamics. The discussion includes an emphasis on understanding and utilizing the thermodynamic quantity table efficiently, especially in exam scenarios. This is important for ensuring that students can apply their knowledge of chemical processes, particularly during exams where timing and safety are crucial.
• 37:00 - 43:00: Free Energy and Work The chapter 'Free Energy and Work' discusses the evaluation of thermodynamic quantities using given values during examinations. It mentions that the only opportunity to test students' ability to find correct values will be in an upcoming quiz on thermochemistry. A PDF will be provided or linked, containing thermodynamic quantities for selected substances, which will serve as a reference material.
• 43:00 - 47:00: Free Energy of Ideal Gas In the chapter titled 'Free Energy of Ideal Gas,' it discusses the standard references for thermodynamic quantities relevant for an upcoming quiz. The problem posed in the chapter involves using these thermodynamic values to estimate the boiling point of benzene in its liquid state, with the answer to be expressed in degrees Celsius.
• 47:00 - 47:00: Problem Solving on Ideal Gas Problem Solving on Ideal Gas: In this chapter, the concept of boiling point is discussed. The boiling point of a substance is defined as the temperature at which the vapor pressure of the liquid equals the pressure of its surroundings. This means that at the boiling point, the liquid is in equilibrium, where the rate of the forward reaction (evaporation) equals the rate of the reverse reaction (condensation). The understanding of this equilibrium is crucial in solving problems related to ideal gases.
• 47:00 - 48:00: Conclusion and Closing Remarks The chapter 'Conclusion and Closing Remarks' discusses the concept of chemical equilibrium, particularly focusing on the forward and reverse reactions being equal. It emphasizes that in an equilibrium state, the Gibbs free energy change (delta G) is zero.

Free Energy (B) Transcription

• 00:00 - 00:30 hello everyone this is jovi laksa once again your lecturer in chemistry for engineers so this is the second part of our discussion under free energy despite a continuation of our first part okay so let's continue all right so please take note that gibbs free energy is composed both of the enthalpy and the entropy of the
• 00:30 - 01:00 system it means that it also embodies the properties of the two hence the gibbs free energy is also an extensive property and a state function and since the enthalpy and the entropy of the system is highly dependent on the temperature pressure and state of the reactants and products in the chemical reaction it is also important to take note that its corresponding gibbs free energy is also dependent on
• 01:00 - 01:30 the same thermodynamic parameters so when the enthalpy and entropy of the systems are at in their standard states it means that its corresponding gibbs free energy is also at its standard state now let's recall that a standard state pertains to one atmospheric pressure to 98 kelvin temperature and that the reactants and products are in its pure states moreover the gibbs free energy of reaction can also be calculated
• 01:30 - 02:00 by combining the values of the standard energy of formation or standard free energy of formation of the reactants and products the standard free energy is another thermodynamic quantity that gives us constant values which can be accessed in different sources okay so by the way for quiz number two i need to correct my prior statement that i'm going to give the values of
• 02:00 - 02:30 your thermodynamic quantities because um i need to check on your skill on on whether or not you know how to use this thermodynamic quantity table or not because during the exam i will be the one who will give you the values to save time because remember during the midterm exam we have chemical safety that thermodynamics and nuclear chemistry asked the
• 02:30 - 03:00 coverage so mahabasha so during the exam i'm going to give you values and i do not have any other chance or platform in which i can test your skill whether or not you know how to look for correct values or not except for the upcoming quiz on thermo chemistry so i will be uploading or will be giving the link for this pdf file for the thermodynamic quantity for the selected substances which will serve as our
• 03:00 - 03:30 standard reference for the thermodynamic quantities that we are going to use during the third quiz okay so let us solve another problem use the thermodynamic values given to estimate the boiling point of benzene which is in its liquid state express your answer in degrees celsius okay when we talk about
• 03:30 - 04:00 boiling point um of a substance it it means that it is when the vapor pressure of the liquid is in equilibrium with the pressure of its surrounding so if you have an equilibrium it means that the rate of uh where the forward reaction and the reverse reaction are equal so you have here
• 04:00 - 04:30 c6 c6 h6 okay a forward and reverse reaction equal for c6 h6 so it's up to you where is your liquid and your gas okay in reverse all right we also know for a fact that delta g is equal to zero when you are dealing with equilibrium how do we know that we
• 04:30 - 05:00 are in equilibrium from the definition of the boiling point okay delta g is equal to our delta h minus t delta s right so we have this as zero is equal to delta h minus can i change the t into t b to signify that we are looking for a boiling point so your tb or your boiling point is equal to delta h over delta so
• 05:00 - 05:30 transpose and then okay so your delta h is your age of products minus your age of reactants all over s of products minus s of
• 05:30 - 06:00 reactants so meaning we need to look for the delta h of formation of your liquid and gaseous benzene same is true with their entropy entropy values this is s okay so let's look for the table
• 06:00 - 06:30 okay so there we're looking for benzene here c6h6 that is gaseous and liquid so you may first write them in your paper so you have 82.9 for the delta heat delta h information rather
• 06:30 - 07:00 for the gaseous benzene and 49.0 for the liquid benzene and then for the s you have 30 200 wait here 269.2 jolt per mole kelvin for the gaseous benzene and 172.8 joule per mole kelvin for our um liquid benzene so lagrangito
• 07:00 - 07:30 so for the liquid benzene we have here 49 kilojoules per mole and this is 82.9 kilojoules per mole
• 07:30 - 08:00 for s we have 172.8 so this joule per mole kelvin and this one you have 269 joule per mole kelvin so let's substitute can i move it here so your tb is equal to the tus
• 08:00 - 08:30 product 82.9 kilo joule per mole minus 49 kilojoule per mole divided by 269.2 joel per mole kelvin
• 08:30 - 09:00 minus 172.8 joules per mole kelvin okay but as you can see you have kilojoules at the numerator so we might as well change the joule into kilojoules so that it will cancel out all right so let's check on our units napa temperature la la bass the final
• 09:00 - 09:30 unit okay the kilojoules will cancel out same is true with the mole including the joules in the denominator so you have the kelvin going up so you have a temperature unit so let's solve 269.2 minus 172.8 divided by 1000.
• 09:30 - 10:00 so memory in minus 49. now for your memory call 268.2 minus 172.8 divided by 1000. you have 82.9 minus 49. okay so i got two 351 point 66 kelvin
• 10:00 - 10:30 but we are required to express our answer in centigrade so tb is 351.66 kelvin minus 273.15 to make it celsius we have 78.51 78 51 degree
• 10:30 - 11:00 celsius 78.5 now let's ignore the zero there 49 and 49 so 79 equals zero 79 degrees celsius
• 11:00 - 11:30 so 79 degrees celsius that is 78.5 yeah let's move on to another problem calculate delta h standard delta h standard delta s and standard delta g at 298 kelvin for the reaction below and
• 11:30 - 12:00 show that delta g is equal to delta h minus t delta s and we are we are also required to report if the process is spontaneous or not okay so hindi na delta s of the universe young acting basis but of delta g okay of course um the standard heat of reaction delta h is equal to the summation
• 12:00 - 12:30 for letter a we have here okay delta h of the reaction is equal to the summation of n the delta h of formation standard of your products here minus the summation of n the delta h of formation of our
• 12:30 - 13:00 reactants okay so this is equivalent to n moles of c to h six the gas times the delta h of formation of vrc2h6 nagas plus n moles of your h2o times the delta h of formation
• 13:00 - 13:30 of your h2o minus sorry after and moles of your ch3oh times a delta h of formation of your ch3oh plus n of h2 times the delta h of formation of your
• 13:30 - 14:00 h2 sorry about that in the fascia [Music] okay so this means that we know we need
• 14:00 - 14:30 delta h of formation of all the reactants there so now all right so we're looking for a 2ch3oh tasha gashu so you have 201.2 for the delta h of formation 161.9 for the gibbs free energy of formation and 237.64 delta s so see
• 14:30 - 15:00 we're going to solve then for delta g n delta s based on the problem so we might as well get everything that we need and we have c2h6 that is gaseous you have negative 84.6 my then you have negative 32.89 for gibbs free and then 229.5 for the entropy then move on let's move on to h2
• 15:00 - 15:30 so this is alphabetically arranged so here for h2 the heat of formation for enthalpy of formation is zero delta g is zero for s you have thirty one thirty point fifty eight and then your h2o by the way your h2o is under oxygen okay not hydrogen it's under oxygen so gaseous ionic liquid so you have negative 241.82
• 15:30 - 16:00 and then negative 228.57 for delta g and then 188.83 for the entropy so i hope you were able to erode those figures down okay so here for the delta h of formation ch3oh you have
• 16:00 - 16:30 here negative 201.2 so so let's delta h of formation and it is in kilojoule per mole okay for h2 this is zero and this is negative 84.68 and then it was negative 241.82 so let us substitute the values
• 16:30 - 17:00 so this is difficult okay so you have there one mole times negative 84.68 kilojoules per mole plus 2 moles times negative 241.82
• 17:00 - 17:30 kilojoules per mole minus 2 moles times negative 201.2 kilojoule per mole plus one mole times zero
• 17:30 - 18:00 kilojoule per mole during the quizzes and exam if i don't see zero here i actually subtract from the points on the process zero not because it is a free element you just write zero without explaining why okay
• 18:00 - 18:30 should be complete okay so so negative eighty four two four point sixty eight plus two times negative 241.82 minus 2 times negative 201.2
• 18:30 - 19:00 so 165.92 let's repeat negative 84.6 plus 2 times negative 241.82 minus 2 times negative 201 point what this is 165.92
• 19:00 - 19:30 or let's omit the zeros of our sig figure so it is 1 65.9 negative 165.9 kilojoules because all the mole will cancel out another moan mold mold mold so we are left with kilojoules okay
• 19:30 - 20:00 so here for delta s so from the values that we took from the thermodynamic uh value table so have your s there in joules mole per kelvin so ch3oh from my notes is this is 237.6 this one is 130 point
• 20:00 - 20:30 58 and this is 229 this one is negative 188.83 okay so the standard entropy of reaction is equal to the summation of n s of your products minus the summation and of the
• 20:30 - 21:00 is of your reactants okay so similar to what we did with the standard heat of formation so let us substitute our values so you have here one mole times 229.5 jol per mole kelvin
• 21:00 - 21:30 plus okay two moles of water times negative 188.83 okay so what happened there sorry whoops pen pen pen please kilo this is kilo no joel per mole kelvin
• 21:30 - 22:00 okay minus you have two moles here times 237.6 joe per mole kelvin plus one mole times have their one thirty point fifty eight joule per mole
• 22:00 - 22:30 kelvin okay so notice that all of the moles will cancel out again okay so let's solve delta not delta s it's as long
• 22:30 - 23:00 of your reaction yes six plus one set and fifty
• 23:00 - 23:30 negative 188.83 so it's 1.38 joule per kelvin 2 times 237.6 plus thirty point fifty eight two nine point five plus two times
• 23:30 - 24:00 negative one eighty eight point eighty three it's negative this negative or positive it's nothing positive sorry
• 24:00 - 24:30 let us check keep h2o h2o positive 188.3 what is positive 188.3 1.5 plus 2 times 188.83 so it's 138 point this is positive one point
• 24:30 - 25:00 38 joel kelvin for a sig fig okay next we will solve for the delta g we'll use all right
• 25:00 - 25:30 so same linear procedure the delta g of formation this is kilojoule per mole so you have here negative 161.9 this is zero and this is 32.89 and this one is negative 2 to eight point
• 25:30 - 26:00 six no five fifty seven so this is 57 all right so the delta g standard delta g of reaction pattern the summation of n times the delta g of formation
• 26:00 - 26:30 of all the products minus the summation of m the delta g of formation of the reactants okay so for the delta g of the reaction you have [Music] one mole times negative 32.89 kilojoule per mole
• 26:30 - 27:00 plus 2 moles for the water times negative two to eight point fifty seven kilo joule per mole okay minus two moles times negative 161.9 kilojoule per mole
• 27:00 - 27:30 plus one mole of your hydrogen times zero kilojoule per mole okay so let's solve right so let's solve all together
• 27:30 - 28:00 negative 32.89 plus 2 times negative 2 to 8.57 minus 2 times negative 161.9 so i got 166.2 negative 166.2 kilojoules okay so your
• 28:00 - 28:30 delta g of reaction is negative 166.2 kilojoules that is for letter a and then show that delta g is equal to delta h minus t delta s okay so we know that delta g is equal to delta h minus t delta s at 298 kelvin
• 28:30 - 29:00 so the delta g of reaction is equal to what is our delta h plus it is negative 165.9 kilojoules that is negative 1 6 165.9 kilo joules what is negative 165.9
• 29:00 - 29:30 kilojoules minus the temperature is 298 kelvin times the s 1.380 it is 1.380 joule per mole no jolt per kelvin joel per
• 29:30 - 30:00 kelvin so the unit is joe per kelvin or energy per unit temperature for the entropy kelvin okay let's make this into kilojoules so that we can subtract it so that 1000 joules is to 1
• 30:00 - 30:30 out we have kilojoule kilojoule there so let's solve 298 times 1.38 divided by not 1000
• 30:30 - 31:00 is negative 166.3 kilojoules so basically they're almost the same so 166.2 n 0.3 sometimes it's on the calculation errors okay you want to round off linear all right is the process spontaneous or not since delta g is negative the
• 31:00 - 31:30 process is spontaneous because delta g is negative right what's next
• 31:30 - 32:00 okay here an unspontaneous step can be driven off by a spontaneous process to occur such as the spontaneous process of fuel combustion inside an internal combustion engine to run a non-spontaneous car so we need to have a spontaneous process to drive it off um this is true
• 32:00 - 32:30 for chemic this is also true for chemical reactions so for chemical reaction we have a coupling process okay that is being used we're in a spontaneous process here drives off a non-spontaneous process so if you can see to produce copper from this copper to o which is not a spontaneous process we need to copper couple it with carbon plus oxygen process so
• 32:30 - 33:00 when which is uh which is a spontaneous process so when you add them together just like in the has law the one half o2 will cancel out you have now a spontaneous process as the net reaction now um let us see the relationship between free energy and work okay when delta g is
• 33:00 - 33:30 zero it means that there is a spontaneous process in here your delta g is the same as the maximum useful work obtainable from the system as the process takes place but when it is greater than zero it means that the process is non-spontaneous which uh in here your calculated delta g will be the minimum work that must be done to the system to make the process uh yeah to keep that process
• 33:30 - 34:00 taking to make the process okay take place okay so please take note of this two different definitions right so let's solve a problem what is the maximum useful work that can be accomplished under standard conditions by the combustion of acetylene gas assume that liquid water is produced okay so maximum useful work so we are
• 34:00 - 34:30 here um dealing with most probably a spontaneous process anyway we are dealing with the combustion process we know that all combustion processes are spontaneous so you have the combustion of acetylene gas c2 h2 gas plus oxygen gas to give you your carbon dioxide gas plus your
• 34:30 - 35:00 water and subverting to assume it to be liquid so is it balance okay no h2 5 4 5 so this is five over two so what do we need in order to know the maximum useful work okay the value of work max is equivalent to your delta g of reaction if the process is is
• 35:00 - 35:30 spontaneous so we can check actually so this is your summation of n times the delta g of formation of your products minus the summation of m the delta g of formation of your reactants okay so again we need the delta s of this values
• 35:30 - 36:00 let's check here we have acetylene c2h2 c2h2 there c2h2 gaseous you have 209.8 so it's a column you saw formation and then for the oxygen oh p here you have zero for carbon dioxide co2 gas
• 36:00 - 36:30 negative 394.4 kilojoule per mole and then for water move back to oxygen gaseous liquid water you specify so it is 237.13 okay so let me write this those values here delta g's of formation which is in kilo joule per mole for aceto h2 you
• 36:30 - 37:00 have 209.2 for oxygen you have zero you have negative 394 point four here and this is negative two thirty seven point thirteen so this is 3 94. so you have your w max
• 37:00 - 37:30 is equal to the delta g of your reaction which is two moles of carbon dioxide times negative 394.4 kilojoule per mole plus one mole of your water times negative 237.13
• 37:30 - 38:00 kilojoules per mole minus one mole of your acetylene times 209.2 kilojoule per mole plus five over two mole of our oxygen which is zero kilo joules per mole okay
• 38:00 - 38:30 w max which is the same as your delta g of reaction is equal to hopefully this is negative okay 2 times 394.4 that is negative sorry 2 times negative 394.4 plus negative 237.13 minus 29.2 so i got negative
• 38:30 - 39:00 negative 1235 point 13 kilojoules of course work done by the system going out so it is negative so w max is 1235.13 kilojoules the negative sign only denotes
• 39:00 - 39:30 that a work is done by the system so we might as well add it to our answer okay what are we going to do when we're given problems about the free energy for an ideal gas so first we need to derive for the working equation that we're going to use
• 39:30 - 40:00 okay so this is our life and thermodynamics derive and derive derived all right so from the definition of the gibbs free energy you have delta g is equal to your delta h minus t delta s okay but can i just write it here but delta h from
• 40:00 - 40:30 the definition of enthalpy is equal to the internal energy plus your delta pv so expanding that further you have delta h is equal to delta u plus p delta v plus v delta p okay for an isothermal process your delta u is equal to zero okay for an isothermal process
• 40:30 - 41:00 your delta u is equal to zero so please take note of that here you have delta h is equal to p delta v plus v delta p we're going to substitute this equation there later on but let's have a other equation for delta s okay delta s
• 41:00 - 41:30 is equal to reversible work over t right for your system but since we are dealing with an isothermal process you know that delta u is equal to q plus w and since this is zero your q is equal to negative w this is zero again it's an isothermal
• 41:30 - 42:00 process and that w is equal to negative b dvr that's this triangle increment so that it's consistent with because so that it is consistent with the ones that were written above okay so this delta b with that delta s is equal to negative
• 42:00 - 42:30 times and negative p delta v so your delta s is equal to p delta v so substituting everything in your delta g equation you have delta g is equal to p delta v plus v delta p so this is your delta h term and then you have your delta s there minus
• 42:30 - 43:00 t okay delta s which is your p delta v over k over t over t here u t okay so the t will cancel out
• 43:00 - 43:30 all right and then this negative p b p v so your delta p your delta g rather is equivalent to v delta p so your delta g is equal to v delta p remember this is at isothermal process so we can write this one as dg is equal to v dp
• 43:30 - 44:00 so we can integrate that one so this is your d g is v this of p but remember we're dealing with an ideal gas therefore your pv is equal to nrt your v is equal to and rt over p therefore your delta g for unideal gas is equal to
• 44:00 - 44:30 n r t d p over p the integral of p1 o2 p2 that is your nrt ln of p2 over p1 so that will be your working equation when you are dealing with the energy free energy for ideal gases at constant
• 44:30 - 45:00 temperature okay so let's solve the problems um 2.0 moles of an ideal gas are compressed isothermally and reversibly so isothermal and reversible meaning that this equation is consistent with the one given the problem therefore we can use it at 100 degree celsius from pressure of 10
• 45:00 - 45:30 to 25 atmospheres find the value of delta g use r is equal to 1.987 calorie per mole kelvin so then your delta g is for ideal gas nrt lon of p2 over p1 so you have a direct substitution process so you have their 2.00 mole times 1.9
• 45:30 - 46:00 calories per mole kelvin temperature is 100 degrees celsius plus 273 273.15 to make it kelvin and then the lawn of 25 atmospheres all over 10 atmospheres so you have there the delta
• 46:00 - 46:30 g it's equal to let's check on the units first small cancels out kelvin cancels out atmospheres cancels out so you have calorie which is a unit for energy 2 times 1.987 times 373.15 times the loan of 25 over 10. so the answer is 1
• 46:30 - 47:00 [Music] 358.77 calories so we have three two sig figs there so your delta g is equal to one point four times ten to one two three three calories
• 47:00 - 47:30 okay so i will cut again uh this lecture show because i think we all know we have more than 40 minutes now for this video some requested to say not to have more than one hour okay so we will move
• 47:30 - 48:00 on with the third part okay see you