Free Energy (C)

Summary

In this engaging lecture by Florence Joie F. Lacsa, the focus is on the concept of free energy, particularly its relationship with chemical thermodynamics and equilibrium. As the concluding topic in this chemistry course for engineers, the lecture dives into the basics of chemical equilibrium, including definitions and important equations such as the law of mass action and the distinction between equilibrium constant (K) and reaction quotient (Q). The video highlights these key ideas with comprehensive examples, calculations, and practical problem-solving techniques to illustrate the relationship between Gibbs Free Energy and chemical reactions, making the theoretical concepts more accessible and easily digestible.

Highlights

• Florence Joie F. Lacsa wraps up the topic of free energy by reviewing chemical equilibrium 🌱.
• Key distinction made between equilibrium constant 'K' and reaction quotient 'Q' to determine reaction progress 🔍.
• Insightful explanation on how Delta G is used to understand reaction spontaneity and directionality 💡.
• Detailed problem-solving walk-throughs illustrate practical applications of these complex concepts 🧮.
• The lecture concludes chemical thermodynamics with relatable examples and prepares for the next topic: heat engines 🌡️.

Key Takeaways

• Understanding chemical equilibrium is crucial for mastering free energy concepts ⚖️.
• Equilibrium constant (K) and reaction quotient (Q) are pivotal in determining the direction of a chemical reaction 🔄.
• The relationship between Gibbs Free Energy and equilibrium helps predict reaction spontaneity and direction 🚀.
• Delta G indicates if a reaction is spontaneous (negative) or non-spontaneous (positive) 📉📈.
• Practical examples and calculations make complex concepts understandable through relatable problem-solving 🧪.

Overview

Florence Joie F. Lacsa engages her chemistry for engineers class with the final topic of free energy, blending fundamental principles of chemical equilibrium. She sets the stage by reviewing critical concepts necessary to grasp the relationship between equilibrium and free energy. These foundational ideas are key for students to understand more advanced discussions later in their chemical studies.

Throughout the lecture, Lacsa emphasizes the comparison between equilibrium constant (K) and reaction quotient (Q), which are essential for understanding reaction directions. She explains how Delta G values indicate reaction spontaneity, providing students with practical problem-solving examples to reinforce these theoretical concepts. Her insightful walk-throughs offer clarity to potentially confusing areas, such as calculating Delta G under standard and non-standard conditions.

As Lacsa wraps up the section on chemical thermodynamics, she ensures her students are ready for the upcoming topic of heat engines. The lecture is packed with examples and detailed explanations, making complex ideas accessible and laying a solid foundation for further exploration into chemical kinetics and beyond.

Chapters

• 00:00 - 00:30: Introduction to Free Energy This chapter introduces free energy in the context of chemistry for engineers. Jovi Laksa, the lecturer, mentions that this is the final topic under free energy, which is also the concluding topic in the chemical thermodynamics section. Before advancing to further discussions, there's a brief review of fundamental concepts related to equilibrium, focusing only on those concepts essential for understanding the topic.
• 00:30 - 05:00: Equilibrium Concepts and Basic Review The chapter introduces the concept of free energy, with an emphasis on chemical equilibrium and chemical kinetics. It suggests that if students haven't had the opportunity to study these topics in high school, they should review the material. There are references to general chemistry textbooks for in-depth study of these concepts.
• 05:00 - 10:00: Chemical Equilibrium and Kinetics Chemical equilibrium is described as the state where reactant and product concentrations in a chemical reaction cease to change, meaning the forward and reverse reaction rates are equal. Various resources such as textbooks, online chemistry websites, and YouTube tutorials can offer further insights if needed.
• 10:00 - 20:00: Relationship between Free Energy and Equilibrium This chapter discusses the relationship between free energy and equilibrium in chemical reactions. It emphasizes the concept of reaction kinetics, explaining that the rate of a reaction reaches equilibrium when it becomes equal in both forward and reverse directions. Though kinetics is linked to general chemistry, the focus here is specifically on its relevance to understanding equilibrium without delving deeply into chemistry concepts.
• 20:00 - 35:00: Problem Solving: Gibbs Free Energy Calculation The chapter focuses on problem-solving related to Gibbs Free Energy calculation, using a hypothetical chemical reaction as an example. It differentiates between the forward reaction, moving towards the right, and the reverse reaction, moving towards the left. It introduces the concept of the equilibrium constant, which quantifies the ratio of equilibrium concentrations of products and reactants at a specific temperature.
• 35:00 - 50:00: Problem Solving: Calculating Equilibrium Pressure This chapter covers the concept of equilibrium pressure in the context of chemical reactions. It explains the significance of concentration, specifically molarity, which is the number of moles of solute per liter of solution. The chapter suggests consulting a general chemistry book for more information on molarity if needed, indicating that understanding this basic concept is essential for grasping equilibrium pressure calculations.
• 50:00 - 61:00: Conclusion and Next Topic Preview The chapter discusses various methods of expressing concentrations in a chemical reaction. It highlights the role of coefficients (represented by letters C, D, A, and B) in a balanced chemical equation. At a given temperature, it is noted that a chemical system can reach a state where there is a specific ratio of reactant and product concentrations.

Free Energy (C) Transcription

• 00:00 - 00:30 hello everyone this is jovi laksa once again your lecturer in chemistry for engineers we are now down to the last topic under free energy which is also the last topic under chemical thermodynamics okay before we proceed to the next part of the lesson let us first have a brief review on the basic concepts of equilibrium so this is just the basic concept i only choose those that we need in our
• 00:30 - 01:00 study of free energy so there is a separate chapter in general chemistry wherein you will learn in depth about what chemical equilibrium is and chemical kinetics so if you didn't have the chance to study this one during your high school for any reason there is okay kindly review this one it is available in different textbooks general chemistry
• 01:00 - 01:30 textbooks and there are some general chemistry websites and some tutorials in youtube discussing about equilibrium so please do if you are a bit lost in this part of review okay so equilibrium is defined as a state wherein the reactant and product concentrations and a chemical reaction stops changing because the forward rate and the reverse
• 01:30 - 02:00 rate of the reaction becomes equal so this amount talks about the kinetics of the reaction so kinetics is also a part of your general chemistry so please review if you yeah it's okay to review but we don't need much chemistry all right so if we have
• 02:00 - 02:30 this hypothetical reaction here the forward reaction is the one towards the right and the reverse reaction is the one towards the left okay a number that gives the ratio of the equilibrium concentrations of the products and reactants at a particular temperature is called your equilibrium constant okay take note that
• 02:30 - 03:00 this bracket here represents the concentration of the different substances composing the chemical reaction okay this concentration is expressed in molarity so it should be number of moles of solute over liters of solution again if you are not aware about molarity look for a general chemistry book and try looking for the chapter wherein
• 03:00 - 03:30 it discusses about the different means of expressing concentrations okay the small letters here c d a and b are the coefficients of the balanced chemical reaction it was also observed that at a given temperature a chemical system reaches a state in which a particular ratio of reactant and product concentrations has a
• 03:30 - 04:00 constant value this is the expression of the law of mass action okay the ratio that represents here the concentration on the components of the system at any time during the reaction is known as your reaction quotient q so what is the difference between the k and the q so you can actually calculate them in the same way so this is still the molarity okay of the different
• 04:00 - 04:30 substances comprising the system and your small letters here are still the coefficients of the balance reaction the only difference is q refers to the concentrations at equilibrium and k rather again k refers to the concentrations at equilibrium while q refers to those concentrations at any time during the reaction so when substances
• 04:30 - 05:00 in the reaction are gases the reaction quotient may be expressed in terms of partial pressure of the gaseous components of the chemical reaction instead of concentrations okay this partial pressure here is actually your n over v from your ideal gas law okay now let us take a look on the
• 05:00 - 05:30 relationship between free energy and equilibrium so here when you have q is equal to k this means that the reaction has already reached equilibrium since the concentrations available okay at the time is the same as the concentration of the components at the equilibrium when q is less than k it means that the concentrations uh it means that the chemical reaction
• 05:30 - 06:00 rather did not yet reach the equilibrium hence the chemical reaction will proceed to the right okay the left chateal years principle will have that explained so again if you're lost please review your equilibrium okay the chateau years principle is under the discussion on equilibrium
• 06:00 - 06:30 in general chemistry when q is greater than k on the other hand the reaction proceeds to the left okay because it means that the reaction is already shifted to the reverse reaction we have here the equation that relates the gibbs free energy and equilibrium this is delta g is equal to rt ln of q over k here if you have q over k is equivalent to 1
• 06:30 - 07:00 and have this expression substituted to this lon of one is equal to zero that's that that is the reason why delta g here is zero which means that the reaction is equilibrium so you see the first entries of the two tables are the same but you have q is less than one there any number that is negative has also the loan of any negative number rather gives you a negative
• 07:00 - 07:30 answer so your delta g is less than zero which means that the reaction is spontaneous hence it proceed proceeds to the right okay when q over k is greater than one here the loan of a positive number is always positive this will give you a positive delta g as an answer okay so the reaction means that it is not spontaneous on the forward reaction but
• 07:30 - 08:00 is it spontaneous on the reverse reaction hence the reaction proceeds to the left given those aforementioned relationships it is valid to say then that your delta g is equal to rt lon of q over k okay which can also be written as uh this considering the logarithmic identity elon q over k is the same as loan of q minus lon of k
• 08:00 - 08:30 when delta g is at standard condition what will happen to all of the molarities of the composition uh on the k of the chemical reaction is that it all becomes one molar therefore q will become one log of one is zero therefore this first term is eliminated it becomes zero delta g therefore is equal to rt loan of k when you are under standard conditions
• 08:30 - 09:00 and since reactions that usually begin with all the components in their standard states there must be an available equation that gives us the relationship that that applies to any starting consideration so this is the one okay so delta g is equal to delta g dot plus rt log of q this is basically the same as this one if you can see this uh delta g here is this term then you have our uh rt here
• 09:00 - 09:30 okay so this one uh this is artelon k negative plus this so all of these equations are basically the same right let's solve some problems to illustrate the concept consider the given reaction below so you have two moles of nitrogen dioxide forming one mole of dinitrogen tetra oxide calculate delta g of the reaction
• 09:30 - 10:00 at 298 kelvin and calculate delta g at 298 kelvin if the partial pressures of no2 and n2 for our 0.4 atmosphere and 1.6 atmospheres respectively okay so let's solve [Music] for letter a the delta g of your reaction is equivalent to the summation of n
• 10:00 - 10:30 delta g of formation of your products of your product for your n2o4 minus summation of n delta gf of your reactant which is your no2 so let's look again for the values from our table nitrogen solution and ln nitrogen
• 10:30 - 11:00 so for nitrogen no2 glass delta g is 51.84 kilojoules per mole some delta g is a second column and then for n2o4 gaussians you have 98.28 all right so let's substitute the this values again this is
• 11:00 - 11:30 51.84 kilojoules per mole and the other one is 98.8 alcohol 28.98
• 11:30 - 12:00 28.98 kilojoule per mole right so let's substitute delta g of reaction is one mole of into four times 98.28 kilojoules per mole
• 12:00 - 12:30 minus 2 moles times 51. 84 kilo joules per so your delta g of your reaction is 2 times 51. 84. negative
• 12:30 - 13:00 negative 5.4 kilojoules okay for letter b calculate delta g at 298 kelvin so so we have delta g is equal to standard gibbs free energy this is plus rt
• 13:00 - 13:30 since it is not in equilibrium meaning we're going to use your q okay all right so this is equal to your delta g dot plus rt so what is our q ln of okay the partial pressure of n2o4 raised to 1 all over the partial
• 13:30 - 14:00 pressure of no2 raised to 2 because you have here 2 and 1 as the coefficients right so delta g is equal to negative 5.4 kilojoules plus you have 8.314 joule per mole kelvin
• 14:00 - 14:30 for the r times the temperature is given at 298 kelvin times the lawn of okay so this is according to the given um is this point 40 atmosphere and this one is on a ballista this is 0.4 atmosphere and the other one is 1.6
• 14:30 - 15:00 this is 1.6 atmosphere this is 1.6 so this is loan of 1.6 okay it is raised to 1. atmosphere all over
• 15:00 - 15:30 0.4 raised to 2 sorry raised to two atmospheres because your q is dimensionless okay raise two
• 15:30 - 16:00 to atmosphere okay but you have kilojoules and joel's [Music] so that the jaw will become kilojoules so let's check on our units all right so
• 16:00 - 16:30 we have mole here okay so joel cancels out atmosphere cancels out kelvin cancels out so let's say this is one mole let's assume that
• 16:30 - 17:00 we have one more times one more okay
• 17:00 - 17:30 one more so that the mole will cancel out so that's all eight point three one four times 298 times long of ultimate
• 17:30 - 18:00 [Music] divided by 1000 [Music] it's positive point thirteen so you have their point
• 18:00 - 18:30 thirty kilojoules that is eight point three one point six divided by 0.4 squared 1 times 298 times 8.314 divided by 1000
• 18:30 - 19:00 [Music] i it is 0.304 kilojoules
• 19:00 - 19:30 okay so one more substance kilojoule per mole right so let's solve another wait
• 19:30 - 20:00 okay next we have problem number eight consider the decomposition of biocarbonate calculate the equilibrium pressure of carbon dioxide 298 kelvin and 1100 kelvin
• 20:00 - 20:30 oh what are we going to do with this so we're giving volume carbonate because ba gives you bao and co2 so we are at equilibrium so we're looking for an equilibrium pressure your p co2 298 kelvin and pco2 uh 1100
• 20:30 - 21:00 kelvin okay so how are we going to solve for that one we know that delta g is equal to rt or negative rt rather lon of k k because we are in equilibrium okay and then this k is equal to
• 21:00 - 21:30 pco2 why because what we're going to consider only is the pressure of the gaussian substances so that is rt lon of b co2 for delta g okay but since the this one is what we need
• 21:30 - 22:00 so let me write this one as delta g is equal to rt ln of co2 so long of pco2 what is pco2 [Music] pco2 is equal to
• 22:00 - 22:30 delta g over [Music] rt okay we need to raise both to e paramobala so that we will arrive to the working equation pco2 is equal to delta g over rt
• 22:30 - 23:00 e so e raised to delta g r t so this will be our work equation therefore we need to solve for the delta g and then the r t okay so let's solve so how are we going to solve for delta g so delta g of reaction is equal to delta h of the reaction minus t delta s so we need
• 23:00 - 23:30 the heat of formation and the delta s because we are we have the two temperatures 298 and 1100 kelvin so we need to look for the values of the delta heat of uh formation and the t delta s here okay so
• 23:30 - 24:00 see barium carbonate we have we all need the three values or the two values rather negative one two one six point three and one hundred twelve point one and then for barging oxide we have negative one five hundred fifty three point five
• 24:00 - 24:30 and seventy point forty 42 okay carbon dioxide we have negative 393.5 and 213.6 so i hope you were able to take note of those values right so let's see
• 24:30 - 25:00 okay so here the delta h of formation in kilojoule per mole so you have negative one two and six point three nine
• 25:00 - 25:30 point three and then you have negative fifty five i think that five five three point five for carbon dioxide negative three ninety three point okay and then for s in
• 25:30 - 26:00 joule per mole kelvin it's one one two point one and then 70. 42 and 213.6 so first let's solve for the delta h of reaction
• 26:00 - 26:30 okay for the delta h of reaction we have this equivalent this is equivalent to the summation of n times the delta h of formation of your products minus the summation of m moles
• 26:30 - 27:00 times the delta heat of formation of our reactants okay so heat of reaction is equivalent to one mole for your barium oxide times negative five five three point five kilojoules per mole plus one mole for carbon dioxide times
• 27:00 - 27:30 negative three ninety three point five kilo joule per mol minus one mole for barium carbonate times negative one two one six point three kilo joule per mole okay all the mole will cancel out
• 27:30 - 28:00 so let's solve negative five five three point five plus negative two ninety three point five minus negative 1 2 6.3 so this is 269.3 so delta heat of reaction is
• 28:00 - 28:30 269. kilojoules all right so let's solve for the s this is joel per mole kelvin from all of this so we have here 112.1
• 28:30 - 29:00 70.42 and 213.6 okay so the delta s of the reaction is equal to the summation of n moles times s of the products minus the summation times the standard entropy of the reactants so
• 29:00 - 29:30 delta s of reaction is equivalent to one mole for the barium oxide times seventy point forty two joule per mole kelvin plus one mole of carbon dioxide times two one 213.6 joel per mole kelvin
• 29:30 - 30:00 minus one mole of barium carbonate times 112.1 joule per mole kelvin okay so your delta s of reaction is
• 30:00 - 30:30 171.92 joel per kelvin okay let's check on the units and the mole canceling out we'll cancel out more cancel cell oops
• 30:30 - 31:00 no so we need more space i should hire okay so
• 31:00 - 31:30 let's solve for the delta g at 298 kelvin first so so delta g uh 298 campaign
• 31:30 - 32:00 um all right so delta h of your reaction minus t delta s of reaction okay so your delta g at 298 kelvin is what is our delta h 269.3 so that is 269.3
• 32:00 - 32:30 this is kilo joules per mole minus temperature is 298 kelvin times what our delta s is 171 point 92 joel per mole or more joel per kelvin rather
• 32:30 - 33:00 kelvin chopper kelvin and you have 1000 joules is to one kilojoule so this is
• 33:00 - 33:30 [Music] 218 0.07 kilojoules therefore the pco2 is equal to e raised to the delta g negative rt so this is 218.07
• 33:30 - 34:00 kilojoules all over negative eight point three joel per mole kelvin times one mole times a temperature of 298 kelvin times 1000 joel is to one
• 34:00 - 34:30 kilojoule okay 6e so this will give you e raised to c that will be 0.314 goes to 98 divided by 1 000 218.07 where the e raised to negative let me change the location of my e and
• 34:30 - 35:00 lio so that is e raised to negative 88.02 and this is equal to this e raised to e x five point five point ninety five times ten to negative
• 35:00 - 35:30 thirty-nine atmospheres that is 5.95 that sent a negative 39 atmosphere so this is your e raised to x in your calculators okay e raised to x next a delta g at 1 000 100
• 35:30 - 36:00 okay our delta g at 2 100 kelvin is equivalent to the delta g of your reaction minus your t delta s of your reaction so this will give you the same quantity [Music] 269.3 kilojoule per mole minus
• 36:00 - 36:30 your temperature now is 1100 kelvin times 171.92 joule per mole kelvin times okay 1000 joules is to one
• 36:30 - 37:00 foreign
• 37:00 - 37:30 [Music] more let's remove this per mole
• 37:30 - 38:00 sorry for that joe per kelvin for entropy
• 38:00 - 38:30 okay all right so this is joe per kelvin so the kelvin will cancel out so we have kilojoules in kilojoules there is 4
• 38:30 - 39:00 so we have 80. 19 kilojoules all right so the partial pressure of carbon dioxide is equal to e raised to the delta g over negative
• 39:00 - 39:30 rt so this is e raised to 80.19 kilojoules all over negative 8.314 joel per mole kelvin okay times one mole times one thousand one hundred kelvin then one thousand joules is to one
• 39:30 - 40:00 kilojoules okay so more cancels so solve negative eight point negative eight point point three one
• 40:00 - 40:30 four times one thousand one hundred divided by one thousand d two point nine where the answer is you have e raised to negative eight point seven to seven which is the answer 1.56 one point fifty six times ten to negative four atmospheres
• 40:30 - 41:00 okay so that ends our lesson in free energy and also is the end of our lesson in chemical thermodynamics up next is our introduction to hip tensions