I overlooked the fact that all concentrations were 1 in standard conditions and was breaking my brain over this equation. This video just cleared it all up in 10 minutes Thanks a lot for doing what you do!
The explanation was very clear. This was not in the syllabus when I studied but now as a teacher, I have to teach it. So I was trying to figure out how to best explain to my students and this was extremely helpful. Thank you so much.
I have just spent the last three hours on a problem and only just found correct answer at the end of this video when I converted joules to kJ. thank you lol
keq is the one you get from the pressure(kp) or from concentration(kc)? because if the variation of gas moles is not equal to 0 Kc and kp are different. Greetings from spain!
Im just wondering what the ICE table would look like for a standard condition reaction. If I use Hess's law to calculate standard free energy from a chemical equation, in my ICE table in the "initial concentration" row i would have 1 M for products and reactants, then in my "change concentration" the changes would be the coefficients times the variable "x". From the way i use hess's law it would make sense that the variable x=1 since in hess's law i just use the coefficients from the chemical equation as they are to multiply by the standard free energy of formations. But then if the coefficients in my chemical equation are not simply just ones i can end up with negative final reactant concentrations. something doesnt add up.
Hello, In here you used delta G for the examples with the formula (Formula for Delta G @ stand con. and Delta G @ non-stand con.) Would it also apply with Delta S or Delta H? (meaning I could just use the exact same formula just change the symbols/letters)
What does standard free energy mean in this context? Is it free energy exchanged starting from the initial concentrations? Starting with reactants and no products?
hi chad! your videos are so helpful! I had a question about delta g standard state. so i know you said at equilibrium delta g will equal zero so you can just rearrange the equation and solve for delta g standard. but I thought this was a fixed value in a textbook? so then does this new value of delta g standard tell us if the reaction is spontaneous at equilibrium at a certain temperature? thank you!
Hi Devika! Delta G standard is simply the delta G value under standard conditions (not at equilibrium). And those standard conditions are where all aqueous reactants and products are present at 1M concentrations and all gaseous reactants and products are present at 1 atm partial pressures. So when delta G standard is negative then we know that the reaction will be spontaneous under standard conditions and will therefore end up at equilibrium having more products than reactants and an equilibrium constant greater than 1 (If delta G standard < 0, then K > 1). And if delta G standard is positive then we know that the reaction will be nonspontaneous under standard conditions (or spontaneous in the reverse direction) and will therefore end up at equilibrium having more reactants than products and an equilibrium constant less than 1 (If delta G standard > 0, then K < 1). Hope this helps!
@@ChadsPrep Hey Chad! Yes I understood that part! But what confuses me is when at equilibrium the new formula becomes delta g standard = -RTlnK so when we solve for this delta g standard does this tell us spontaneity even though we're not at standard conditions? I'm just trying to clarify for myself when to use which formula! Thanks for your help!!
@@devikanair44 So the non-intuituive part of this is that the Equilibrium Constant is mathematically related to the Delta G value under standard conditions. This is not intuitive at all as they correspond to two different conditions: standard conditions and equilibrium. But I think the key is realizing that the equation doesn't imply that they are the same conditions, but just that the Delta G at one condition (standard conditions) is related to the ratio of products to reactants at another condition (Equilibrium). I don't know if this helped untangle it at all for you but I gave it a shot! 🙂
Hello Harshit, standard conditions refers to the concentration of aqueous reactants being 1 Molar or gaseous reactants being present with a partial pressure of 1atm, and this is completely independent of any of the coefficients in the balanced reaction. Keep in mind that the coefficients in a balanced reaction NEVER tell you how much of the reactants you have, but simply the ratio in which they react and are produced. Take for example: N2 + 3H2 --> 2NH3 The balance reaction doesn't mean that we have 1 mole of N2 and 3 moles H2. It simply means that N2 and H2 will react in a 1:3 ratio. For every mole of N2 consumed, 3 times as many moles H2 will be consumed at the same time. Hope this helps to clarify!
A reaction will always proceed to a minimum in free energy which is why for a spontaneous reaction delta G is negative. But once a reaction reaches that minimum point it has reached equilibrium and can go no lower in free energy. So at that point the free energy is no longer changing and delta G is zero. Hope this helps!
For this to be true you'd be looking for a reaction with an equilibrium constant of 1.0, but you'd also have to manipulate the equilibrium concentrations of the reactants and products to be 1M for it to be considered standard conditions. So while it is possible it would be a somewhat unique reaction under a very unique set of circumstances. Hope this helps!
You explained the link between these concepts well mathematically but not intuitively. How actually (at the physical/chemical level) does reactant/product concentrations (Q) affect delta G? E.g., is there some sort of modulation of entropy? I have no idea! Maths has never "explained" anything to me, just rules for a numbers game.
Where have you been during my last 3 semesters of chemistry?! Wonderfully explained!
I've been here the whole time...where have you been? Glad you finally came across these Breezle and glad they're helpful!
You're an absolute legend. We have this in our metabolism module for uni and it looked quite daunting at first. Thanks so much Chad!
You're welcome and Thank You.
This guys need some serious recognition. OP teaching keep it up
Thank you.
I overlooked the fact that all concentrations were 1 in standard conditions and was breaking my brain over this equation. This video just cleared it all up in 10 minutes
Thanks a lot for doing what you do!
Excellent, AR - glad the video helped you out!
im in college taking biochem and this was a real good explanation helps you connect everything
Good lecture I wish my professor was as passionate!
how lucky your students are! I wish you were my teacher!. God bless you
Thanks for saying so!
Great explanation, it really helped me. I was confused between standard del G and del G. Thank you
Glad it was helpful!🙂
I learned more in the first 3 minutes than 3 whole days of lecture.
😂😂😂 Yikes!
@@ChadsPrep im not kidding I dropped CHM 2. Now I passed Chem 2!!! Thank you soo much ♥
@@robhousehold You're very welcome and congrats on passing Chem 2! Glad I could be a part of the process!
The explanation was very clear. This was not in the syllabus when I studied but now as a teacher, I have to teach it. So I was trying to figure out how to best explain to my students and this was extremely helpful. Thank you so much.
Awesome Angela and I love it! Congratulations on being in the best profession ever!
couldve watched 2 hours of my maniac instructor but i thankfully found your videos, cheers
Glad you found us, RR - thanks for the comment!
Well explained! Thank you Mr. Chad, Appreciate it.💓💕🥰🥰🥰
You're welcome and Thank You!
I have just spent the last three hours on a problem and only just found correct answer at the end of this video when I converted joules to kJ. thank you lol
Excellent! I do my best to include the common places where errors are made. Glad you found it helpful (but sorry it took 3 hours)!
dude. you make things so clear. thank you.
You're very welcome dude...er I mean Eric!
Wonderful lecture.
Great video. I’m on the college level and this video is so helpful
Awesome Yaasin! Thanks for the compliment!
Such a good explanation! Thank you so much!
Thanks Qing! Glad you found it helpful!
This is very helpful. THANK YOU.
great explanation. subscribed
Glad to hear it - thanks!
Such a good explanation! Thank you so much!
regards
KSA
You're welcome and Thank You.
This is gold!
Worth it's weight!
Wow, really cool lecture, thank you very much!
You're welcome Alex! Thanks for the feedback
Wow great teachers really exist? who would have known
"Hiding" in the vastness of TH-cam - you found me! Thanks for the kind words, kwaku appiah.
NICELY EXPLAINED
Glad you enjoyed it, Chadmuditha - thanks for saying so.
Very clear. Thank you
Welcome 😊
keq is the one you get from the pressure(kp) or from concentration(kc)? because if the variation of gas moles is not equal to 0 Kc and kp are different.
Greetings from spain!
Very nice lecture! One question: where did the equation ∆G=∆G0+RTlnQ come from?
Hey Karl - you mean how it is derived or who derived it?
@@ChadsPrep Thanks for replying! Yes, how or from what basis was it derived?
I like your perspicacity
Glad you think so, Brandon - I like your word usage!
Im just wondering what the ICE table would look like for a standard condition reaction. If I use Hess's law to calculate standard free energy from a chemical equation, in my ICE table in the "initial concentration" row i would have 1 M for products and reactants, then in my "change concentration" the changes would be the coefficients times the variable "x". From the way i use hess's law it would make sense that the variable x=1 since in hess's law i just use the coefficients from the chemical equation as they are to multiply by the standard free energy of formations. But then if the coefficients in my chemical equation are not simply just ones i can end up with negative final reactant concentrations. something doesnt add up.
Hello, In here you used delta G for the examples with the formula (Formula for Delta G @ stand con. and Delta G @ non-stand con.) Would it also apply with Delta S or Delta H? (meaning I could just use the exact same formula just change the symbols/letters)
What does standard free energy mean in this context? Is it free energy exchanged starting from the initial concentrations? Starting with reactants and no products?
Thank you
You're welcome!
hi chad! your videos are so helpful! I had a question about delta g standard state. so i know you said at equilibrium delta g will equal zero so you can just rearrange the equation and solve for delta g standard. but I thought this was a fixed value in a textbook? so then does this new value of delta g standard tell us if the reaction is spontaneous at equilibrium at a certain temperature? thank you!
Hi Devika! Delta G standard is simply the delta G value under standard conditions (not at equilibrium). And those standard conditions are where all aqueous reactants and products are present at 1M concentrations and all gaseous reactants and products are present at 1 atm partial pressures. So when delta G standard is negative then we know that the reaction will be spontaneous under standard conditions and will therefore end up at equilibrium having more products than reactants and an equilibrium constant greater than 1 (If delta G standard < 0, then K > 1).
And if delta G standard is positive then we know that the reaction will be nonspontaneous under standard conditions (or spontaneous in the reverse direction) and will therefore end up at equilibrium having more reactants than products and an equilibrium constant less than 1 (If delta G standard > 0, then K < 1). Hope this helps!
@@ChadsPrep Hey Chad! Yes I understood that part! But what confuses me is when at equilibrium the new formula becomes delta g standard = -RTlnK so when we solve for this delta g standard does this tell us spontaneity even though we're not at standard conditions? I'm just trying to clarify for myself when to use which formula! Thanks for your help!!
@@devikanair44 So the non-intuituive part of this is that the Equilibrium Constant is mathematically related to the Delta G value under standard conditions. This is not intuitive at all as they correspond to two different conditions: standard conditions and equilibrium. But I think the key is realizing that the equation doesn't imply that they are the same conditions, but just that the Delta G at one condition (standard conditions) is related to the ratio of products to reactants at another condition (Equilibrium). I don't know if this helped untangle it at all for you but I gave it a shot! 🙂
@@ChadsPrep thank you so much, you're amazing!!
hello, how can the standard Gibbs energy be defined for 1 mol of reactants and products when the stoichiometric coefficients can be diff
Hello Harshit, standard conditions refers to the concentration of aqueous reactants being 1 Molar or gaseous reactants being present with a partial pressure of 1atm, and this is completely independent of any of the coefficients in the balanced reaction. Keep in mind that the coefficients in a balanced reaction NEVER tell you how much of the reactants you have, but simply the ratio in which they react and are produced. Take for example:
N2 + 3H2 --> 2NH3
The balance reaction doesn't mean that we have 1 mole of N2 and 3 moles H2. It simply means that N2 and H2 will react in a 1:3 ratio. For every mole of N2 consumed, 3 times as many moles H2 will be consumed at the same time.
Hope this helps to clarify!
Why change in free energy is zero at equilibrium? And why it should be at reversible process?
A reaction will always proceed to a minimum in free energy which is why for a spontaneous reaction delta G is negative. But once a reaction reaches that minimum point it has reached equilibrium and can go no lower in free energy. So at that point the free energy is no longer changing and delta G is zero. Hope this helps!
@@ChadsPrep Thank you sir
Love from India 🇮🇳
IIT JEE & NEET mark your attendance.😂
Thanks from U.S.A. - you are counted!
Thank u.. ❤️❤️❤️
You're welcome 😊 - thanks for the comment!
Is there a reaction for which standard G is the G at equilibrium?
For this to be true you'd be looking for a reaction with an equilibrium constant of 1.0, but you'd also have to manipulate the equilibrium concentrations of the reactants and products to be 1M for it to be considered standard conditions. So while it is possible it would be a somewhat unique reaction under a very unique set of circumstances. Hope this helps!
Chad's Prep It was very helpful. Thank you so much!
i am studying this in class 12th India for jee.
Happy Studying!
You explained the link between these concepts well mathematically but not intuitively. How actually (at the physical/chemical level) does reactant/product concentrations (Q) affect delta G? E.g., is there some sort of modulation of entropy? I have no idea! Maths has never "explained" anything to me, just rules for a numbers game.
great
thanks.
shi made perfect sense 🙏🙏
Glad you found the channel!
🤩🤩
👍 👍
Hai sir, from kerala
Hello, from U.S.A.
nice explanation, might want to change the profile pic to something other than CP though Thanks
You're welcome
No derivation?
Which part of the video is this about?
Johnny bhaiya
Happy Studying!
I thought u r jonny😅😅😅😅
Chad's the name - welcome to the channel!