This guy is so good at emphasizing points. Thank you!
@sami_daye3 жыл бұрын
He emphasized the entire video
@fatimahabib65372 ай бұрын
You are so good at this. Thank you for repeating until it's drilled in my head.
@alexvoegele66763 жыл бұрын
thank you! your reiteration helped me to better understand this topic .
@kitkatcass85093 жыл бұрын
I wonder if he purposely did that lol. It does help! But the random shifts in the screen irks me for some reason. 🤣
@neighborhoodstudent66344 жыл бұрын
I love the energy and the way you explain!
@FS-me8mj4 жыл бұрын
The only part that annoys me is when you write something and it disappears.
@Thebestcoldkilla2 ай бұрын
Thank you so much for relating these two, I finally understand oh my gosh
@edsonacosta2967 Жыл бұрын
I was having problems understanding this topic and you toyally helped me
@trainwreck15293 жыл бұрын
Wow what a great explanation! Dude you rock!
@KauaiDee5 ай бұрын
I like the enthusiasm, but this is very hard to listen to. Slow down.
@spartanrabbit3 жыл бұрын
thanks a lot! it helped me so much! still struggling with these concept and their differences after many years, even after having passed .. i wanna understand, not just apply formulas.
@sciencesimplified38903 жыл бұрын
Yeah... the way I see it is “standard state delta G” is a way that chemistry can compare different reactions from one another.... while “non-standard state delta G” is how in real world conditions and real world applications we find the energetics of a specific chemical reaction
@adriangoleby6 ай бұрын
why is this man yelling at me?! 🤣🤣
@mehrshadgafarzadeh294411 ай бұрын
Perfect explanation
@janaelkateeb55452 жыл бұрын
can QK be applied here?
@RA-pu9jo7 ай бұрын
Can you PLEASE help me answer this question: how do you know when Gibbs and enthalpy are in there units, vs there units per mole? For instance "molar enthalpy" is represented by q/n, so this emplies that enthalpy (H) = q (heat)? same thing here, you sated Gibbs is Joules/mol, but on google it says Gibbs is in just Joules. THANK YOU !!!
@sciencesimplified38907 ай бұрын
Gibbs free energy is a type of energy so the units will be joules… however the amount of joules release depends on number of mols… the more mols you produce of a product, the more energy released (this should makes sense). for example a reaction has a Gibbs free energy of 10 joules/mol…. Then if you produce 1 mol it release 10 joules… if you produce 10 mols it release 100 joules… so Gibbs free energy is energy so it’s in unit joules but the amount of joules produced depends on how many mols of product is produced… it literally tells you the exact same information and when you’re doing a question on the MCAT it will be clear based on what the question provides you with …
@RA-pu9jo7 ай бұрын
THANK YOU SO MUCH MAN, YOUR VIDEOS ROCK!!!!@@sciencesimplified3890
@Aml5875 жыл бұрын
Great video - thanks a lot!
@hha7562 жыл бұрын
Under standard state conditions if both reactants and products are 1 M wouldn’t K = 1? How can under standard state K can be > or < 1 if we have the same conc. of reactants and products?
@andelos4632 жыл бұрын
No, if reaction is under standard state conditions that mean that on temperature of 273K (and some others factors) we have same amount of products and reactants. It doesnt mean that reaction is in balance(equilibrium). So if we just leave it going it would come to same different ralation of product and reactants, actually it will be going untill it comes to relation were same amount of product will go to reactants and reactants to products (that is K). So with standard delta G we are just looking how much work reaction can make moving from state where it has same amount of products and reactants to state where it will be in balance(equilibrium).
@pink_floyd1 Жыл бұрын
A pressure of 1 atm for all substances means that substances A and B of 1 atm partial pressure were taken before the reaction and that each product of 1 atm pressure was also formed after the reaction?
@NhuHuynh-kj5dv Жыл бұрын
For the standard state condition, since all reactants and products are 1 mol wouldn't Keq = 1/1 and ln of 1 is 0, then making standard state delta g = 0 and not -500j/mol? Just trying to understand where the -500 j/mol came from...
@sciencesimplified3890 Жыл бұрын
So the reaction in the video, when it reaches equilibrium, the concentrations of reactants and products will not be 1M rather it will be different numbers… you plug in those numbers into the Keq equation to get the Keq value…. The Keq will most likely not be 1 it will be some other number…. Then you plug in the Keq into the standard state delta G equation to get the standard state delta G value which would give you -500 j/mol
@sciencesimplified3890 Жыл бұрын
You’re getting topics confused To determine the Keq, you let a reaction react and once it reaches equilibrium you plug in the reactants and products into the Keq equation to get the Keq… it will not equal 1 However, for standard state delta G… that’s how much free energy is released when all reactants and products are 1M… for this equation you need to know the Keq number…
@sarazandy44304 жыл бұрын
sooo helpful! thank you!
@alfredjackson1620 Жыл бұрын
if i use hess's law to calculate gibbs free energy for a reaction that does not have only ones as coefficients does the value of the gibbs free energy i get still mean that the initial conditions are 1 mol for all reactants and products?
@sciencesimplified3890 Жыл бұрын
It doesn’t matter how you calculate standard state Gibbs free energy, it represents Gibbs free energy at 1M
@alfredjackson1620 Жыл бұрын
@@sciencesimplified3890 Thanks for the prompt response. I always thought the standard gibbs, enthalpy, and entropy changes i calculated with hess's law were for if i only had reactants and zero products and I mixed the reactants together to react them. I guess I was wrong. I appreciate the help.
@thegreenskittle2 жыл бұрын
7:09 - can someone help me understand why Q=8? I thought each concentration would raised to the power of its coefficient. So why isn't Q=(2^2)(8^8)/(2^2)(1)=64??
@ayan82332 жыл бұрын
In the balanced equation we have A+B -> C+D i.e the stoichiometric coefficients are 1. I understand your confusion. The 2M 1M 2M 8M that's given in the second reaction are the *concentrations* of each reactant/product that we're dealing with and not the molar ratio (stoichiometric coefficient). So when we apply the expression for reaction quotient, Q =[C]^1[D]^1/[A]^1[B]^1 we plug in the concentrations of each. That's Q= 2^1 * 8^1 / 2^1 * 1^1 = 8. There you go. Q=8
@bazyltay2 жыл бұрын
OMG thank you man
@bonganikato57513 жыл бұрын
Under none standard conditions where you use the complete equation.. G=StandarddeltaG+RTln(Q) let's say the temperature is 700K...would you use the standard deltaG at room temperature or would you have to find a new standard deltaG at 700K. I can't find a single video on KZbin that addresses this.
@sciencesimplified38903 жыл бұрын
Standard state conditions don’t change they are convention humanity has chosen to use... standard state delta g will never change so you would not use 700K for calculating standard state delta G... if temperature were 700k then by definition it is not standard state... by definition standard standard state conditions are 273 kelvin along with other parameters like pressure molarity etc.. So if you were finding non standard state delta G, you would use standard state 273K delta G