This guy is so good at emphasizing points. Thank you!

@sami_daye

2 жыл бұрын

He emphasized the entire video

The only part that annoys me is when you write something and it disappears.

I love the energy and the way you explain!

Wow what a great explanation! Dude you rock!

Great video - thanks a lot!

thank you! your reiteration helped me to better understand this topic .

@kitkatcass8509

3 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. 🤣

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?

Thanks for explaining!

sooo helpful! thank you!

can QK be applied here?

I was having problems understanding this topic and you toyally helped me

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?

@andelos463

2 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).

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.

OMG thank you man

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 !!!

@sciencesimplified3890

4 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-pu9jo

4 ай бұрын

THANK YOU SO MUCH MAN, YOUR VIDEOS ROCK!!!!@@sciencesimplified3890

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??

@ayan8233

2 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

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…

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.

@sciencesimplified3890

3 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

Perfect explanation

why is this man yelling at me?! 🤣🤣

goated

I like the enthusiasm, but this is very hard to listen to. Slow down.

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 KZread that addresses this.

@sciencesimplified3890

3 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

@bonganikato3016

3 жыл бұрын

@@sciencesimplified3890 Okay thank you. 👌

Could you speak faster?

Really loud and drawn out. i hate chemistry.