you are God sent (cries in organic chemistry tears). I finally understand thank you, and God bless you. AMEN. This was made 8years ago and it's still very helpful.
thank you so much!!!! So far I didn't understand why starting presure is P2, not P1 in irreversible process. your exain make me understanding. But stiill could not imagine a situation that starting presure, p1, in the piston is going down very quickly even the volume is still v1.
Deviation is clear, but thing I don't understant is the Gibbs Free Energy relation to maximum work that the system can do on surroundings and how this equation represents that. Could you help me with that?
Mam one specific question Which book to study for better understanding of thermodynamics? specifically how you explain first 2 mins I don't have that type of understanding regarding Reversible and Irreversible process. Hopefully get a positive response from you mam🙏🙂
You are the best chemistry teacher on RU-vid. You have a God's gift to teach. Please post videos regularly. You can post videos for AP Chemistry or Gen Chemistry 1 and 2.
You are the best chemistry teacher on RU-vid. You have a God's gift to teach. Please post videos regularly. You can post videos for AP Chemistry or Gen Chemistry 1 and 2.
You are the best chemistry teacher on RU-vid. You have a God's gift to teach. Please post videos regularly. You can post videos for AP Chemistry or Gen Chemistry 1 and 2.
You teach far better than my Organic Chemistry I professor. I am glad you posted this video. You are making the lives of students like me easier. Please post on a regular basis. You'll be helping a lot of students. Thank you and God Bless you for helping me and other students.
Something I've never been able to understand is that under standard conditions, by definition, everything (products & reactants) is at 1 M, which would make Q (K) ALWAYS 1, so why isn't delta G naught ALWAYS 0 ???
When we talk about the standard Gibbs free energy change for a reaction, we're looking at the difference between the Gibbs free energy of the products at standard state and the reactants at standard state. So instead of thinking of a mixture of reactants and products all at 1M, think about having products (only) at standard state and the corresponding Gibbs free energy. Then consider the reactants (only) at standard state and the corresponding Gibbs free energy. The difference between these two values is what we're referring to when we talk about the standard Gibbs free energy change for a reaction. Also keep in mind that, mathematically, ΔG° = -RTlnK (ΔG° does NOT = -RTlnQ) and so ΔG° = 0 when K = 1 (not necessarily when Q = 1). Does this help?
You're absolutely correct - you can use the concentration values within the log. Because this is a buffer, the acid and the base are in the same solution, so we would use the same volume to calculate the concentration for the acid as we would to calculate the concentration for the base. We aren't told the volume here, so I'll put it in as X L. We could calculate the log component as log [(0.10mol/X L)/(0.40mo/X L)]. Because the X L component is the same in the numerator and denominator, we can cancel them out such that this term can simplify to log (0.10/0.40). That is, for a buffer you can use EITHER the ratio of moles or the ratio of concentrations and you will get the same answer, because the ratio is the same.
@@t.c.hayalivatandas1453 Yes, you should use the base (CH3COO-, 0.10 mol) divided by the acid (CH3COOH, 0.40 mol) in the Henderson-Hasselblach equation. This is exactly how the calculation was done in the video.
CH3NH3+ is a weak acid and HCl is a strong acid. In this question your are asked about CH3NH2 (ie the conjugate base of CH3NH3+), which is a weak base, and HCl, which is a strong acid. Technically CH3NH2 could act as either an acid or a base, but when mixed with HCl (a strong acid) it will act as a base (since it is a much much weaker acid than HCl).
You're making an illicit logical jump at the beginning of the video: 1. You first say that a spontaneous process is one which increases entropy. This is true. 2. Then you affirm the reciprocal: that if a process increases entropy, then it is spontaneous. This is false. And everyone makes this illicit jump. Could you please explain to me how you make this logical jump? After all, not all irreversible processes (i.e., those who increase entropy) are spontaneous.
To reply to your point 2: are you perhaps confusing the chemical understanding of 'spontaneous' with the casual use of the word spontaneous in English (in non-chemistry contexts)? When we speak/write in English and use the word spontaneous, it usually brings to mind something that happens (right there and then!). We often think about this thing happening quickly / suddenly / by itself without help. This is not the same as how the word spontaneous is used in the world of chemistry. In chemistry, spontaneous means that the process is 'thermodynamically favourable' (i.e. the ending point of the process corresponds to higher entropy of the universe than the starting point of the process). That is, if the process were to occur, it would result in an increase in the entropy of the universe. This does not mean that the process DOES occur, just that IF it did then the result would be a more thermodynamically stable state (higher entropy) for the universe. You may be wondering why a process that is classified as 'spontaneous' and ultimately takes the universe to a more stable state does not necessarily occur. Spontaneity (in chemistry) relates to thermodynamics. In thermodynamics analyses like those here we're considering only how the beginning state (e.g. reactants in a chemical reaction) relates to the ending stage (e.g. products in a chemical reaction). If the ending state corresponds to higher entropy of the universe than the starting state then we say that it is a spontaneous process. BUT, we have not considered HOW this process occurs (i.e. how do reactants become products?) and whether the change from our beginning to ending state is possible / feasible under our given conditions. When we study kinetics we are looking at HOW chemical processes/reactions happen, and it is this kinetics piece that can tell us why a process might be spontaneous (on paper when we consider thermodynamics), but not functionally occur. For example, perhaps a chemical reaction requires the reactants to collide together in an incredibly unlikely orientation or the reactants must collide together with an amount of energy that is unfeasible at the given temperature. Both these factors could make a reaction so slow that it doesn't functionally happen, even if it is technically 'spontaneous' when we look at the thermodynamics. An example is that you may have heard advertisements for diamond rings that say "diamonds are forever". However, from a thermodynamics perspective, the process of diamond converting into graphite is spontaneous. We know, though, functionally that the diamonds of the world aren't all turning into graphite. While this process is spontaneous, it is so kinetically slow (very large activation barrier) that it doesn't functionally happen. That is, this process is spontaneous (IF it occurred the entropy of the universe would increase) but does not occur (the activation barrier for reactants to become products is too high for it to occur). Does this help?
@@Chemistry123, Wow, thanks a lot. I knew it was just a terminological problem but I read countless books and they weren't clear in their use of terms. Thanks again.
Hello I'm French speaking, I've been searching this stuff on French vidéos for almost a whole week because I did not get a shit of what was said in class, and finally few hours before my test I find answers to my questions, this...this is a message from God ohhhh thanks
5:20 why can you presume that this process happens isothermal ? Isn't it already happening isobaric and since p * V = n * R * T with constant T and constant p, V would also be constant?
The process described at 5:20 is neither isothermal nor isobaric. Calling this process isothermal was a verbal typo. I should have said "no overall change in temperature" rather than isothermal. For the irreversible process described at 5:20, the starting and ending temperature is the same, so overall the change in T is 0, but we would expect the temperature to change during the process (during the expansion the temperature would initially decrease until we give the system enough time to re-establish thermal equilibrium and rise back up to the initial temperature). For a process to be isothermal, the temperature would have to stay constant throughout the process (not just the same for the initial and final states). The irreversible process here is not isothermal but the reversible version of this state change is isothermal. I can see why it seems that the process might be isobaric (constant pressure), but this is not the case. For a process to be isobaric, the pressure of the system would have to be the same throughout the process. Here, the external pressure is constant, but the pressure of the system is changing, so the process is not isobaric.
