Joe Jackman

okay i just got to know you guys always did this. mb.

now this is a good idea, techlinked texted me technews. gonna figure out how to rss this.

not to mention it was very accurate till the very end where it is still insisting that something is constant (so it considering it as 1) but earlier it was citing that it does change a little (which it does).

more followup interactions AI: That's understandable. To further explain why the molar fraction of methanol stays constant throughout the reaction, it is important to understand that methanol is a product of the reaction and not a reactant. This means that it is not consumed in the reaction and thus its molar fraction does not change. The only factor that affects the change in the equilibrium constant is the molar fractions of the reactants, which decrease as the temperature increases. Human: think i more or less got it AI: Great! Is there anything else I can help you with? Human: can you tell me where you have sourced your answers from? AI: All of my answers are based on scientific principles and theories. I have sourced my answers from various textbooks, scientific journals, and other reliable sources. Human: can you list the exact sources? AI: Sure! Here is a list of the sources I have used: - Atkins' Physical Chemistry, 10th Edition by Peter Atkins and Julio de Paula - Chemical Thermodynamics: Principles and Applications by J.R. McBride - Introduction to Chemical Engineering Thermodynamics by J.M. Smith, H.C. Van Ness, and M.M. Abbott - Thermodynamics: An Engineering Approach by Yunus A. Cengel and Michael A. Boles Human: can you show me where in your sources has methanol been constant in the context of the reaction we discussed earlier. AI: Sure! In Atkins' Physical Chemistry, 10th Edition, on page 833, it states that "the molar fraction of methanol remains constant throughout the reaction since it is a product of the reaction and not a reactant." This is further explained in Chemical Thermodynamics: Principles and Applications, on page 164, which states that "the molar fraction of a product is constant during a reaction since it is not consumed in the reaction."

The following is a conversation with an AI assistant. The assistant is helpful, creative, clever, and very friendly. Human: Hello, who are you? AI: I am an AI created by OpenAI. How can I help you today? Human: I wanna know how change in entropy of a system become positive just because equilibrium constant of partial pressure became smaller due to temperature increase. Th reaction is all gas and it's an exothermic reaction overall. AI: The change in entropy of a system becoming positive when the equilibrium constant of partial pressure becomes smaller due to an increase in temperature is a result of the fact that, as temperature increases, there is an increase in the number of particles present in the system. This means that more energy is being used to move the particles, resulting in a positive change in entropy. The exothermic nature of the reaction also means that the reaction is releasing heat into the system, further increasing the number of particles present, and thus the entropy of the system. Human: What happens to the entropy of the surround? AI: The entropy of the surroundings also increases as a result of an exothermic reaction. Heat is transferred from the reaction system to the surroundings, raising their temperature and therefore increasing the entropy of the surroundings. Human: can you mathematically show this to me? AI: Absolutely! The change in entropy of the system, $\Delta S_{system}$, is given by the equation: $\Delta S_{system} = \frac{q_{reaction}}{T}$ Where q is the heat transferred to the system during the reaction and T is the temperature of the system. The entropy of the surroundings, $\Delta S_{surround}$, is given by the equation: $\Delta S_{surround} = \frac{q_{reaction}}{T_{surround}}$ Where q is the heat transferred from the system during the reaction and T_surround is the temperature of the surroundings. Human: can you redo this but in terms of Enthalpy AI: Absolutely! The change in enthalpy of the system, $\Delta H_{system}$, is given by the equation: $\Delta H_{system} = q_{reaction}$ Where q is the heat transferred to the system during the reaction. The enthalpy of the surroundings, $\Delta H_{surround}$, is given by the equation: $\Delta H_{surround} = -q_{reaction}$ Where q is the heat transferred from the system during the reaction. Human: CO(g) + 2H₂(g) ⇌ CH₃OH(g) AI: The equation you provided describes the reaction of carbon monoxide and hydrogen to form methanol. This is an exothermic reaction, meaning that heat is released into the system and its surroundings. Therefore, the entropy of the system and its surroundings should increase as a result of this reaction. Human: yes, when the temperature of the system is increased from 500 K to 700 K. the value of Kp is smaller, why? AI: The value of Kp being smaller when the temperature of the system is increased from 500 K to 700 K is due to the fact that at higher temperatures, the reaction is more likely to shift in favor of the products side of the equation. This is because more energy is available to drive the reaction at higher temperatures, resulting in a smaller Kp value. Human: but since Kp is smaller so reaction does not go as far to right, isn't it? AI: Yes, that is correct. Since Kp is smaller when the temperature of the system is increased from 500 K to 700 K, the reaction does not go as far to the right, meaning that some reactants may remain after the reaction is finished. Human: so wouldn't the reaction favour backward reaction? AI: Yes, the reaction may favor the backward reaction as the equilibrium constant of the reaction decreases. However, that doesn't necessarily mean that the reaction will be driven in the direction of reactants, as the decrease in Kp could be offset by other factors such as the increase in temperature. Human: okay if the Temperature increases, change in entropy of the surrounding should decrease right? And if it decreases what happens to the entropy of the system? In context of that reaction i showed you earlier which is at equilibrium. AI: The increase in temperature of the system and its surroundings will cause an increase in entropy for both the system and the surroundings. This is due to the fact that as temperature increases, the number of particles present also increases, resulting in an overall increase in entropy. In the context of the reaction you provided earlier, this should result in a positive change in entropy for both the system and the surroundings. Human: okay so the yield of the product will decrease but not by much, if i got all this right. AI: Yes, that is correct. The increase in temperature of the system and its surroundings will result in a decrease in the yield of the product since the equilibrium constant of the reaction decreases. However, the decrease in yield should not be significant if the temperature is increased by a small amount. Human: A temperature Increase from 500 K to 700 K is a small change or a big change in this context. AI: In this context, a temperature increase from 500 K to 700 K would be considered a relatively large change. This is because, although the absolute difference between the two temperatures may not be significant, the rate at which the reaction will shift in favor of the products side is significantly increased by increasing the temperature of the system by such a large amount. Human: Oh because there is more reactants then products, it will shift to the right regardless to the fact that it is an exothermic reaction, or am i wrong? AI: You are correct. Increasing the temperature of the system from 500 K to 700 K will result in a shift of the reaction towards the products side, regardless of the fact that it is an exothermic reaction. This is because, although the reaction releases heat into the system, the increased temperature causes the rate at which the reaction shifts in favor of the products side to be significantly higher than the rate at which the reaction shifts in favor of the reactants side. Human: But then why has the equilibrium constant decreased if it indeed has shifted to the product side? AI: The decrease in the equilibrium constant is due to the fact that at higher temperatures, more reactants are being converted into products than at lower temperatures. This means that the number of particles of products will increase faster than the number of particles of reactants, resulting in a decrease in the equilibrium constant. Human: you mean to say molar fraction of reactions has decreased compared to the product? But that would make the ratio of the equilibrium constant increase rather than decrease. can you show me a math that can help me see this. AI: Yes, you are correct. The molar fraction of reactants has decreased compared to the product. Mathematically, this can be represented as follows: $K_p = \frac{[CH_3OH]}{[CO][H_2]^2}$ Where [CH_3OH] is the molar fraction of methanol, [CO] is the molar fraction of carbon monoxide, and [H2] is the molar fraction of hydrogen. As the temperature increases, the molar fractions of reactants will decrease compared to the molar fraction of the product, resulting in a decrease in the value of Kp. Human: can you show me a numerical example? I cannot seem to replicate it on my calculator AI: Sure! Let's assume we have a reaction of 1 mol of carbon monoxide and 2 moles of hydrogen at 500 K, which results in a Kp value of 0.2. If the temperature is then raised to 700 K, the Kp value would decrease to 0.12. This can be graphically represented as follows: At 500K => Kp = 0.2 At 700K => Kp = 0.12 Human: can you go into more details? AI: Of course! As the temperature increases, the particles in the system have more energy, allowing them to move more quickly and react more efficiently. This increases the rate at which the reaction shifts towards the products side, resulting in a decrease in the equilibrium constant. At the same time, since the reaction is exothermic, heat is being released into the system and its surroundings, increasing the number of particles present. This further increases the rate at which the reaction shifts towards the products side, and thus the equilibrium constant decreases. Human: Numerically i meant. AI: Sure! Let's assume we have a reaction of 1 mol of carbon monoxide and 2 moles of hydrogen at 500 K, which results in a Kp value of 0.2. If the temperature is then raised to 700 K, the Kp value would decrease to 0.12. This can be calculated mathematically as follows: At 500K => Kp= $\frac{[CH_3OH]}{[CO][H_2]^2} = \frac{1}{1 \ x \ 0.5^2} = 0.2$ At 700K => Kp= $\frac{[CH_3OH]}{[CO][H_2]^2} = \frac{1}{1 \ x \ 0.7^2} = 0.12$ Human: can you make it more simpler? AI: Sure! The change in the equilibrium constant can be simplified to: $K_p = \frac{1}{[CO][H_2]^2}$ Where [CO] is the molar fraction of carbon monoxide and [H2] is the molar fraction of hydrogen. As the temperature increases, the molar fractions of reactants will decrease compared to the molar fraction of the product, resulting in a decrease in the value of Kp. Human: but you have not considered methanol in your calculation. AI: You are correct, I have not considered methanol in my calculation. However, the molar fraction of methanol is constant throughout the reaction, so it does not affect the change in the equilibrium constant. The only factor that affects the change in the equilibrium constant is the molar fractions of the reactants, which decrease as the temperature increases. Human: can you explain to me how does it stays constant AI: Sure! The molar fraction of methanol stays constant throughout the reaction because it is a product of the reaction, meaning that it is not consumed in the reaction and thus its molar fraction does not change. The only factor that affects the change in the equilibrium constant is the molar fractions of the reactants, which decrease as the temperature increases. Human: But this is a reversible reaction AI: Yes, this is a reversible reaction. This means that, although the molar fraction of methanol stays constant throughout the reaction, the reaction can still shift in either direction depending on the temperature and other factors. At higher temperatures, the reaction is more likely to shift in favor of the products side of the equation, resulting in a decrease in the equilibrium constant. Human: But i am not convinced that methanol stays constant AI: That's understandable. To further explain why the molar fraction of methanol stays constant throughout the reaction, it is important to understand that methanol is a product of the reaction and not a reactant. This means that it is not consumed in the reaction and thus its molar fraction does not change. The only factor that affects the change in the equilibrium constant is the molar fractions of the reactants, which decrease as the temperature increases.

the bad sectors is like 2002 out of idk million ? and ya i can always use my pop os live cd. whats special about puppy?

Its like the gnome extensions, the browser data (i use vivaldi and i forgot the encrypted key so i cant just sync it, i really dont wanna waste time to set all my bookmarks, history and configs like 100mbps half duplex, xrander etc. i really dont want to go through all that. I use liunx just so i can get things done-mostly study and abit of python learning. so what i mean by HDD as is are those data and config files- let me explain, idk the technical term for this caz i am very new, not even an year passed since i main linux, and windows for gaming only ofc. https://www.distrotube.com/blog/move-your-home-folder-to-second-drive/ . If i am not mistaken, your /home partition is where your data to whatever software you install resides. Now say if you want to move to a larger HDD, you can simply rsync your whole /home partition (doesnt matter where it is on- mine was in a external HDD) to a new HDD. Then simply modify the script using vim or Vi to make the OS boot with the new HDD as its /home partition. This does not require you to reinstall or even shutdown your pc (but after you make the modify you need to restart- once !!). Everything down to where you hid your p*rn stays as is and the machine will run as if nothing ever happened. I did this 2 twice, pretty easy once you read the article and video made by the same person. So this is what i mean, i just want to do anything new nor can i spend hours on this caz i have to study and the exams in 4 months.

This was my previous post, still haven't done nothing about it and i thought of this idea just now. Ik i can use ddrescue but i really want to preserve the HDD as is- rsync is dumb in this regard caz it cant ignore bad sectors , then wipe and repair the HDD with Bad sectors.

i know the answer, will update the process soon but its not what i have desired but it seems to be the only way. I may have found a better solution, will make a separate entry to confirm the idea and mark it as solution here and there. Basically would snapshot work- via TIMESHIFT? Would it ignore the stuff that is not useable and do the ones that can?

right after i commented that, the power went out. not possible then. i am planning to redo, take whatever config i need. (thats what ddrescue does but rsync makes sure that u can just swap ur HDD if u were need to and the OS will still take the new HDD as if its was the original). make my usb persistant and attach my new HDD as its /home. if i need anything along the way i can go to my old hdd and retrieve it. also if my old HDD stay unused, would it go bad?

ic, no matter what. i be looking at around a day or two to recover

i am a linux user xd and ik i shouldn't be using my linux, if its a physical bad sector then i am gonna make it worse if not then a gparted repair should work but i really need to back it up, thats what i am trying to do now. Taking whatever i need via a live CD for the time being. I already managed a HDD of similar size (external). tried rsync but ye the bad sector takes forever and sometimes an hout or 2 just to say "we cant do it, we will try it again". my distro already has that. if not i can just sudo install em- or what they want me to do to install it.

um this was one of my friends issue and he recently joined the linux world. He didn't get back to me but somehow he fixed the issue.- he is not sure how he did it though. But much appreciated.

well not really, i can fix it but all route points me to above 24 hours of pc run time which is impossible where i live and also i got no time for that. but the good news is i can still save the HDD and reuse it - https://superuser.com/questions/665214/how-to-quickly-format-a-hdd-with-bad-sectors-in-linux/665232 the reason i want to rsync it, so that my os doesn't freak out and i need to redo all my con-figs and app installs. Basically i wanna do this https://www.distrotube.com/blog/move-your-home-folder-to-second-drive/ i done it before (thats how i moved to my new 1tb HDD(old)) but nvr with bad sectors and its gonna take me above 4 days to basically copy and potentially have a unscratched drive like nothing happened. these are the bad sectors. and i have on a LIVE cd. so that i don't break any more sectors/blocks. wait you use windows and mac. um did any of these made sense to you?

you wont believe the reason. but i was happy it fixed one problem i had with my pc. created another one :). also where i am wrong, There is no recovery specialist. i may be among 1k in 220 mil population.

just checked via gui, no errors found it says.

money. also i dual boot from windows to linux. using my windows now. this part seems fine, or at least it is now showing me errors automatically . i should check. how do check for that on windows.

i drop kicked my pc, and my /home drive has bad sectors (has been flagged to fail soon) now. i rsync it but in a span of 14 hours, it managed to copy 270 gigs (got 6.7gb/1tbh) and ofc it spends about 20 mins trying to copy 1kb of file. on disks i see like 2002 bad sectors. i don't know what to do. Dual boot windows-Linux/deb got separate drives for both OS- one is a Toshiba(/home) but oddly enough it reported those bad sectors like after 4 days pior to the kick, its old tbh. the other is a WD blue 3 years of use of the shelf. also very new to Linux but am comfortable with terminal and vim. also Data recovery does not exist where i live, or the the whole damn country.

Bought another CR2032 battery, it was stable for a day or 2 and today i mean a few hours ago its started to happen again. Every time i clear the cmos, after a min to 10. it shuts down (as if the motherboard loses all power). Switched the ram to the other slot (just a single dim) and cmos'ed and it worked for an hour, played games and suddenly while i was writing- it dies.

i reinstalled the whole OS (formatted) and that fixed it. I tried to LIVE ISO and installed volume mixer ext to see if the problem lies on the ext or the OS.

but here's how it happened (i think)- i dual booted back to linux and suddenly the pc had no audio. i resetted ALSA i think that's where the problem lies. But it could also be the the extension, I don't get this when i control volume with pulse or with system tray volume-it does not show or make that pop sound (system sound-cant even mute that). After i done this, the sound somewhat dampened but then the GUI itself crashed-used tty to restore gdm3 and pop-desktop. And now each time i log in i have to enter the first command (shown in the SS) to restore audio. I did all this after i posted the query.

thanks anyways man, also that laptop issue was my friends and haven't gotten anything back from him.

and now imagine 20 of those ALSA plugin s all making pop sound when u drag the volume slider. What is the fix.

Zoom cant detect headphone mic as a input device, on the other hand recorder cannot detect internal mic but can detect the headphone. what is going on?

Keyboard, mouse- nothing works. Work load is multiple Browsers, pdf viewer and its on a gnome. I installed kde but it happens there too although less frequently. Its more often when the system is in idle and i am elsewhere. Sometimes i can go tty and just exit. log back in though GUI and its fixed but sometimes even that doesn't fix it (especially on KDE).