Thermodynamics L13 | Enthalpy of Formation (And More Processes) | JEE & NEET 2022 | Pahul Sir

[@CatalysisbyVedantu]

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[@Scointist]

Wow

[@ishithm7636]

Sir sometime u feel that ur class is not energetic....pls dont sir....we love u always......when you are there....there is infinite amount of energy

[@avrpavankumar842]

What makes catalysis special is pahul sir .He is an amazing teacher who simplified thermodynamics

[@abhiramt1862]

Standard enthalpy of GLUCOSE= -1260.4KJ/MOL

[@Kabir_Vines]

Sir mai online sir aapko thanks bolne aaya ur awesome 🙏🙏🙏🙏

[@AjayKumar-lt7io]

Wonderful teacher for chemistry for organic chemistry and physical chemistry....yo pahil sir🔥♥️

[@venkatakrish2060]

ans -1260.4kj sir and for comburstion of propane previous qs ans 2217kj/mol you are the approx king sir

[@prakhar4174]

Sir's math is like my chmeistry. Very relaxed to know.

[@prakhar4174]

# op

[@deepansingh390]

Sir can I get prepared for KVPY and JEE if I take the subscription of vedantu pro CBSE course .

[@radhikaraniverma6655]

Best channel

[@SURENDERKUMAR-fv7hc]

HEAT OF FORMATION OF C6H12O6 WILL BE ( -1260.4 )

[@amanadleshorts7429]

Exllent explanation👌

[@punitatripathi844]

*1*

[@disha-cz8fy]

sir you are amazing

[@atulrana0209]

Ans of the question : -1260.4 Kj /mol

[@kgagan8639]

Answer to home work question -1260.4 KJ/mol

[@hahhaharsh]

HOF -1260 kJ.. Love you sir plz take 1 batch on paid platform.. a humble request.. I won't let you down

[@venkatasathyaanandvarmadan6694]

3rd 6c+6H2+3O2

[@yoginipathak6881]

Ans : -1260.4 kJ/mol

[@sangeethasivakun1100]

Sir ans: delta H 0F c6h12o6= -1260.4 KJ EXPLANATION: C6H12O6 = 6CO2+ 6H2O delta Hcom=sigma H0F (reactant) from -2816=(6×H0F(Co2)+6×H0F(h2o)) -(H0Fc6H20+O6) FROM: -2816=(6×-393.5) + (6×-285.9) - H0F + C6H1206 THE ANSWER IS -1260.4KJ (FOR CONSTION REACTION DeltaH0F(O2) IS CONSIDER =0)

[@aryandhaka3629]

Ans . 1136.6kj/mole

[@Sidharth-wy8uo]

Sir standard state Ka MATLAB yeh hota h na ki uski room temperature per kya state h

[@eshi6241]

pahul sir.. you are my favourite chemistry teacher ever!! Sir when will we start s-block? please reply...

[@aviralmishraofficial1626]

Sir, there's one question, you calculated the enthalpy for the sublimation of ice if wee provides that much heat to ice, then will that sublime directly i.e without entering its liquid state?

[@CatalysisbyVedantu]

Nope. That was just a theoretical calculation. Look for phase transition diagrams on google - that might give you some more insight

[@aviralmishraofficial1626]

@@CatalysisbyVedantu Ok! thanks sir!

[@ishithm7636]

@@CatalysisbyVedantu Sir sometime u feel that ur class is not energetic....pls dont sir....we love u always......when you are there....there is infinite amount of energy

[@Scointist]

Amaazing

[@sanjaykumar-bl6tp]

Answer of homework question: --1260.4 kJ

[@rudrabhandari5934]

Sir has showcased a bottle three times and each time its a different bottle

[@punitatripathi844]

CORRECT ANSWER IS *-1260.4 kj/mol*

[@susannimisha9832]

H.W answer = -1261 kJ/mol

[@sohamgursale6720]

28:45

[@bhageerathswaraj5689]

ANS:1269.4☢

[@CSEjn-zj4ed]

28:13 sir to use hess law for enthalpy the temp must be same na sir and vapourisation of water and fusion of ice dont take place at same temp na sir so we cont be using hess law directly na sir using kirchoffs laws we should change enthalpy at desired temp and then add na sir

[@mehuldarak8927]

SOMEONE plz tell me. Are these lectures complete for JEE Adv. Can I rely on them completely?

[@ishithm7636]

Ofcourse... some problems more

[@manishborah18]

mila jee ?

[@piran5314]

Sir I can't find Biomolecules in playlist.....You completed the lesson or not? If completed pls provide the playlist....#pahulfan❤

[@hahhaharsh]

Watch it on vedantu JEE channel check it out..

[@piran5314]

@@hahhaharsh ok buddy thanks

[@kaushiksengupta6511]

Answer is -1255.4 KJ

[@sbalaji8794]

ANSWER FOR HOF QUESTION: -1260.4Kj

[@rajrani5051]

Minus 2725

[@jadiprabhakar3942]

Answer: - 1260.4kj/ mol

[@dillidonka]

kj bro not kg

[@deepaka5633]

Answer 1260 kj

[@jakejoanna4040]

-1260.4

[@blazerchicken1619]

Sir answer is coming 1260 kj/mol for the homework question.

[@allonetechnicalgugloo]

correct answer=-1260.4kj/mol

[@hrishi7570]

That's -1260.4kj sir

[@Sovralin]

"You might be living in Antarctica. If you do... Why are you watching my videos?!" Well, watching you and listening to you is pleasant. ^^

[@Sidharth-wy8uo]

Sir black Phosphorus Ko kyu nhi consider karte jabki yeh to white se intervonvertible nhi h

[@hahhaharsh]

40:50 Your answer may watch class carefully

[@lakshyakotwani9570]

Sir if I have not studied chemical bonding and states of matter can I directly jump to thermodynamics after studying periodic properties?

[@nikhilpathak6465]

Yes

[@gursimrnjitsingh6192]

-1260 kj ans

[@edha4831]

Sir how many more videos are to be there for completion of thermodynamics??

[@edha4831]

@Naman it'll be great if that happens, hope you are sure about it.

[@sushantkumar6035]

16

[@sunandarayudu9206]

Sir what about your cooking channel.... I tried peanut butter, it was soo tasty 🤩🤤

[@raidubvs2277]

Bro what's the name of the channel ??

[@sunandarayudu9206]

It's "cook it out"

[@divyansh_kala]

Homework Question Answer: ΔH°f of C6H12O6 = - ( - 2816 ) + 6*(- 393.5) + 6*(- 285.9) = - 1260.4 kJ / mol

[@anasshaik0510]

kzbin.info/www/bejne/rKKWl42Gj7BrkKs Link to the Beats at the start of the Video

[@bhavneetkaur3811]

Sir can yoy make video in hindi plss

[@AjayKumar-lt7io]

Guys plz like if u are a student....😊

[@gourisahoo7086]

Just gili gili akka class 😅😅😅😅😅😅😅😅😅😆😆😆😆😆😆😇😇😇😇😇

[@ishithm7636]

Sir sometime u feel that ur class is not energetic....pls dont sir....we love u always......when you are there....there is infinite amount of energy

[@koustavroycommercial7259]

AT 30:20 WHAT ABOUT OZONE??? IT IS NATURAL DOES EXIST IN NATURE COMPOSED OF OXYGEN ATOMS ONLY THEN HOW COME ITS NOT AN ELEMENTAL FORM OF OXYGEN

[@visalaj767]

Bro its just assumption that only they are at standard forms or else many can be in standard forms like diamond its just convention for us

[@maddyi7545]

Bro OZONE form hone ke liye kaafi conditions chahiye hote hai. o2 se ozone banta hai due to action of uv rays. Nature men mainly o2 hi hota hai. isiliye it is the standard form.

[@anonymous-lx6kt]

🎩 😁 👕👍Great! 👖

[@aviralmishraofficial1626]

HALL OF FAME ANSWER = 1260.4 kJ

[@lavanya3443]

Ans- 1260 KJ

[@mohammedmusthafa1573]

Hi sir

[@koustavroycommercial7259]

Standard enthalpy of GLUCOSE= -1260.4KJ/MOL