:) But I'm not sure of your answer too. More replies please.
This Question came in my Boards. I solved it, but my friends are getting other answer. So I'm afraid that I lost 4 marks. (Since I know they have studied Chemical Equilibrium more than me and other reason is that his answer was much higher than me)
I'm posting exact wording of question :
Equilibrium constant of Reaction : A_{2} + B_{2} \Leftrightarrow 2AB @ 100Â°C is 50
A 1L flask containingg 1 mole of A_{2} gas is connected to a 2L flask containing 2 Mole of B_{2} gas. Calculate the no. of mole of AB at 100Â°C at equilibrium.
My friends' answer : 10 Mole
Even a correct answer would suffice.

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15 Answers
@vivek how on earth did u even FAINTLY think ur friend knew chemical eqb. better?? :P i cn say he does not even know mole concept correctly[3][3].......1 mole of A can produce at max 2 mole of product!!!...........10 moles OMG!!!!!
@vivek bhai bata de ye sahi hai ya nahin...my knowledge of chemical eqb. is under JEOPARDY!
no idea dude how u got that[1][infact if u see the k value is 50......so answer seems too less [3][3]]
but atleast ur answer sounds better than the one by his "friend"!
Board exam mein solution thodi milta hai.. waise agar tum apna solution post karo main sure ho sakta ho..
this must open eyes of many students i guess:
19:40
mohit_5990: @kunal ... isse stoiciometry se relate mat karo
mohit_5990: ye formula based ques hai
20:18
mohit_5990: hw can u say that k eqil is 50
mohit_5990: these are not the actual data
20:20
arunavsi.._8640: dont u think mohit that 2 moles of b2 wud have some effect on the rate law expression
20:25
mohit_5990: y r u guys r talking about the no of moles
20:26
mohit_5990: in equil it is all about molarity
20:28
arunavsi.._8640: oh yeah skipped it
i don't have time to say much so just wrote this and pasting it here too[1]
kunl_8072: ok mohit i will give u 1 mole A
20:45
kunl_8072: and 2 mole B
20:45
kunl_8072: and 1 L flask
20:45
kunl_8072: and after jee give me 15*root 2 mole product ok?
20:46
kunl_8072: i hope by that time ur reaction would have aatained equilibrium!.and as u say its molarity that matters let me see those 15*root 2 moles and i would "no,minate u for nobel prize"[1]
a2 + b2 =2ab
1 2 0 moles intially
1x 2x 2x moles at eq.
1x/3 2x/3 2x/3 conc at eq.
k = (2x/3)^{2}/ (1x/3)(2x/3) =50 => x=0.93
total no. of moles of ab = 1.86
this question came in IIT 1990
k = (2x/3)2/ (1x/3)(2x/3) =50 => x=0.93
One Doubt : Is the expression like this :
k = (2x/3)2/ (1x3)(2x3) =50 ??
I solved this!! and if not then why?
I think the intended expression is :
K = \frac{\left( \frac{2x}{3}\right)^2 }{\left(\frac{1x}{3} \right)\left(\frac{2x}{3} \right)}=50