Industrial Chemistry Equilibrium question (2 Viewers)
- Thread starter Steven88
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tabularasa
Member
i got 6. something but i reckon i got it wrong or something.
Um, i have a different answer to you all 
equation of form 2A ----> B to simplify it
given info:
0.132 moles of A INITIALLY
0.04 moles of B at EQUILIBRIUM
from 2:1 mole ration, 0.08 moles of A must have reacted to form 0.04 moles of B
therefore, A at equilibirum = .132-.08
k = B/A^2
= 14.8
equation of form 2A ----> B to simplify it
given info:
0.132 moles of A INITIALLY
0.04 moles of B at EQUILIBRIUM
from 2:1 mole ration, 0.08 moles of A must have reacted to form 0.04 moles of B
therefore, A at equilibirum = .132-.08
k = B/A^2
= 14.8
desperate cat
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the smartest in our school got the same result, 14.sth
wrong_turn
the chosen one
using the ICE method you should get about 14.something as well
Chicago I think you need to re-read the question. It said find the equilibrium constant and, as shown in my above post the correct answer is 14.8. Secondly, this new constant is at 25 degrees, a LOWER temperature to the constant they give you (2.08). Therefore you can conclude that the eaction is exdothermic (at lower temperatures more products are formed).
banana_monkey
Better than you
And that is correct. I got your same conclusion, using the ICE method others have used.Xenocide said:Um, i have a different answer to you all
equation of form 2A ----> B to simplify it
given info:
0.132 moles of A INITIALLY
0.04 moles of B at EQUILIBRIUM
from 2:1 mole ration, 0.08 moles of A must have reacted to form 0.04 moles of B
therefore, A at equilibirum = .132-.08
k = B/A^2
= 14.8
Anyone who thinks 14.8 is wrong really needs to read this section on equilibrium constants here
http://hsc.csu.edu.au/chemistry/options/industrial/2761/Ch952.htm#c3
Read that, then redo the question. I think you'll find 14.8 is the correct answer.
Oh, and it was an exothermic equation, like you said. If K increases with decreasing temperature, it means that the equilibrium lies to the products at lower temperatures, and thus is exothermic.
angelic_sammy
New Member
just shut up and accept the answer is 14.8
banana_monkey
Better than you
Yeah, you start with NO2 ONLY in the reaction. There is no N2O2 in the container initially. AFTER the reaction takes place, then some N202 is created. Xenocide already went through and explained his reasoning, so I think it's fruitless to do it again. HE IS CORRECT.
I went through many of these types of questions to prepare for a question like this, and none were wrong using the ICE method. I think maybe you thought there was N2O2 in the flask as well as NO2 initially? Because that is wrong, like I said before, ONLY NO2 was in the flask initially. Go ask your chem teacher.
I went through many of these types of questions to prepare for a question like this, and none were wrong using the ICE method. I think maybe you thought there was N2O2 in the flask as well as NO2 initially? Because that is wrong, like I said before, ONLY NO2 was in the flask initially. Go ask your chem teacher.
fadykozman
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Conquiring Chem, page 315
2NO2 <....> N2O4 is endothermic, I got the same answer as Chicago, 1.8 or 1.4 can't remember, but it was less that the value of K at 100c
chicago, you and me all the way dude!!!
2NO2 <....> N2O4 is endothermic, I got the same answer as Chicago, 1.8 or 1.4 can't remember, but it was less that the value of K at 100c
chicago, you and me all the way dude!!!
banana_monkey
Better than you
Endothermic which way, reverse or forward reaction path? Remember that if the forward reaction is exothermic, the reverse reaction path becomes endothermic, and vice versa. And although Conquering Chemistry is a good book, BOS examiners can pretty much make shit up and change equations people know are exothermic into endothermic if they want. Thats why they give you the numbers, since it wasn't a question of who remembered if the reaction was endo or exo, but who could get the correct equilibrium constant and make a conclusion about the reaction.
Anyone got the question so they can type it up? I really want to do it again, but I didn't take my paper since I was in a rush to get out as it was my last exam.
Anyone got the question so they can type it up? I really want to do it again, but I didn't take my paper since I was in a rush to get out as it was my last exam.
fadykozman
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bana_monkey
you are soo sad!!! Whata hell are you doin here when u finished yesterday?
by the way, the reaction IS ENDOTHERMIC, that is becasue the reaction in conq chem is in the reverse and is exothermic, please go do something else beside checking the website, u r free to go,, gooooooooooooo
you are soo sad!!! Whata hell are you doin here when u finished yesterday?
by the way, the reaction IS ENDOTHERMIC, that is becasue the reaction in conq chem is in the reverse and is exothermic, please go do something else beside checking the website, u r free to go,, gooooooooooooo
fadykozman
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4 marks can affect you heaps from a 100% to 96%, basterma