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November 01, 2025, 12:23:03 pm
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Author Topic: Some guidance? =[  (Read 720 times)  Share 

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methodsboy

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Some guidance? =[
« on: November 05, 2009, 10:25:02 pm »
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hey
Im baffled on this one ppl:
[IMG]http://img683.imageshack.us/img683/1521/equial.jpg[/img] [IMG]http://img683.imageshack.us/img683/equial.jpg/1/w470.png[/img]

Answer says D. I got A.
how do you know which line is NO2 and N2O4?
« Last Edit: November 05, 2009, 10:30:20 pm by methodsboy »
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TrueTears

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Re: Some guidance? =[
« Reply #1 on: November 05, 2009, 10:26:47 pm »
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I think the top line is NO2 since it starts off with twice the concentration of N2O4
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methodsboy

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Re: Some guidance? =[
« Reply #2 on: November 05, 2009, 10:30:07 pm »
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^exacltly, that's how i thought.
Answer:
1 mol of N2O4 will be produced for every 2 moles of NO2 reacting. Therefore, top line must be the conc of N2O4 b/c when it changes conc, it changes to half the extent of the bottom line.

Translation?
 :o
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Mao

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Re: Some guidance? =[
« Reply #3 on: November 05, 2009, 10:31:30 pm »
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To determine which line is which, the important thing is not about the starting value, but their changes.

Compare the change from t=5 min to t=10min when equilibrium is re-established. The bottom line has increased by 2x, and the top line has decreased by x. This is the stoichiometrical relationship between the two reactants, i.e. N2O4 --> 2NO2, hence the bottom line is NO2 (for every mole of N2O4 reacting, 2 moles of NO2 is produced).

Hence N2O4 is added at t=5min, and temperature is decreased at t=10min (since reaction is exothermic, from practical knowledge)
« Last Edit: November 05, 2009, 10:34:44 pm by Mao »
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methodsboy

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Re: Some guidance? =[
« Reply #4 on: November 05, 2009, 10:35:54 pm »
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OH!
thanks i understand now
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Re: Some guidance? =[
« Reply #5 on: November 05, 2009, 10:39:31 pm »
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At the start we have . One doesn't need to be twice the other.
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