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Topic: Making productive assumptions on rate: 2 H2O -> 2 H2 + O2  (Read 4199 times)

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Offline snoobies

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Making productive assumptions on rate: 2 H2O -> 2 H2 + O2
« on: January 24, 2011, 04:01:27 PM »
I really don't feel like this should be so difficult, but I can't seem to reason with it.  The only thing I am given is basically the evaporation of water equation 2 H_2O <-> 2 H_2 + O_2 and I am asked for the rate.  I'm guessing I have to make some assumptions SOMEWHERE and plug them in, but the answer is supposed to come from (delta [H_2O]/delta time) and equal -0.5.  Could someone please explain this to me?  Thanks.

Offline DevaDevil

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Re: Making productive assumptions on rate: 2 H2O -> 2 H2 + O2
« Reply #1 on: January 24, 2011, 04:08:53 PM »
... the evaporation of water equation 2 H_2O <-> 2 H_2 + O_2 ...

electrolysis of water I hope you meant?

in general form:
for reaction:  x A + y B --> z C
r = - 1/x * d[A]/dt = -1/y d[.B]/dt = 1/z d[Z]/dt

Offline snoobies

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Re: Making productive assumptions on rate: 2 H2O -> 2 H2 + O2
« Reply #2 on: January 24, 2011, 04:19:27 PM »
I'm slightly familiar with that equation, but without any concentration or time values given, how am I supposed to use it?  I have the coefficient x obviously, but d[A] and dt for reactant/products is not given.  All that is given is the balanced equation, and I'm supposed to find the rate.  I could do 1M and 1s, which would give me -0.5 (the correct answer), but is that a standard assumption to use 1M/s when nothing else is given?  I just don't understand how to deduce the correct assumptions when substituting for the variables d[A] and dt.  Thanks for the quick reply by the way.

Offline DevaDevil

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Re: Making productive assumptions on rate: 2 H2O -> 2 H2 + O2
« Reply #3 on: January 24, 2011, 04:52:27 PM »
well, I believe you may assume that in a large quantity of water the rate at which water reacts away remains constant throughout the reaction, so d[H2O]/dt = constant. Why this would be 1 escapes me at the moment

I assume they would just like to see that you know how to apply the stoechiometric coefficient in there (the -1/2).

Offline snoobies

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Re: Making productive assumptions on rate: 2 H2O -> 2 H2 + O2
« Reply #4 on: January 24, 2011, 04:54:10 PM »
Great, that makes sense, and actually I vaguely remember something said in lecture to that effect now that you mention it of course :P.  Thanks!

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