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by mttorii » Fri Jul 06, 2007 4:10 pm
One more question. Tks in advance!
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Last edited by mttorii on Fri Jul 06, 2007 4:32 pm, edited 1 time in total.
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by givemeanid » Fri Jul 06, 2007 4:29 pm
I can't see any file attachment.
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by mttorii » Fri Jul 06, 2007 4:43 pm
Sorry ....
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by givemeanid » Fri Jul 06, 2007 4:46 pm
I can see the file now.

Rate of the reaction is proportional to A^2/B.
R = k*A^2/B

Now, B is increased by 100%
Lets call it B1. So, B1 = 2B
New rate R1 = k*A1^2/2B

To keep R = R1
k*A^2/B = k*A1^2/2B
A1^2 = 2A^2
A1 = 1.414 * A
That is a little over 40% increase.


Answer is (D).
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by mttorii » Fri Jul 06, 2007 5:43 pm
Thanks ... Very didatic!!!
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by moneyman » Sat Jul 07, 2007 8:04 am
Amazing technique...But can u pls tell me hw did u get A^2/B pls because I dont know what exactly proportional means in this prob..
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by givemeanid » Sat Jul 07, 2007 4:02 pm
moneyman wrote:Amazing technique...But can u pls tell me hw did u get A^2/B pls because I dont know what exactly proportional means in this prob..


The question says chemical reaction is directly proportional to 'square of concentration of A' and 'inversely proportional to the concentration of B'.

When x is proportional to y, it means x = k*y where k is a constant.
When x is inversely proportional to y, it means x = k/y where k is a constant.

So, rate = k * A^2/B
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by moneyman » Sat Jul 07, 2007 9:55 pm
Ok..Now I understand..Thanks a lot dude..
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