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The rate of a certain chemical reaction is directly

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The rate of a certain chemical reaction is directly [#permalink] New post 04 Apr 2006, 02:38
The rate of a certain chemical reaction is directly proportional to the square of the concetration of chemical A present and inversely proportional to the concetration of chemical B present. If the concetration of chemical B is increased by 100 percent, what is percent change in the concentration of chemical A required to keep the reaction rate unchanged?


Thanks
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 [#permalink] New post 04 Apr 2006, 03:00
rate=A^2/B

B increases by 100

rate=A^2/2B, this means that A have to be increased by sqr(2)

rate = (sqr(2)A)^2/2B=A^2/2B

sqr(2)A/A=sqr(2)

(1-sqr(2))*100 = 41percent
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 [#permalink] New post 04 Apr 2006, 03:35
Thanks

What is the logic behind this

rate=A^2/2B, this means that A have to be increased by sqr(2)
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 [#permalink] New post 04 Apr 2006, 03:51
jodeci wrote:
Thanks

What is the logic behind this

rate=A^2/2B, this means that A have to be increased by sqr(2)


I meant multiplied by sqrt(2) :)

Well, the question states:...what is percent change in the concentration of chemical A required to keep the reaction rate unchanged

so the formula according to the stem is:
rate=A^2/B

if the rate is to be the same as before, the new value for A has to be sqrt(2)*A.

(A will now be sqrt(2)*A).

So,
rate = (sqrt(2)*A)^2/2B = 2A^2/2B = A^2/B

Hope this helps!
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 [#permalink] New post 04 Apr 2006, 06:16
Rate = k(CA)^2/CB

k = constant

If CB increases by 100%, we have Rate = k (CA)^2/2CB. So we need To multiply by 2 to keep the rate constant. Thus CA needs to increase by sqrt(2) and that's about 1.4. Thus have to increase by about 40%
  [#permalink] 04 Apr 2006, 06:16
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