Keeping things in proportion

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Keeping things in proportion

by shoot4greatness » Thu Jun 02, 2011 11:35 am
Hi ya'll,

Ran into a very insightful article on btgm.

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
(a) 100% decrease
(b) 50% decrease
(c) 40% decrease
(d) 40% increase
(e) 50% increase

After reading though the article, I did a 'plug in the number.' Can anyone check it out please?

Chem rate is proportional to A and inversely proportional to B

If r=100 A=100 B=100

B increase of 100% = 200, then r=50 and A=25

let x be a multiplier

50(x)^2 = 100

x^2 = 2

sq. root (2) = 1.4

40% increase necessary
Source: — Problem Solving |

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by rparvathaneni » Thu Jun 02, 2011 12:14 pm
I guess it is correct.

Lets say rate of reaction is R.

as per the problem we can say (A^2/B)*k = R , where k is a constant.

so we need to have (A1ˆ2/2B)*k = R, denominator is 2B as B is increased 100%

=> Aˆ2/B = A1ˆ2/2B
=> A1^2 = 2 * Aˆ2
=> A1 = √2 Aˆ2
=> A1 = 1.4 Aˆ2

hence 40%

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by cans » Thu Jun 02, 2011 7:10 pm
let rate=r
as r proportional to A^2 and inversely proportional to B
r=a^2 *k/b where k is any constant.
Now 100% increase in b means b becomes 2b
Thus to keep r same, A^2 has to become double
or A has to become root(2) = 1.41
Thus 40% increase in A
IMO D
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