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Prep question

by hnature0704 » Tue Nov 09, 2010 2:55 am
Please help! Thank you :) (From Powerprep)

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 incresed 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
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by Rahul@gurome » Tue Nov 09, 2010 3:09 am
Let r be the rate of the chemical reaction.
Then r = kA^2 / B, where k is any constant.
When the concentration of chemical B is incresed by 100 percent, let the new concentration of A = A1 and that of B = B1
Then kA^2 / B = kA1^2 / B1
Now chemical B is incresed by 100% means B1 = 2B
kA^2 / B = kA1^2 / 2B
A^2 = A1^2 / 2
2(A^2) = A1 ^ 2
A1 = A√2
A1 = 1.4A implies A1 = A + 0.4 A, which is 40% increase.
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by beat_gmat_09 » Tue Nov 09, 2010 3:12 am
hnature0704 wrote:Please help! Thank you :) (From Powerprep)

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 incresed 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
R = k (Ca)^2/(Cb)
Can-new Ca
Cb increased by 100% i.e. Cb is doubled.
New R = k (Can) ^2/2Cb = k (Ca) ^2/Cb
cancel out Cb.
Can ^2 = 2. Ca ^2
Take square root of both sides.
(New Ca)Can = (2)^(1/2) Ca
(2)^1/2 is = 1.41
Therefore % increase in Concentration of chemical A = 40%
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