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The rate of a certain chemical reaction is directly [#permalink]

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07 Aug 2005, 20:00

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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 present is increased by 100 percent, which of the following is closest to the percent change in the 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

Let's say here is the formula of how A and B affects the reaction speed:

rate = A^2/B

After the concentration B is increased by 100%, the percentage of B
become 2B, to keep 'rate' the same, need to have A^2 being doubled.
Which means A increased by 2^(1/2) times.

Let's say here is the formula of how A and B affects the reaction speed:

rate = A^2/B

After the concentration B is increased by 100%, the percentage of B become 2B, to keep 'rate' the same, need to have A^2 being doubled. Which means A increased by 2^(1/2) times.

In closest percentage, that is 40% increase

D

plz QPoo could explain why doubled A^2 is to increase it by 2^(1/2) times.
thanks
regards
mandy

Re: The rate of a certain chemical reaction is directly [#permalink]

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03 Dec 2013, 08:36

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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 present is increased by 100 percent, which of the following is closest to the percent change in the 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

NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).

We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means \(x^2=2\) --> \(x\approx{1.41}\), which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}\)

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 present is increased by 100 percent, which of the following is closest to the percent change in the 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

NOTE: Put directly proportional in nominator and inversely proportional in denominator. \(RATE=\frac{A^2}{B}\), (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).

We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means \(x^2=2\) --> \(x\approx{1.41}\), which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}\)

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