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don't understand the question..

by bakhshaliyev » Wed Dec 01, 2010 11:18 pm
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which one of the following is the closest to the percentage 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 Night reader » Wed Dec 01, 2010 11:55 pm
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Last edited by Night reader on Fri Dec 03, 2010 1:29 pm, edited 2 times in total.
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by The Jock » Thu Dec 02, 2010 12:54 am
Let the rate of the reaction be R
Let concentration of chemical A be A
Let concentration of chemical B be B

Then R is proportional to A²
R is also proportional to 1/B
Hence, R is proportional to A²/B
If C is a constant, R=C*(A²/B)
If the concentration of B is increased 100% B becomes 2B ( B+(100/100)*B = 2B)

Let A2 be the new concentration of chemical A for the rate to be constant
Then R=C*(A²/B) = C*(A2²/(2*B))
Hence, A² = A2²/(2) So A = A2/√2

A2 = √2 * A = 1.41 * A
Hence A becomes 1.41 * A If the concentration of B is increased 100%

So, there is a 41% increase in A.
Answer is D
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by bakhshaliyev » Thu Dec 02, 2010 1:04 am
The Jock wrote:Let the rate of the reaction be R
Let concentration of chemical A be A
Let concentration of chemical B be B

Then R is proportional to A²
R is also proportional to 1/B
Hence, R is proportional to A²/B
If C is a constant, R=C*(A²/B)
If the concentration of B is increased 100% B becomes 2B ( B+(100/100)*B = 2B)

Let A2 be the new concentration of chemical A for the rate to be constant
Then R=C*(A²/B) = C*(A2²/(2*B))
Hence, A² = A2²/(2) So A = A2/√2

A2 = √2 * A = 1.41 * A
Hence A becomes 1.41 * A If the concentration of B is increased 100%

So, there is a 41% increase in A.
Answer is D
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by beat_gmat_09 » Thu Dec 02, 2010 1:04 am
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by Night reader » Thu Dec 02, 2010 1:54 am
The Jock wrote:Let the rate of the reaction be R
Let concentration of chemical A be A
Let concentration of chemical B be B

Then R is proportional to A²
R is also proportional to 1/B
Hence, R is proportional to A²/B
If C is a constant, R=C*(A²/B)
If the concentration of B is increased 100% B becomes 2B ( B+(100/100)*B = 2B)

Let A2 be the new concentration of chemical A for the rate to be constant
Then R=C*(A²/B) = C*(A2²/(2*B))
Hence, A² = A2²/(2) So A = A2/√2

A2 = √2 * A = 1.41 * A
Hence A becomes 1.41 * A If the concentration of B is increased 100%

So, there is a 41% increase in A.
Answer is D
Jock can you explain the above highlighted expression in your answer. I revised the problem, it really looks for 40% increase and yes it must be approximately 1.40*A for raising into Power (^2) leave no change on 'R'.

Why A2/(Sqr.root 2) and not A*(Sqr.root 2)-?
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by The Jock » Thu Dec 02, 2010 6:08 am
Just a calculation jumble..nothing more than that.....
it is A2 = square root of 2 * A only.
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