reaction

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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

by shibal » Tue Jun 30, 2009 4:30 pm

The rate of a reaction is directly proportional to the square of the concentration of A and inversely proportional to concentration of B. If B increases by 100%, which of the following is closest to the % change in concentration of A required to keep the rate unchanged?

OA [spoiler]40% increase[/spoiler]

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Source: Beat The GMAT — Problem Solving | Tue Jun 30, 2009 4:30 pm

ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

Re: reaction

by ssmiles08 » Tue Jun 30, 2009 5:24 pm

shibal wrote:The rate of a reaction is directly proportional to the square of the concentration of A and inversely proportional to concentration of B. If B increases by 100%, which of the following is closest to the % change in concentration of A required to keep the rate unchanged?

OA [spoiler]40% increase[/spoiler]

r---> directly proportional to a^2
r---> inversely proportional to b

b now becomes 2b
which means r become r/2.
if r becomes r/2 ----> a^2 becomes a^2/4 [(a/2)*(a/2)]
to change r/2 back to r; we have to multiply (a^2/4)*4

so we are increasing a 40% for r to remain the same.

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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

by shibal » Tue Jun 30, 2009 6:06 pm

could you do the second part step-by-step pls?

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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Tue Jun 30, 2009 6:45 pm

shibal wrote:could you do the second part step-by-step pls?

Lets start with rate and B.

B has a concentration of x. It increases 100% = so 100/100*(x) + x = new B.

B now is 2x.

if B becomes 2x, the rate decreases by a 1/2.

Now we can forget about B and 2x and deal with r and A.

r :: A^2.
now r becomes 1/2. to maintain the same proportion, you have to make A (1/2)^2

but we want to maintain r as it was and not 1/2r. so we have to increase r by 4 to bring back A to 1.

so we are increasing r by 40%.

The best way to look at this is that the concentrations are not dependent on the rate, but the rate is dependent on the concentrations.

-hope this helps. This problem is sort of hard to explain.

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tryin700: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sun Jul 13, 2008 6:23 pm

by tryin700 » Tue Jun 30, 2009 11:13 pm

My take on it --

Let rate be R

Given
R propotional to A^2
propotional to 1/B

In other words we can say

R = [K (A^2)]/B -- (1) where K is some constant

now B is increased to 100% ==>2B

New Rate R' = [K (A'^2)]/2B -- (2) (A' is the new concentration)

Required from Question stem R=R'

[K (A^2)]/B = [K (A'^2)]/2B
==> A^2 = (A'^2)/2
==> (A'^2)/(A^2)=2
==> A'/A= _/2
==> (A'-A)/A = _/2 - 1 (using properties of ratios)

This gives me ~ 41% hence answer is 40% increase

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shanmugam.d: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Thu Apr 30, 2009 1:32 pm; Thanked: 2 times

by shanmugam.d » Wed Jul 01, 2009 7:53 am

Since the problem states that the reactions are same after modification, just equate them:
-> a²/b = (ax)²/2b (the constants are canceled out)
-> a²/b = a²x²/2b
-> x = √2 =1.4

Hence % increase = (1.4-1)x100 = 40%

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