BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Proportions

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 73
Joined: Tue Sep 16, 2008 4:33 pm
Thanked: 2 times

Proportions

by relaxin99 » Thu Feb 26, 2009 5:16 pm
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical B present. If the concentration of checmical B is increased by 100%, which of the following is closest to the percent change in the concentration of the checmical A required to keep the reaction rate unchanged?


100% decrease
50% decrease
40% decrease
40% increase
50% increase
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 221
Joined: Wed Jan 21, 2009 10:33 am
Thanked: 12 times
Followed by:1 members

by krisraam » Thu Feb 26, 2009 8:01 pm
Some information is missing i guess.

What is th relation between concentration of A ans rate of reaction.

Thanks
raama
Top
Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Aug 07, 2008 5:32 am
Thanked: 16 times

by x2suresh » Thu Feb 26, 2009 8:10 pm
GMATPrep: chemical reaction

--------------------------------------------------------------------------------

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%, 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


Yeap question is missing info. Here is the complate question.



Answer would :D

r --> Ca^2/Cb = sqrt(2)^2 *Ca^2/2Cb

Ca *1.4 --> 40% increase
Top
Junior | Next Rank: 30 Posts
Posts: 10
Joined: Sun Jan 11, 2009 10:47 pm

chemical reaction answer

by NareshCS » Thu Feb 26, 2009 8:34 pm
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 chemical A is x and Chemical B is y and rate is r
chemical A is incereased to 2x and rate remains same r
chemical B is changed to Y

r*(4x^2)*y= r*x^2*Y

new concentration of Chemical B Y =4y
there shoud be 300% increase in concentration of B to keep the rate same

I hope 300% increase is the answer
Top
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

Re: chemical reaction answer

by bluementor » Fri Feb 27, 2009 1:48 am
x2suresh has provided the easiest solution to the problem. Another way to look at this is to conceptually 'guestimate' your way through.

Step 1: Decide if the rate will increase or decrease.

-Concentration of B increased by 100% means the concentration of B has just doubled. Since the rate is inversely proportionate to B's concentration, the rate must go down by half.
-Concentration of A now has to change in order to raise the rate back the original. Since the rate is proportional to the square of A's concentration, A's concentration must increase in order to increase the rate.

So far, you can kill the options that show a decrease in the concentration (choices A, B and C). You are now left with D and E.

Step 2: Evaluate the increment

-Concentration of B increased by 100% means the concentration of B has just doubled. Since the rate is inversely proportionate to B's concentration, the rate must go down by half.
-The rate must now double (multiply by 2) to be back at its original value.
-Try choices D and E.
-If A's concentration is increased by 50% (A -> 1.5A), the square of 1.5A is 2.25A^2 (a multiple of 2.25)
-If A's concentration is increased by 40% (A -> 1.4A), the square of 1.4A is approximately 2A^2 (a multiple of ~2)
-So, a 40% increase in A's concentration is equivalent (approx.) to a multiple of 2 of the rate. Choose D.

-BM-

P.S. I agree this solution is a bit wordy, but when I attempted this question, this was how I thought it through and I think it took me just slightly more than a minute to answer.
Top
Post new topic Post Reply
• Page 1 of 1