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The rate of a certain chemical reaction is directly proporti

by Guest Fri Jul 11, 2008 9:17 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 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

GMAT PREP EXAM 1
WHAT'S THE BEST WAY TO SOLVE THIS PROBLEM.
THANKS

ANSWER IS D
guest
 
 

by guest Sat Jul 12, 2008 3:35 am

step1: ratio should be constant
step2: the denominator increased by 2 times - hence numerator should also increase by the same amt.
step3: since the numerator has to increase in squared terms- ((sqrt 2)*A)^2 gives the right increase.
step4: sqrt 2= 1.414, i.e. 40% increase.
RonPurewal
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by RonPurewal Tue Jul 29, 2008 5:11 am

good explanation above.

if you have trouble with this problem, the issue may well be that you don't have an adequate grasp of the meaning of "direct proportion" and "inverse proportion". if this is the case, please advise, and i (or one of the other moderators) can type up a quick and dirty guide to direct and inverse proportions.

you can also solve this problem via number plugging (which i can also illustrate, if you like), although that number plugging still requires a decent understanding of what direct and inverse proportions are.
Guest
 
 

ratios

by Guest Thu Jul 31, 2008 1:46 pm

I am having trouble with the ratio concept with respect to this problem. Ron, would you please give your "quick and dirty" overview?

Thanks.
Guest
 
 

by Guest Sat Aug 02, 2008 4:16 pm

When I plug in numbers, I get

A= 3
B = 10

So, r = 3^2/10 = 9/10 = r

Then, when B is increased by 100%, hence also increase A by 100% *2)

r = 2(9)/20 = 18/20 = 9/10 = r

I am confused about how A going from 3 to 18 is a 40% increase.

BTW I did understand the first reply explaination, I just don't intuitively think to increase something in "square terms". Is there another way to do this?

Wikipedia was also very helpful in understanding direct and inverse proportions:

http://en.wikipedia.org/wiki/Direct_proportion
esledge
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by esledge Sun Aug 10, 2008 6:40 pm

OK, here's my "quick and dirty" method, although it is similar to the first explanation above.

r is proportional to the square of A's concentration:
r = c(A^2), where c is some constant.

r is proportional to the inverse of B's concentration:
r = k(1/B), where k is some constant.

So, the rate expressed as a function of both concentrations is:
r = ck(A^2)(1/b), where c and k are constants.

The "quick and dirty" occurs here--I ignore the constants from here on. For one thing, they are what they are--they can't be changed to make the equation "balance" when I start messing with the concentrations.

So, pretend those constants are 1, and r = A^2/B. From there, I would think of it as the first poster did.

If you want more proof, though, you could do this: Replace B with 2B (we doubled it's concentration). Replace A with xA (we want to know what to do to A's concentration). The rate should stay the same.

r = original rate = rate after changes
r = A^2/B = (xA)^2/(2B)
A^2/B = (x^2)(A^2)/(2)(B)
1 = [(x^2)(A^2)/(2)(B)] * [B/A^2]
1 = x^2/2 (the A^2's and the B's cancelled)
x^2 = 2
x = 1.4

1.4A indicates a 40% increase in A's concentration.
Emily Sledge
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ManhattanGMAT
lpelaez
 
 

missing 1 step

by lpelaez Sun Aug 24, 2008 11:23 pm

I follow the math and logic up until you combine both equations:


So, the rate expressed as a function of both concentrations is:
r = ck(A^2)(1/b), where c and k are constants.



The problem sounds like it is expressing two equations. So I would have expected to see:
c*A^2 = k*(1/b) since both equations equal r and you can replace r for the second equation.

Could you please let me know what I am missing here? Thanks in advance!
esledge
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by esledge Sat Sep 27, 2008 5:12 pm

Oops, I made a small but confusing error. I should have said this:
If r = c(A^2), where c is some constant,
and r = k(1/B), where k is some constant,
Then r = j(A^2)(1/B), where j is some other constant.

Looking back, that was actually where the "quick and dirty" began!

Let's start from those formulas:
r = c(A^2), where c is some constant.
r = k(1/B), where k is some constant.

In words, this says that r is cA^2 assuming B is present in some concentration that doesn't change. The constant c is based on some fixed concentration of B. Likewise, r is k(1/B) assuming A is present in some concentration that doesn't change. The constant k is based on some fixed concentration of A. You can see my point about c and k by solving your equation for either c or k.

In reality, we can mess with the concentrations of A and B simultaneously, but the proportionality for each substance should stay the same: r is proportional to both A^2 and 1/B, so r is proportional to A^2/B. It's therefore right to say r = j(A^2/B), but I was wrong to imply that j = ck.
Emily Sledge
Instructor
ManhattanGMAT