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The rate of a certain chemical reaction is directly

by mitzwillrockgmat » Sun Jun 27, 2010 7:37 am

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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 B present. If the concentration of chemical B is increased by 100 percent, 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

Hi I need HELP with this question please! This is what I understand so far....

lets call rate, R
........concentration of chem a, A
..........concentration of chem b, B
C is a constant

Given that concentration of chem a SQUARED is directly proportional to rate , we get ... [ R/A^2 ]= C

Given that concentration of chem b is inversely proportional to rate ... R*B= C

If B is increased by 100%, it means it is doubled so to maintain the same constant, C we must halve the rate, R

i.e. 4*2 = 8, if 4 is halved then 2 needs to be doubled to get 8 & vice versa.


& When R is halved , A^2 will have to be reduced (maybe even halved) to maintain the same, C

i.e. given 14/2 = 7 if 14 is halved it becomes 7, then 2 will have to be halved to become 1 to get 7.

So clearly A^2 needs to be reduced by some percent

BUT ! the answer is D - which means A^2 has to be increased!! I don't understand pls help!!!

I need as simple a solution though!! thank you veeeeery much in advance! :)
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by selango » Sun Jun 27, 2010 8:20 am
Rate =KA^2/B

Now B is increased by 100%(ie) B+B=2b

Rate=KA^2/2B

To keep the rate unchanged,we need to increase A.

A will become V2 A so that rate is unchanged.

Rate=K(V2)^2/2B

We are increasing A.So percentage increase in A

(V2A-A)=(1.414-1)A=(.414)A=41%

Approximately 40% increase.

Hence D
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by kvcpk » Sun Jun 27, 2010 8:27 am
Hi,

You are confused a lot. Understand the main Point in the question: keep the reaction rate unchanged
The increase in conc of A should be nullified by decrease in conc of C

This means that the rate of reaction should be same throughout. Let me explain this:

Let R1,R2 be the rate of reactions
A1,A2 be the conc of chem A
B1,A2 be the conc of chem B

now,

(R1/R2) = (A1/A2)^2 * (B2/B1)

We know that B2=2B1, Also, R1=R2(since rates should be same)

Therefore we get, (A1/A2)^2 = 1/2
A2 = root(2) * A1

Therefore approx 40 percent increase.

Hope this helps!!
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by shankysainik » Sun Jun 27, 2010 9:21 am
Let Rate R=A^2/B
A- Concentration of A, B similar
Now according to the next statement R1 = A^2/2B
Now for R1 to equal R, the value of a needs to be v2 so that it balances the 2 in the denominator.
v2 = 1.414 which signifies an increase of approx 41%.

I am not very good at explaining stuff.
Hope this was helpful !
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by reply2spg » Sun Sep 26, 2010 8:46 pm
I got the same question and I got it wrong. However, after confirming answer I came to know that I was wrong.

Equation is very simple R = A^2/B. You need some constant but let's say constant is 1.

Now let's say A = 10 and B = 100

R = 1.

B increases to 200, you need R nearly equal to 1.

How will you get. which square is near to 200?

It is 196.

so 196/200 = will be approximately 1.

Sqrt(196) = 14. increase in A is 4, means 40% increase.
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by GMATGuruNY » Mon Sep 27, 2010 6:12 am
The trick is translate the chemical relationship into the following equation:

R = (A^2)/B

Here's why the equation above works:

1)The problem states that R is directly proportional to A^2. Directly proportional means that as one value increases, the other value also increases by a proportionate amount. In the equation above, if we increase R, we'll have to increase A^2 by a proportionate amount in order for the equation to remain valid.

2) The problem states that R is inversely proportional to B. Inversely proportional means that as one value increases, the other value decreases by a proportionate amount. In the equation above, if we increase R, we'll have to decrease B by a proportionate amount in order for the equation to remain valid.

Now let's plug in values.
Let A = 10 and B = 2.
R = (10^2)/2 = 100/2 = 50.
If we increase B by 100%, new B = 4.
R = 50 must be unchanged.
50 = (A^2)/4
A^2 = 200
New A = √200 = 10√2 ≈ 14

% change in A = Difference/(Original A) * 100 = (14-10)/10 * 100 = 4/10 * 100 = 40%.

The correct answer is D.
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by goyalsau » Mon Sep 27, 2010 8:31 am
reply2spg wrote:I got the same question and I got it wrong. However, after confirming answer I came to know that I was wrong.

Equation is very simple R = A^2/B. You need some constant but let's say constant is 1.

Now let's say A = 10 and B = 100

R = 1.

B increases to 200, you need R nearly equal to 1.

How will you get. which square is near to 200?

It is 196.

so 196/200 = will be approximately 1.

Sqrt(196) = 14. increase in A is 4, means 40% increase.
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by fangtray » Thu Apr 05, 2012 4:11 pm
GMATGuruNY wrote:The trick is translate the chemical relationship into the following equation:

R = (A^2)/B

Here's why the equation above works:

1)The problem states that R is directly proportional to A^2. Directly proportional means that as one value increases, the other value also increases by a proportionate amount. In the equation above, if we increase R, we'll have to increase A^2 by a proportionate amount in order for the equation to remain valid.

2) The problem states that R is inversely proportional to B. Inversely proportional means that as one value increases, the other value decreases by a proportionate amount. In the equation above, if we increase R, we'll have to decrease B by a proportionate amount in order for the equation to remain valid.

Now let's plug in values.
Let A = 10 and B = 2.
R = (10^2)/2 = 100/2 = 50.
If we increase B by 100%, new B = 4.
R = 50 must be unchanged.
50 = (A^2)/4
A^2 = 200
New A = √200 = 10√2 ≈ 14

% change in A = Difference/(Original A) * 100 = (14-10)/10 * 100 = 4/10 * 100 = 40%.

The correct answer is D.
Sorry for restarting an old thread, but as i as i was solving this question using those plug-ins, i got (100x/4)=50. And then I would solve for X giving me 2. Could you show me what was wrong with this approach?

I multiply A by a constant X, to show me the increase, just like we multiplied 2 by a constant (100%, or 2) to show an increase in B also so why not A?
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by fskilnik@GMATH » Sat Nov 17, 2018 3:00 pm
mitzwillrockgmat wrote: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 B present. If the concentration of chemical B is increased by 100 percent, 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
\[rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\]
\[\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)\]
\[?\,\, \cong \,\,k - 1\]
\[{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
{k^2} = 2 \hfill \\
\sqrt 2 \cong 1.41 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2 - 1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2 - 1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)\]


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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by Scott@TargetTestPrep » Tue Apr 02, 2019 6:09 pm
mitzwillrockgmat wrote: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 B present. If the concentration of chemical B is increased by 100 percent, 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
We can let n = the rate of a certain chemical reaction, a = the concentration of chemical A, and b = the concentration of chemical B. We are given that 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, so, for some positive constant k, we have:

n = ka^2/b

When b is increased by 100 percent, b becomes 2b. To keep the reaction rate unchanged, we can let a become c, so we have:

ka^2/b = kc^2/(2b)

2bka^2 = bkc^2

2a^2 = c^2

c = √(2a^2)

c = a√2

Since √2 ≈ 1.4, c ≈ 1.4a or approximately 140% of a, i.e., a 40% increase in the concentration of chemical A.

Answer: D

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