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# The rate of a certain chemical reaction is directly

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05 Feb 2010, 12:06
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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 chemical 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
[Reveal] Spoiler: OA
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05 Feb 2010, 12:16
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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 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

NOTE: Put directly proportional in nominator and inversely proportional in denominator.
$$RATE=\frac{A^2}{B}$$, (well as it's not the exact fraction it should be multiplied by some constant but we can ignore this in our case).

We are told that B increased by 100%, hence in denominator we have 2B. We want the rate to be the same. As rate is directly proportional to the SQUARE of A, A should also increase (nominator) by x percent and increase of A in square should be 2. Which means $$x^2=2$$ --> $$x\approx{1.41}$$, which is approximately 40% increase. $$R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}=\frac{2A^2}{2B}$$

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20 Aug 2010, 22:14
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CR = (A^2) * ( 1/B)

put any values

625 = 25*25

B increases by 100% so it becomes 50

CR remains same at 625
1/B becomes 50

so A^2 should be

625/50 = 12.5 or say decreased by 50% (b)
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23 Aug 2010, 18:06
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Given equation = $$A^2/B$$

Also given that the concentration of B is increased by 100 %. Hence B becomes 2B.

New concentration = $$A^2/2B$$. In order to keep the new concentration the same as the original concentration, A should be equal to $$\sqrt{2}A$$ so that A^2/2B becomes $$A^2/B$$

$$\sqrt{2}A$$ == 1.414 * A or approximately 141% of A.

141% of A is 41 % increase of A.

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14 Sep 2010, 18:40
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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 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?

Given: Rate - $$(A^2/B)$$

B increased by 100%. New B is 2B.

Rate - $$(A^2)/2B$$. In order for the initial rate of chemical reaction to be the same as the final rate. A should be increased such that it should cancel out the 2 in the denominator.

Hence A should be $$A\sqrt{2}$$ and hence final rate of reaction is $$(A\sqrt{2}))^2/2B$$

$$2A^2/2B$$ which becomes the same as the initial rate of reaction. Hence A should be $$A\sqrt{2}$$.

Or final concentration of A is $$1.41A$$ or an increase of approximately 41%.

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19 Oct 2010, 01:37
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satishreddy wrote:
the rate of certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of the 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.

100% decrease
50% increase
40% decrease
40% increase
50% increase

Let the rate of the reaction is r and concentration of chemical A is A and concentration of chemical B is B.

According to the question , r is proportional to A^2
and r is also proportional to 1/B

Combining these two condition , r =K (A^2/B) , where K is proportionality constant.

Now in the changed condition B has increased by 100 % means is has become 2B.

So for keeping the rate constant we have to change the concentration of A, from A to ((Square root of 2)*A)

As K (((square root of 2)A)^2)/2B = K (A^2/B) =r

Now, let after increase of x% ,A has become ((square root of 2)A) i.e 1.414*A

A+(xA/100)=1.414 A
Hence x=41.4%

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Re: % increase , proportions [#permalink]

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31 Jan 2011, 06:40
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The rate of a 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 percent , which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?

100% decrease
50% decrease
40% decrease
40% increase
50% increase

Let r be the rate of chemical reaction and A and B be the concentration of chemical A and chemical B respectively.

r is proportional to A^2
r is inversely proportional to B

thus, r = k(A^2/B), where k is some constant.

now B becomes 2B (100% increase). Let x be the new concentration of A.

The new equation becomes

r=k(x^2/2B)

Since r remains the same, we have

k(A^2/B)=k(x^2/2B)

so, x=1.4A which is an increase of 40%

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Re: Direct and Inverse Proportion Q [#permalink]

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18 Mar 2012, 20:26
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Rate= r = k (A^2) (1/B) [where k is a constant]

If B is replaced by 2B, then
r = k (A^2) (1/2B)
=> k (A'^2) (1/2B) = k (A^2) (1/B) [where A' is the new concentration of A]
=> A'^2 = 2(A^2)
=> A' = sqrt (2) * A = 1.414 *A

Therefore the concentration of A must increase by ~40% [option D]
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Re: Direct and Inverse Proportion Q [#permalink]

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19 Mar 2012, 00:16
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A-> (sq rt 2)*A

as R=K A^2/B

HENCE
((sq rt 2)-1)/1 =41% increase

hence D
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Re: Direct and Inverse Proportion Q [#permalink]

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19 Mar 2012, 00:37
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dddmba2012 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 chemical B present. If the concentration of chemical B present is increased by 100 percent, which of the following is closest to the percent change in the 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

Also discussed here: the-rate-of-a-chemical-reaction-is-directly-proportional-to-76921.html

the-amount-of-coal-a-train-burns-each-mile-is-directly-93667.html

a-is-directly-proportional-to-b-when-a-8-b-88971.html
in-a-certain-formula-p-is-directly-proportional-to-s-and-80941.html

Hope it helps.
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Re: The rate of a certain chemical reaction [#permalink]

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24 Oct 2012, 07:33
danzig 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 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

From Q Stem Chemical reaction R =K (A^2/B)
Now if B is increased by 100%----> it becomes R=K (A^2/2B)
Now To bring the relationship to its initial condition we need to cancel out 2 in denominator. So my multiplying root 2 we can do that. So root2 = 1.4 approx.
That means we need to increase 40% of A.

Hence D.
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Re: The rate of a certain chemical reaction is directly [#permalink]

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04 Nov 2012, 01:38
Hello Bunuel
I saw this question coming around couple of times.
But doesn't the answer depend on how you express the concentration reaction?
If I said CR = A^2 - B, (or in general a*A^2 + b*B, where a >0 and b <0) the answer will still be D?
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Re: The rate of a certain chemical reaction is directly [#permalink]

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08 Jul 2013, 00:09
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Re: The rate of a certain chemical reaction is directly [#permalink]

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04 Jan 2014, 00:49
I am unable to understand this question.

Rate= a^2
Rate= 1/b

Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?

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Re: The rate of a certain chemical reaction is directly [#permalink]

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04 Jan 2014, 04:12
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theGame001 wrote:
I am unable to understand this question.

Rate= a^2
Rate= 1/b

Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?

Lets make your life more easy.

Let us say x = 4, y = 2 then R = 16/2 = 8

Now keep R same and double y to see what is the change in X.

R = X^2/Y
8 = X^2/4
X = 4\sqrt{2}

X got multiplied by \sqrt{2} which means a 40 % increase

OR

R = X^2/Y
If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase
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Re: The rate of a certain chemical reaction is directly [#permalink]

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04 Jan 2014, 04:19
PerfectScores wrote:
theGame001 wrote:
I am unable to understand this question.

Rate= a^2
Rate= 1/b

Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?

Lets make your life more easy.

Let us say x = 4, y = 2 then R = 16/2 = 8

Now keep R same and double y to see what is the change in X.

R = X^2/Y
8 = X^2/4
X = 4\sqrt{2}

X got multiplied by \sqrt{2} which means a 40 % increase

OR

R = X^2/Y
If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase

Thank you but what I am unable to understand is why are we multiplying?

x=4 and y=2
Squared 4= 16 ----- I understood this
Inversely proportional = 1/2 ------ Understood this
16*1/2 ------Unable to understand this part
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Re: The rate of a certain chemical reaction is directly [#permalink]

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04 Jan 2014, 04:22
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theGame001 wrote:
PerfectScores wrote:
theGame001 wrote:
I am unable to understand this question.

Rate= a^2
Rate= 1/b

Where is constant K coming from? And why are we multiplying all the there i.e. K, a^2 and 1/b?

Lets make your life more easy.

Let us say x = 4, y = 2 then R = 16/2 = 8

Now keep R same and double y to see what is the change in X.

R = X^2/Y
8 = X^2/4
X = 4\sqrt{2}

X got multiplied by \sqrt{2} which means a 40 % increase

OR

R = X^2/Y
If Y got multiplied by 2 X will have to get multiplied by \sqrt{2} to keep R same. Still 40 % increase

Thank you but what I am unable to understand is why are we multiplying?

x=4 and y=2
Squared 4= 16 ----- I understood this
Inversely proportional = 1/2 ------ Understood this
16*1/2 ------Unable to understand this part

R is directly proportional to X^2 and inversely proportional to Y which means that R is directly proportional to X^2/Y

So if x = 4 and y = 2 R is directly proportional to (4^2)/2 which is equivalent to X^2/Y
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The rate of a certain chemical reaction is directly [#permalink]

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09 Apr 2014, 13:26
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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 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

$$Rate = \frac{k*A^2}{B}$$

$$Rate = \frac{k*NewA^2}{2B}$$

$$\frac{k*NewA^2}{2B} = \frac{k*A^2}{B}$$

$$NewA = sqrt{2} * A$$

$$%change = \frac{sqrt{2}*A - A}{A}*100$$

$$%change= 0.414*100=41.4% increase$$

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Re: The rate of a certain chemical reaction is directly [#permalink]

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03 May 2014, 01:50
Ratio A to B : A^2 = B

Say B = 100
Then A = 10^2

B Increases 100% ==> 200
x^2 = 200 ?
14^2=196 (almost 200, so x is a slightly bigger)

10 ==> 14 ==> increase of 40%.

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The rate of a certain chemical reaction is directly [#permalink]

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21 Dec 2014, 07:40
$$\sqrt{2}$$= 1.414
Option D)
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The rate of a certain chemical reaction is directly   [#permalink] 21 Dec 2014, 07:40

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