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!
The rate of a certain chemical reaction is directly
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- selango
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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
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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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!!
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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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 !
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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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.
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.
Sudhanshu
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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.
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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Easiest among all,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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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?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.
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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\[rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\]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
\[\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)\]
\[?\,\, \cong \,\,k - 1\]
\[{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
{k^2} = 2 \hfill \\
\sqrt 2 \cong 1.41 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2 - 1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2 - 1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)\]
This solution follows the notations and rationale taught in the GMATH method.
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Fabio.
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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: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
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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