Problem Solving

Quite baffled...

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

OA: D

Given,

R = k A^2 / B

If B is increased 100%, we have 2B but the A^2 / 2B ratio should be the same for the rate to be equal.

Since there is a extra 2 in denominator, we should look to increase the Num to compensate for it (or cancel out the 2)

A is squared, so Sqrt2 A, when squared will give 2A^2 and the 2 cancels out.
Sqrt2 is 1.414, so thats a 41% increase. Approx D

R(original) = k*A^2/B
Let A' be the new value of A required to keep the Rate same
R(New) = k*A'^2/2B

As we require R(new) = R(original), we have A'^2/2B = A^2/B or A'^2 = 2A^2 or A' = sqrt(2)*A
Percent change in concentration of A is (sqrt(2)*A -A)/A = 40%increase

razorback wrote:Quite baffled...

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

OA: D

R is directly proportional to A² means that as R increases, A² increases by a proportional amount:
R/A² = k.
In the equation above, if R doubles, then A² must double, so that the value of R/A² is always equal to the constant k.

R is inversely proportional to B means that as R increases, B decreases by a proportional amount:
RB = k.
In the equation above, if R doubles, then B must be halved, so that the value of RB is always equal to the constant k.

Thus, the problem above describes the following relationship:
R = A²/B.
In the equation above:
R/A² is always equal to B and RB is always equal to A².

Given R=A²/B, we can plug in values.

Plug A=10 and B=1 into R=A²/B:
Then R = 10²/1 = 100.

If B doubles and R is unchanged, we get:
100 = A²/2
200 = A²
A = √200 = 10√2 ≈ 14.

A increases from 10 to 14.
Percent increase in A = Difference/Original = 4/10 = 40%.

The correct answer is D.

It would be wise to remember the following:

x is directly proportional to y and inversely proportional to z means x = y/z.

Thanks! Makes sense now.

In all the quant prep materials Ive come across I havent seen a proportionality question this difficult.

Also, not sure why I labeled the post "Probability", if a mod can and wants to fix it to "Proportionality" that would be sweet. Thanks again.

Kindly check out the below link for video solution:

https://www.youtube.com/watch?v=KNR3R78SHPA


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The expression will be

R = K (Ca)^2/(Cb)

let Ca= 4
and Cb = 1
This will give R's value= 16K ---(1)

As per the condition, Cb is increased by 100% i.e. the new value of Cb =2

put this value in the equation (1)

We will get the new value of Ca --

16k= k *(Ca)^2/2 or Ca = 4*1.41

i.e. the value is increased by 41% approx

Answer D