Striver 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 percent, which of the following is closer to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?

A) 100% increase
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

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.

Hi Striver

My approach for this question goes like this:

let, Initial conc of A = Ai ; New conc of A = An
Initial conc of B = Bi ; New conc of B = Bn
and proportionality constant = k

equations:
Rate(initial) = (k * Ai^2)/Bi
Rate(new) = (k * An^2)/Bn
= (k * An^2)/2Bi since Bn = 2Bi

Given: Rate(initial) = Rate(new)

therefore, (k * Ai^2)/Bi = (k * An^2)/2Bi
=> An = √2 Ai = 1.414 Ai = (1 + 41.4%)Ai => 40% increase(approx)

Ans: D

Hope it helps:)