The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical $A$

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Sun Apr 11, 2021 2:59 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 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

Answer: D

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Source: Beat The GMAT — Problem Solving | Sun Apr 11, 2021 2:59 am

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Re: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical \(A\

by Scott@TargetTestPrep » Mon Apr 26, 2021 6:52 am

Gmat_mission wrote: ↑
Sun Apr 11, 2021 2:59 am
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

Answer: D

Solution:

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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