Bunuel 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 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=~1.41, which is approximately 40% increase. \(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Answer: D.
To elaborate more: \(a\) is directly proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets bigger too, so there is some non-zero constant \(x\) such that \(a=xb\);
\(a\) is inversely proportional to \(b\) means that as the absolute value of \(b\) gets bigger, the absolute value of \(a\) gets smaller, so there is some non-zero constant constant \(y\) such that \(a=\frac{y}{b}\).
So, when we are told that the rate (R) is directly proportional to the square of A and inversely proportional to B we can write \(R=\frac{(A^2x)y}{B}\).
Hope it's clear.
Great explanation
BunuelOne question not sure why new and old rate are being equal to each other in below? Could you kindly help clarify? Thanks
\(R=\frac{A^2}{B}=\frac{(1.4A)^2}{2B}\)
Check the highlighted part in the question above.