dferm 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

Thanks for your help in advance!

Interesting question!

At the very least, we can quickly give ourselves either a 1/3 or 1/2 chance by figuring out the direction in which A needs to go.

We know 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 B present.

So, an increase in B will lower the rate of the reaction. To bring the rate of reaction back up to where it was, we need to increase the amount of A.

Accordingly, eliminate (a), (b) and (c).

At this point, we can solve using concepts or by picking numbers. Let's try both approaches. (As an aside, if you're practicing and see multiple ways to attack a question, try them all - that's the only way you'll figure out what works best for you on test day.)

We can view the relation between A and B as a fraction:

(A^2)/B

will give us our rate of change of the reaction.

Therefore, to keep the reaction constant, we need to increase A by less than we increase B. Since the numerator is a square function, we actually need to increase A by the root of the increase of B.

So, if we 2*B, we (root2)*A to keep the ratio constant.

B is increasing by 100%, so we're multiplying B by 2. Therefore, we need to multiply A by root2 = approx 1.4. So, A needs to increase by approximately 40%: choose (d).

Unless that made perfect sense to you and you saw it quickly, picking numbers (with a bit of backsolving) would have been the better choice on test day.

Let's start with A = 10 and B = 100, so the original ratio is 10^2/100 = 1.

If we double B, we have 10^2/200.

Now let's backsolve:

If we increase A by 50% (choice e), the new ratio is 15^2/200 = 225/200.

If we increase A by 40% (choice d), the new ratio is 14^2/200 = 196/200.

The 40% increase is much closer to our 1:1 ratio - choose (D).

Another explanation:

R1=KA^2/B
R2=KA1^2/2B Since B increases by 100%



Given: R1=R2

KA^2/B=KA1^2/2B

(A^2 * 2B)/B=A1^2

A^2*2=A1^2

taking the sqrt

A1= Sqrt (A^2*2)

A1= A*sqrt2

Sqrt 2 is approx 1.4

A1= 1.4 A

which is approx 40% increase in A.