Problem Solving

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 is d. Please explain, completely lost

sk8ternite 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

Please explain, completely lost

When you have really complicated questions, you have to spend extra time understanding the situation. This question is a case in point.

With a basic understanding we can give ourselves a quick 50/50 shot at the point:

"The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present"... so when A goes up, the rate of reaction goes up.

"and [is] inversely proportional to the concentration of chemical B present"... so when B goes up, the rate of reaction goes down.

"If the concentration of chemical B is increased"... we predict, this will slow down the reaction.

"which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?"... we predict, to counteract the effects of B, we need to increase the amount of A.

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

If we want to actually solve, we could pick numbers to do so. We have a percent question, so let's use 100 to keep things simple.

Original: 100A and 100B.

Rate of reaction is proportional to a^2 and inversely proportional to B, so original rate of reaction is:

(100*100) * 1/100 = 100

New: B increases by 100%, so we now have 200B.

Without adding A, new rate of reaction is:

(100*100) * 1/200 = 50

We want to get back to our reaction rate of 100. Let's just plug in a choice to see which one is correct:

If new A is 150 (increase by 50%, choice (e)), new rate of reaction is:

(150*150) * 1/200 = 150*150/200 = 15*15/2 = 225/2 which is greater than 100. Therefore, a 50% increase is TOO MUCH.. eliminate (e), choose (d).

sk8ternite 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

Answer is d. Please explain, completely lost

I solved it this way...

Given the information in the statement

A^2 = 1/B

Say B = 4, then A = 1/2

If B is increased by 100% then B = 8

and A = 2 * sqrt(2) = 2 * 1.41 = 2.82

.82 needs to be added to A so...

.82/2 = 41%

40% increase.

D.

Of course, there's always an algebraic solution as well.

If the rate of reaction is inversely proportional to the concentration of B, then doubling B will 1/2 the reaction.

To counteract 1/2, we need to double the reaction rate due to A. So:

(1 + x/100)^2 = 2

in which x is the percent change of A.

At this point we could use some complex math to solve for x, or again we could just plug in x=40 or x=50 and see which one works better.

If x=40, then we get:

(1 + 40/100)^2 = 2

(1.4)^2 = 2

1.96 = 2

which is pretty darn close.. pick (D).

If we had plugged in x=50, then we get:

(1 + 50/100)^2 = 2

(1.5)^2 = 2

2.25 = 2

which isn't very close at all... eliminate (E), pick (D).

Hi Stuart,

How abt this way?
Rate = Constant*A^2/B
The 100% increase in B means B has become 2B. To keep the rate constant A^2 also has to be 2A^2. So A needs to be increased by Root 2 times, i.e. 1.372 times. This means 37.2 % increase or approx. 40% increase.

Is there any logical flaw? I mean, am I lucky with this sum or will this theory work well in other sums of same kind as well.

Regards,

ananda271181 wrote:Hi Stuart,

How abt this way?
Rate = Constant*A^2/B
The 100% increase in B means B has become 2B. To keep the rate constant A^2 also has to be 2A^2. So A needs to be increased by Root 2 times, i.e. 1.372 times. This means 37.2 % increase or approx. 40% increase.

Is there any logical flaw? I mean, am I lucky with this sum or will this theory work well in other sums of same kind as well.

Regards,

Yup - definitely another way to solve it. You basically took:

(1 + x/100)^2 = 2

and solved:

1 + x/100 = root2

then knowing the value of root 2 as just over 1.4 (it's not 1.372, not sure where you got that number), you get:

x/100 = 1.4 - 1
x/100 = .4
x = 40

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?

rate doesn't change or its a constant

Rate = K*A^2/B so if we keep the rate constant and increase the con. of B by 100% that means we double the con. of B

then we take the square root , the square root of 2 we get 1.41..

so the answer is 40% increase

No need to solve any equations

The question test the square root of 2 if you can identify that the its very simple