Hey selango,

So, let's set variables for each of the quantities:

r = rate of reaction
A = concentration of A
B = concentration of B

We know that r goes up proportionally as A^2 goes up and B goes down. So you can create an equation like this:

r = k * (A^2 / B)

where k is some constant.

Now, saying that the concentration of chemical B is increased by 100% is really a fancy way of saying that it's doubled (i.e. B becomes 2B). Now, we have to figure out what has to happen to A in order for the rate to remain the same...

r = k * (A^2 / B)

If B becomes 2B, then the denominator has been multiplied by 2, which means numerator has to be multiplied by 2 as well to keep the rate equal. So the quantity A^2 has to be become 2A^2.

You get this by multiplying A by sqrt(2):

A^2 / B

= [sqrt(2)*A]^2 / 2B

= 2A^2 / 2B

= A^2 / B

Multiplying by sqrt(2) is multiplying by roughly 1.414. If you didn't know that offhand, you could estimate it by using perfect squares. Since 14^2 = 196, that means 1.4^2 = 1.96. Since 2 is just slightly above 1.96, you know that sqrt(2) is just slightly above 1.4

Multiplying by 1.414 represents an increase of about 41.4 percent. Choose D

rate = (A^2/B) * k
r1/r2 = (a1 ^2/b1) / (a2^2/b2)
r1=r2, b2 = 2b1

therefore

-> a1^2 = a2^2/2

a2 = sqrt(2)a1

so increase = sqrt(2)-1 * 100
41 percent increase nearly

option D

Should be D

R= k. A^2/B (k is some constant)
R' = k.A^2/2B

For R' to be equal to R, the equation should be k.2.A^2/2B which means A should increase sq.root 2 times because
(sq.root 2.A)^2=2.A^2

Therefore A needs to be increased [(1.41A-A)/A]100 or appx 40%