Problem Solving

Thanks to everybody in advance for replying to this post.

See the attached.

see attached. hope this helps

Since the rate of the reaction is directly proprotional to square of A and inversely proportional to B, when B is increased by 100% in order to keep the rate constant the square of A must also increase by 100%.

i.e initially if A = 10, we have A^2 = 100

Now what should be the value of A such that A^2 = 200.

i.e. A = sqrt(200) = 10 x sqrt(2) =14.12 approximately.

Thus initially A = 10 and now it is 14.

So an increase of 40% would keep the rate constant

Awesome explanations guys, thank you both very much!