Mixture question

by **nk2120** Sat Jan 03, 2009 11:32 am

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%, which of the following in closest to the % change in 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: (D) 40% increase

can anybody explain? THANKS!!

by **Harish Dorai** Sat Jan 03, 2009 5:48 pm

The best way to deal with questions like this is to convert the statement to an equation. In this problem the rate of the reaction can be represented as

(Square of A) divided by B, where A and B represents the concentration of A and B respectively. The reason why we say "divided by B" is because the rate is inversely proportional.

So when we increase the concentration of B by 100%, B will become 2B.

So the new rate of the reaction is (Square of A) divided by 2B

Now in order to keep the rate of reaction the same, the numerator should be multiplied by 2. We can achieve that by increasing the concentration of A to "SquareRoot(2) times A". Square Root of A is approximately 1.41. So that means we need to increase the concentration of A to 1.41A or increase by approximately 41%. The closest answer is (D).

by **RonPurewal** Fri Jan 09, 2009 6:31 am

people, please search the board before posting.

http://www.manhattangmat.com/forums/the ... t3846.html