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, what is the percent increase/decrease in the concentration of chemical A required to keep the reaction rate unchanged?
2006-10-18 10:32:37 · 5 answers · asked by Jason A 1 in Science & Mathematics ➔ Mathematics
The concentration of A needs to increase by 41.42%, or by a factor of squared root of 2.
2006-10-18 10:52:35 · answer #1 · answered by M 3 · 0⤊ 0⤋
Man..... I'm glad I didn't have this question on my GMAT test. But I believe the answer would be that "A" has to be increased by 10%. The rate of the reaction is proportional to A^2, and inversely proportional to B^1. You have to assume then that 1 unit of Chemical B is equal (offsets) a squared unit of Chemical A. If you add units of Chemical B, then the square root of the number of units added have to be added of Chemical A to maintain the reaction rate. Since A^2 = B ^1, A is equal to the square root of B. Since B = 100 in this case, A = 10. Hope this helps!!
2006-10-18 17:49:42 · answer #2 · answered by ljherrz 1 · 0⤊ 0⤋
A - increase 10%
.
2006-10-18 17:46:37 · answer #3 · answered by Zak 5 · 0⤊ 0⤋
It should either be multiplied by .1 or increased by 10000% but this is just guessish, thats just what it sounds like.
2006-10-18 17:43:29 · answer #4 · answered by shenpup 1 · 0⤊ 0⤋
Rate proportional to [A]^2
Rate inversely proportional to [B]
So rate proportional to [A]^2/[B]
Rate = k [A]^2 / [B]
Rate1 = k [A1]^2 / [B1]
Now]
[B2] = [2B1]
[A2] = [nA1]
Rate2 = k [A2]^2 / [B2] = k [nA1]^2 / [2B1]
Rate 1 = Rate 2 if
k [A1]^2 / [B1] = k [nA1]^2 / [2B1]
Simpliflying:
1 = n^2 / 2
n^2 = 2
n = sqrt(2). n positive so [A] will increase.
2006-10-18 17:50:24 · answer #5 · answered by Dr. J. 6 · 0⤊ 0⤋