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Junior | Next Rank: 30 Posts
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by princessss » Fri Oct 08, 2010 2:24 am
The rate of a reaction is directly proportional to the square of the concentraton of A and inversely proportional to concentration of B. If B increases by 100%, which of the following is closest to the % change in concentration of A required to keep the rate unchanged?

A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase
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by shovan85 » Fri Oct 08, 2010 2:29 am
The rate of a reaction(say r) is directly proportional to the square of the concentraton of A and inversely proportional to concentration of B.

So in this problem, r = kA^2 / B (k is some proportionality constant)

Let the new concentration of A as A', and the new concentration of B as B'.
Now the rate(say r') is the same
Thus r' = r

k (A')^2 / B' = k A^2 / B

We know that if B is increased by 100%, that means B is doubled, so B' = 2B.
k (A')^2 / 2B = k A^2 / B

We can multiply both sides by B/k, so
(A')^2 = 2 A^2

This means that A' = A 2^(1/2) = 1.414A. You should have memorized that:

so A' = 1.414 A, or a 41.4% increase nearly 40% increase.

Hope this helps.
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