The rate of a reaction is directly proportional to the square of the concentration 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
Rate of a reaction
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Ahh....seems like i finally solved this question...
correct me if i am wrong..
Rate of Eq, R is proportinal to A^2 and inversly proportional to B
So, R = k (A^2/B)
=> k = (BR)/(A^2)..................(1)
Now if B is increased by 100%. This means B is now 2B.
So to keep R constant, we would need to increased decrease A by some amount, lets say x.
R= k (xA)^2/2B
From (1), subtitue value of k and solve
R = [BR* (xA)^2]/[2B* A^2]
so x= sq root of 2 = 1.414
ie an increase of 40%
correct me if i am wrong..
Rate of Eq, R is proportinal to A^2 and inversly proportional to B
So, R = k (A^2/B)
=> k = (BR)/(A^2)..................(1)
Now if B is increased by 100%. This means B is now 2B.
So to keep R constant, we would need to increased decrease A by some amount, lets say x.
R= k (xA)^2/2B
From (1), subtitue value of k and solve
R = [BR* (xA)^2]/[2B* A^2]
so x= sq root of 2 = 1.414
ie an increase of 40%