Problem Solving

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, which of the following is closest to
the percent change in the concentration of chemical A to keep the reaction
unchanged?

A) 100% decrease
B) 50% decrease
C) 40% decrease
D) 40% increase
E) 50% increase

D

Hi,
Rate, r = ka^2/b where k is constant, and are concentrations of A and B
b is increased by 100% means new concentration of B is 2b
Let a1 be the new concentration of A
So, r = ka^2/b = ka1^2/2b => a1 = a*sqrt(2) = 1.41a
So, Concentration of A increased by 41%

Hence, D

Hi Frankenstein,
Thanks for this. However, can you please expand your answer. I do not understand how you got from
ka1^2/2b to

a1 = a*sqrt(2) = 1.41

???

Thanks

Viper83 wrote:Hi Frankenstein,
Thanks for this. However, can you please expand your answer. I do not understand how you got from
ka1^2/2b to

a1 = a*sqrt(2) = 1.41

???

Thanks

Hi,
Initially rate is ka^2/b
when b is doubled, rate will be k (a1)^2/(2b)
It is given that rate of reaction should be unchanged
So, both rates should be same
ka^2/b = k (a1)^2/(2b)
=> a1^2 = 2a^2.
So, a1 = (sqrt 2)*a = 1.41a
So, new Concentration of A is 1.41 times old concentration. So, concentration of A should be increased by 41%

Got it thanks! Could you also solve this without using the constant K? It cancels anyway...

a better way is to do by POE
eqn A^2/B
new eqn A^2/2B
in order to get original the numerator has to be inc so A,B,C eliminated
E 50% inc---->means 2.25*A^2/2B rounding of one digit gives us 2.3/2=1.15
D 40% inc----> means 1.96A^2/2B rounding of one digit gives us 2/2=1
hence d is closest