The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversly proportional to the concentration of Chemical B present. If the concentration of chemical B is increased by 100%, which of the follwing is the closest to the % change in the concentration of chemical A required to keep the reaction rate unchanged??
100% decrease
50% dec
40% dec
40% inc
50% inc
GMAT Prep percentage
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
Posted: Thu Nov 15, 2007 7:11 am Post subject: GMAT Prep percentage
--------------------------------------------------------------------------------
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversly proportional to the concentration of Chemical B present. If the concentration of chemical B is increased by 100%, which of the follwing is the closest to the % change in the concentration of chemical A required to keep the reaction rate unchanged??
100% decrease
50% dec
40% dec
40% inc
50% inc
first of all if B increase then rate will decrease (inv prop) so to keep the rate unchanged Av should inc
given
R = xA ^2
R = y/ B ( x,y = constanats , A= conc of A , B = conc o-f B)
so when B become 2B
Rd (dec rate) = y/ 2B i.e reduces by 1/2
so R = 2 * y/2B
= 2*xA^2
= x * (1.4)^2
so A should be inc to 1.4 A
i.e by 40%
D
--------------------------------------------------------------------------------
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversly proportional to the concentration of Chemical B present. If the concentration of chemical B is increased by 100%, which of the follwing is the closest to the % change in the concentration of chemical A required to keep the reaction rate unchanged??
100% decrease
50% dec
40% dec
40% inc
50% inc
first of all if B increase then rate will decrease (inv prop) so to keep the rate unchanged Av should inc
given
R = xA ^2
R = y/ B ( x,y = constanats , A= conc of A , B = conc o-f B)
so when B become 2B
Rd (dec rate) = y/ 2B i.e reduces by 1/2
so R = 2 * y/2B
= 2*xA^2
= x * (1.4)^2
so A should be inc to 1.4 A
i.e by 40%
D
Regards
Samir
Samir
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
I've used Rd to signify the reduced rate R
i.r Rd = 1/2 R = 1/2 *y/B
so to bring Rd back to R we need to multiply it by 2
2Rd = R
2 * y/2B = R
so 2 * xA^2 =R
so x*(rt(2)*A)^2 =R
rt(2) =1.4 approx
so rt(2)*A = 1.4 A
i.e 40% more than A
hope this is clear.
i.r Rd = 1/2 R = 1/2 *y/B
so to bring Rd back to R we need to multiply it by 2
2Rd = R
2 * y/2B = R
so 2 * xA^2 =R
so x*(rt(2)*A)^2 =R
rt(2) =1.4 approx
so rt(2)*A = 1.4 A
i.e 40% more than A
hope this is clear.
Regards
Samir
Samir