Reaction rate between two substance A and B is expressed as following: rate = k[A]n[B]m If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be:

Question

Reaction rate between two substance A and B is expressed as following: rate = k[A]n[B]m If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be: (A) m + n (B) n-m (C) (1/2(m+n)) (D) 2(n-m)

Tardigrade Admin · Accepted Answer

Rate1= k[A]n [5]m Rate2=k[2A]n[(1/2)B] ∴ (Rate2/Rate1)=(k[2A]n[(1/2)B]m/k[A]n[B]m)=(2)n((1/2))m =2n. (2)-m=2n-m

Q. Reaction rate between two substance $A$ and $B$ is expressed as following:
rate = $k [A]^{n} [B]^{m}$
If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be:

$m + n$

$n - m$

$\frac{1}{2 ^{(m + n)}}$

$2^{(n - m)}$

$R a t e_{1} = k [A]^{n} [5]^{m}$
$R a t e_{2} = k [2 A]^{n} [\frac{1}{2} B]$
$∴ \frac{R a t e _{2}}{R a t e _{1}} = \frac{k [ 2 A ] ^{n} [ \frac{1}{2} B ] ^{m}}{k [ A ] ^{n} [ B ] ^{m}} = (2)^{n} (\frac{1}{2})^{m}$
$= 2^{n} . (2)^{- m} = 2^{n - m}$

Q. Reaction rate between two substance A and B is expressed as following: rate = k[A]n[B]m If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be:

m+n

n−m

2(m+n)1​

2(n−m)