The rate law for a reaction between the substance A and B is given by rate = k[A]n [B]m. On doubling the concentration of A and halving the concentration of B , the ratio of new rate to the earlier rate of the reaction will be as

Question

The rate law for a reaction between the substance A and B is given by rate = k[A]n [B]m. On doubling the concentration of A and halving the concentration of B , the ratio of new rate to the earlier rate of the reaction will be as (A) (1/2(m+n)) (B)  (m + n) (C) (n -m ) (D) 2(n-m)

Tardigrade Admin · Accepted Answer

Rate1 = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, Rate2 = k[2A]n [B/2]m Ratio between new and earlier rate (k[2A]n[B/2]m/k[A]n[B]m) = 2n × ((1/2))m =2 (n-m)

Q. The rate law for a reaction between the substance $A$ and $B$ is given by $r a t e = k [A]^{n} [B]^{m} .$ On doubling the concentration of $A$ and halving the concentration of $B$ , the ratio of new rate to the earlier rate of the reaction will be as

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

$(m + n)$

$(n - m)$

$2^{(n - m)}$

$R a t e_{1} = k [A]^{n} [B]^{m}$
On doubling the concentration of $A$ and halving the concentration of $B,$
$R a t e_{2} = k [2 A]^{n} [B /2]^{m}$
Ratio between new and earlier rate
$\frac{k [ 2 A ] ^{n} [ B /2 ] ^{m}}{k [ A ] ^{n} [ B ] ^{m}} = 2^{n} \times (\frac{1}{2})^{m} = 2^{(n - m)}$

Q. The rate law for a reaction between the substance A and B is given by rate=k[A]n[B]m. On doubling the concentration of A and halving the concentration of B , the ratio of new rate to the earlier rate of the reaction will be as

2(m+n)1​

(m+n)

(n−m)

2(n−m)