Question

In a reaction, A+B Product, rate is doubled when
the concentration of B is doubled and rate increases by
a factor of 8 when the concnetrations of both the
reactants (A and B) are doubled, rate law for the
reaction can be written as
1) Rate = k [A] [B] (2) Rate = k [A] [B]
3) Rate = k [A] [B] (4) Rate = k [A] [BP​

Answer 1

Rate law for the  reaction can be written as $Rate=k[A]^2[B]^1$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For the given reaction : $A+B\rightarrow product$

$Rate=k[A]^x[B]^y$   (1)

k= rate constant

x = order with respect to A

y = order with respect to B

n = x+y = Total order

From trial 1: $2\times Rate=k[A]^x[2B]^y$   (2)

From trial 2:$8\times Rate=k[2A]^x[2B]^y$    (3)

Dividing 3 by 2 :$\frac{8\times Rate}{2\times Rate=}=\frac{k[2A]^x[2B]^y}{k[A]^x[2B]^y}$

$4=2^x,2^2=2^x$ therefore x=2

Putting x = 2 in (1) and (2) , we get:

$Rate=k[A]^2[B]^y$  (1)

$2\times Rate=k[A]^2[2B]^y$  (2)

Dividing 2 by 1 :$\frac{2\times Rate}{Rate}=\frac{k[2B]^y}{k[B]^y}$

$2=2^y,2^1=2^y$, y=1

The rate law is $Rate=k[A]^2[B]^1$

Learn More about rate law

https://brainly.com/question/13018523

https://brainly.com/question/13301088

#learnwithbrainly

Answer 2

Answer:

this is the answer

please mark brainlist