Question

For the elementary reaction
mA+nB→Products
A is present in excess and when the concentration of B is changed from 0.01 to 0.02 the rate of reaction becomes 4 times,then the rate law determined experimentally can be

Answer 1

Answer:

Correct option is

D  

 

Hence, option D is correct.

Explanation:

Answer 2

Given:

Concentration of A is in excess

Concentration of B is changed from 0.01 to 0.02

Rate of reaction becomes 4 times

To Find:

Rate law of the reaction

Solution:

Rate law equation:

Rate      $r =K [A]^{m} [B]^{n}$

Rate law = r

Rate constant = k

Concentration of A = A

Concentration of B = B

Order with  respect to A = m

Order with respect to B = n

Since, A is in excess so order is independent of concentration A

$\dfrac{r_{1} }{r_{2} }=\dfrac{[B]_{1}^{n} }{[B]_{2}^{n} }$

$\dfrac{1}{4 }=\dfrac{[0.01]^{n} }{[0.02]^{n} }$

$\dfrac{1}{2}^{2} =\dfrac{[1]^{n} }{[2]^{n} }$

$n = 2$

Order of reaction is 2

$\dfrac{dx}{dt} = K [A]^{0} [B]^{2}$