Question

For reaction A+2b-C+2dthe rate law is

Answer 1

Answer:

The given reaction is:

A+2B+C⟶2D+E

According to question,

Rate=K[A]  

1

[B]  

2

[C]  

0

       −(i)

where, K is rate constant

Now, if the concentration of each reactant is doubled then,

(Rate)  

new

​  

=K[2A]  

1

[2B]  

[

2C]  

0

 

=K.2[A]  

1

.4[B]  

2

.2[C]  

0

 

(Rate)  

new

​  

=8.K[A]  

1

[B]  

2

[C]  

0

       −(ii)

Now, Dividing (ii) by (i) :-

Rate

(Rate)  

new

​  

 

​  

=8

So, the final rate will be 8 times to that of initial.

Explanation:

Answer 2

Answer:

dr/dt = K[A]$[B]^{2}$

Explanation: