For the reaction `2A + B rarrC`, the values of ini concentrations are given in the table below. The rate law for the reaction is:

Options

A. Rat `=k[A]^(2) [B]`
B. Rate `=k [A]^(2) [B]^(2)`
C. Rate =k [A] [B]
D. Rate `=k [A] [B]^(2)`

Answer

D

Explanation

From given data
Rate `=k [A]^(a) [B]^(b)`
Now, `0.045 =k [0.05]^(a+b)" ….(i) ", 0.090 =k[0.05]^(a+b) 2^(a)" ….(ii)"`
`0.72=k [0.5]^(a+b)2^(2a+b)" ….(iii)"`
From equation (i) and (ii) we get,
`rArr (1)/(2) =(1)/(2^(a)) rArr a=1`
And from equation (i) and (iii) we get
`rArr (0.045)/(0.72) = (1)/(2^(2+b)) rArr (45)/(720) =(1)/(2^(2+b)) rArr (720)/(45) =2^((2+b)) rArr Z^(4) =Z^(2+b) rArr b=2`