Question

In a reaction , `A + B rarr` Product, rate is doubled when the concentration of `B` is doubled, and rate increases by a factor of `8` when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written as
A. `rate =K[A][b]^(2)`
B. `rate = k [A]^(2)[B]^(2)`
C. `rate =K [A][B]`
D. `rate=k [A]^(2)[B]`

Answer 1

Correct Answer - D
Let the order of reaction with respect to A and B is x and y respectively so the rate law can be given as
`R=k[A]^(x)[B]^(y)` … (i)
when the concentration of only B is doubled ,
the rate is doubled so ,
`R_(1)=K [A]^(x)[2B]^(y)=2R`
if concentations of both the reactants A and B are doubled tha rate increasess by a factor of 8 ,so
`R'=K[2A]^(x)[2B]^(y)=8R`
`implies K2^(x)2^(y)[A]^(x)[B]^(y)=8R`
from Eqs. (i) and (ii) we get
`implies (2R)/(R)=([A]^(x)[2B]^(y))/([A]^(x)[B]^(y))`
`2=2^(y)`
`therefore y=1`
from Eqs.(i) and (iv) we get
`implies (8R)/(R)=(2^(x)2^(y)[A]^(x)[B]^(y))/([A]^(x)[B]^(y)) or 8=2^(x) 2^(y)`
substitution of the value of Y gives.
`8=2^(x)2^(1)`
`4=2^(x)`
`(2)^(2)=(2)^(x)`
`therefore x=2`
substitution of the value of x and y n Eq.(i) gives.
`R=k[A]^(2)[B]`