For a reaction between A and B, the initial rate of reaction is measured for various initial concentration of A and B. The data provided are
`{:("[A]","[B]","Initial reaction rate"),("(a) 0.2 M","0.30 M",5xx10^(-5)),("(b) 0.20 M","0.10 M",5xx10^(-5)),("(c) 0.40 M","0.05 M",1xx10^(-4)):}`
The overall order of the reaction is

Options

A. one
B. two
C. two or a half
D. Three

Answer

A

Explanation

If the order of reation w.r.t A is n and the order of reaction w.r.t. B is m, rate law becomes
`"Rate"=k[A]^(n)[B]^(m)`
From (i)
`5xx10^(-5)=[0.20]^(n)[0.30]^(m)` …. (i)
From (ii) `5xx10^(-5)=[0.20]^(n)[0.10]^(m)` …. (ii)
From (iii)
`1xx10^(-4)=[0.40]^(n)[0.05]^(m)` ..... (iii)
or `10xx10^(-5)=[0.40]^(n)[0.05]^(m)`
From equation (i) and (ii)
`(5xx10^(-5))/(5xx10^(-5))=((0.20)/(0.20))^(n)((0.30)/(0.10))^(m)`
`1=(3)^(m)implies(3)^(0)=(3)^(m)impliesm=0`
From equation (ii) and (iii),
`(5xx10^(-5))/(10xx10^(-5))=((0.20)/(0.40))^(n)((0.10)/(0.05))^(m)`
`(1)/(2)=((1)/(2))^(n)xx((0.10)/(0.05))^(0)`
implies `(1)/(2)=((1)/(2))^(n)implies((1)/(2))^(1)=((1)/(2))^(n)impliesn=1`