The experimental data for the reaction `2A+B_(2)rarr2AB,` is :

The rate equation for the above data is :

Options

A. `rate =k [B_(2)]`
B. `rate =k [B_(2)]_(2)`
C. `rate =k[A]^(2)[B]^(2)`
D. `rate=k [A]^(2)[B]`

Answer

A

Explanation

Consider the following rate law equation .
`(dx)/(dt)=K[A]^(m)[B_(2)]^(n)`
`1.6xx10^(-4)=K[0.50]^(m)[0.50]^(n)` . ..(i)
`3.2xx10^(-4)=K[0.50]^(m)[10]^(n)` . . .(ii)
`3.2xx10^(-4)=K[100]^(m)[1.0]^(n)`. . .(iii)
by dividing Eq.(iii) by (ii) we get ,
`(3.2xx10^(-4))/(3.2xx10^(-4))=(k[1.00]^(m)[1.0]^(n))/(k[0.50]^(m)[1.0]^(n))`
`1=2 ^(m) or 2^(0)=2^(m)`
` therefore m=0`
By dividing Eq(ii) by (i)
`(3.2xx10^(-4))/(1.6xx10^(-4))=([0.50]^(m)[1.0]^(n))/([0.50]^(m)[0.50]^(n))`
`1=2^(m) or 2^(0)=2^(n)`
` therefore m=0`
by dividing Eq.s (ii) by (i)
`(3.2xx10^(-4))/(1.6xx10^(-4))=([0.50]^(m)[1.0]^(n))/([0.50]^(m)[0.50]^(n))`
`2=2^(n)`
`or 2^(1)=2^(n)`
`therefore n=1 `
hence rate .
`((dx)/(dt))=K[A]^(0)[B_(2)]^(1)`
`=K[b_(2)]`