The experiment data for the reaction `2A + B_(2) rarr 2AB` is
`|{:("Experiment",[A] M,[B_(2)] M,"Initial rate" (mol L^(-1) s^(-1)),),(I,0.50,0.5,1.6 xx 10^(-4),),(II,0.50,1.0,3.2 xx 10^(-4),),(III,1.00,1.0,3.2 xx 10^(-4),):}|`
Write the most probable rate equation for the reacting giving reason for you answer.

Options

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

Answer

C

Explanation

When we double th econcentration or `A` (form `0.50M` to `1.00M`), the reaction rate remains unaffected. Thus, order with respect to `A` is zero, hence option `(1)` and `(2)` are ruledout. When we double the concentration of `[B_(2)]` , Keeping the cone. of `A` constant, we notice that reaction rate is doubled. Thus, order with respect to `B_(2)` is one. Hence
Rate law is
`Rate=k[B_(2)]`
Alternatively
`Rate=k[A]^(x)[B_(2)]^(y)`
`Rate1=1.6xx10^(-4)=k(0.50)^(x)(0.50)^(y)`
`Rate3=3.2xx10^(-4)=k(0.50)^(x)(1.00)^(y)`
`Rate2=3.2xx10^(-4)=k(1.00)^(x)(1.00)^(y)`
Dividing `Rate_(3)` by `Rate_(1)` :
`(3.2xx10^(-4))/(3.2xx10^(-4))=(k(1.00)^(x)(1.00)^(y))/(k(0.50)^(x)(1.00)^(y))`
`1=2^(x)`
or `2^(@)=2^(x)`
Thus, `x=0`
Dividing `Rate_(2)` by Rate_(1)`
`(3.2xx10^(-4))/(1.6xx10^(-4))=(k(0.50)^(x)(1.00)^(y))/(k(0.50)^(x)(0.50)^(y))`
`2=2^(y)`
Or `2^(1)=2^(y)`
Thus, `y=1`
Hence, the rate law is
Rate `=k[A]^(@)[B_(2)]^(1)`
Rate `=k[B_(2)]`