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The experiment data for the reaction 2A ...

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.

Text Solution

Verified by Experts

From an examination of above data, it is clear that when the concentration of `B_(2)` is doubled, the rate is doubled .
Hence, the order of reaction with respect ot `B_(2)` is one.
Further, when concentration of A is doubled , the rate remains unaltered. So, order of reaction with respect ort A is zero .
The probable rate law for the reaction will be
`(dx )/(dt)=k[B_(2)][A]^(0)=k[B_(2)]`
Alternatively, Rate =`k[B_(2)]^(alpha)`
`1.6xx10^(-4)=k[0.5]^(alpha)`
`3.2xx10^(-4)=k[1]^(alpha)`
On dividing we get, `alpha=1`
`:.` Rate =`k[A]^(0)[B_(2)]^(1)=k[B_(2)]`
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