For the reaction : `H_(2)+I_(2)to 2HI,` the differential rate law is

Options

A. `-(d[H_(2)])/(dt)=-(d[I_(2)])/(dt)=2(d[HI])/(dt)`
B. `-(d[H_(2)])/(dt)=-2(d[I_(2)])/(dt)=(d[HI])/(dt)`
C. `-(d[H_(2)])/(dt)=-(d[I_(2)])/(dt)=(d[HI])/(dt)`
D. `-(d[H_(2)])/(dt)=-(d[I_(2)])/(dt)=(d[HI])/(dt)`

Answer

B

Explanation

`H_(2)+I_(2) to 2HI`
Rate of reaction `=-(d[H_(2)])/(dt)=(-d[I_(2)])/(dt)=(1)/(2)(d[HI])/(dt)`
or `(-2d[H_(2)])/(dt)=(-2d[I_(2)])/(dt)=(d[HI])/(dt)`