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The difference rate law for the reaction...

The difference rate law for the reaction
`H_(2) +I_(2) to 2HI` is

A

`(-d[H_(2)])/(dt) =(-d[I_(2)])/(dt)=(-d[HI])/(dt)`

B

`(d[H_(2)])/(dt) =(d[I_(2)])/(dt)=(d[HI])/(dt)`

C

`(1)/(2)(d[H_(2)])/(dt) =(1)/(2)(d[I_(2)])/(dt)=(-d[HI])/(dt)`

D

`-2(d[H_(2)])/(dt) =-2(d[I_(2)])/(dt)=(d[HI])/(dt)`

Text Solution

Verified by Experts

The correct Answer is:
D
Rate of appearance of `HI =(1)/(2) (d[H])/(dt), " "` Rate of formation of `H_(2) =(-d[H_(2)])/(dt)`
Rate of formation of `I_(2)=(-d[I_(2)])/(dt)`
`therefore (-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)`
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