For the reaction: `aA + bB rarr cC+dD`
Rate `= (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt)`
For reaction `3BrO^(ɵ) rarr BrO_(3)^(ɵ) + 2Br^(ɵ)`, the value of rate constant at `80^(@)C` in the rate law for `-(d[BrO^(ɵ)])/(dt)` was found to be `0.054 L mol^(-1)s^(-1)`. The rate constant `(k)` for the reaction in terms of `(d[BrO_(3)^(ɵ)])/(dt)` is

Options

A. `0.018 L mol^(-1) s^(-1)`
B. `0.162 L mol^(-1) s^(-1)`
C. `0.036 L mol^(-1) s^(-1)`
D. None of these

Answer

A

Explanation

For the given reaction,
`(1)/(3) (-d[BrO^(ɵ)])/(dt) = (d[BrO_(3)^(ɵ)])/(dt) = (1)/(2)(d[Br])/(dt)`
Since rate `prop` rate constant
`:. (d[BrO_(3)^(ɵ)])/(dt) = (1)/(3) xx 0.054 = 0.018 L mol^(-1) s^(-1)`