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

Explanation

To find the rate constant \( k \) for the reaction \( 3 \text{BrO}^\circ \rightarrow \text{BrO}_3^\circ + 2 \text{Br}^\circ \) in terms of \( \frac{d[\text{BrO}_3^\circ]}{dt} \), we can follow these steps:

### Step 1: Write the rate expressions for the reaction
For the reaction \( 3 \text{BrO}^\circ \rightarrow \text{BrO}_3^\circ + 2 \text{Br}^\circ \), the rate can be expressed in terms of the change in concentration of each species:

\[ -\frac{1}{3} \frac{d[\text{BrO}^\circ]}{dt} = \frac{1}{1} \frac{d[\text{BrO}_3^\circ]}{dt} = \frac{1}{2} \frac{d[\text{Br}^\circ]}{dt} \]

### Step 2: Relate the rates to the rate constant
From the rate expression, we know that the rate of the reaction can be expressed in terms of the rate constant \( k \) and the concentration of the reactants. For the rate law of \( -\frac{d[\text{BrO}^\circ]}{dt} \):

\[ -\frac{d[\text{BrO}^\circ]}{dt} = k [\text{BrO}^\circ]^3 \]

### Step 3: Express the rate in terms of \( \frac{d[\text{BrO}_3^\circ]}{dt} \)
From the rate expressions, we can express the rate of formation of \( \text{BrO}_3^\circ \) in terms of the rate of consumption of \( \text{BrO}^\circ \):

\[ \frac{d[\text{BrO}_3^\circ]}{dt} = -\frac{1}{3} \frac{d[\text{BrO}^\circ]}{dt} \]

### Step 4: Substitute the known rate constant
Given that the rate constant \( k \) for the reaction in terms of \( -\frac{d[\text{BrO}^\circ]}{dt} \) is \( 0.054 \, \text{L mol}^{-1} \text{s}^{-1} \), we can express the rate constant for \( \frac{d[\text{BrO}_3^\circ]}{dt} \):

\[ k' = \frac{1}{3} k \]

### Step 5: Calculate the new rate constant
Substituting the value of \( k \):

\[ k' = \frac{1}{3} \times 0.054 \, \text{L mol}^{-1} \text{s}^{-1} = 0.018 \, \text{L mol}^{-1} \text{s}^{-1} \]

### Final Answer
Thus, the rate constant \( k' \) for the reaction in terms of \( \frac{d[\text{BrO}_3^\circ]}{dt} \) is:

\[ \boxed{0.018 \, \text{L mol}^{-1} \text{s}^{-1}} \]