Question

Given at $10 \text{ AM}$, reaction is started (i) $A \xrightarrow{k} \text{Product}$ ($1^{\text{st}}$ order reaction) (ii) $BrO_3^- + 5Br^- \rightarrow 3Br_2$. At $10:10 \text{ AM}$, rate of disappearance of $Br^-$ was $2 \times 10^{-3} \text{ M/min}$ and concentration of $A$ was $0.1 \text{ M}$, if both reactions were proceed with same rate at this time then value of $k$ will be ?

• A. $10^{-3} \text{ min}^{-1}$
• B. $2 \times 10^{-3} \text{ min}^{-1}$
• C. $4 \times 10^{-3} \text{ min}^{-1}$
• D. $8 \times 10^{-3} \text{ min}^{-1}$

Hint:

The overall rate of reaction $R$ is defined using the inverse of the stoichiometric coefficient for any species: $R = \frac{1}{a} \times \text{Rate of change of } [A]$. Ensure you convert the given disappearance rate to the overall reaction rate before equating $R_i$ and $R_{ii}$.

Answer 1

The Correct Option is C

For reaction (ii): $BrO_3^- + 5Br^- \rightarrow 3Br_2$.
Rate of reaction $R_{ii} = - \frac{1}{5} \frac{d[Br^-]}{dt}$.
Given: $-\frac{d[Br^-]}{dt} = 2 \times 10^{-3} \text{ M/min}$.
$R_{ii} = \frac{1}{5} \times (2 \times 10^{-3}) = 0.4 \times 10^{-3} \text{ M/min}$.
For reaction (i): $A \rightarrow \text{Product}$ ($1^{\text{st}}$ order). Rate $R_i = k[A]$.
Given $R_i = R_{ii}$ and $[A] = 0.1 \text{ M}$.
$k[A] = 0.4 \times 10^{-3}$.
$k (0.1) = 0.4 \times 10^{-3}$.
$k = \frac{0.4 \times 10^{-3}}{0.1} = 4 \times 10^{-3} \text{ min}^{-1}$.