If the rate of disappearance of Br–(aq) at a particular moment during the reaction is 3.5 × 10−4 mol L−1 s−1, what is the rate of appearance of Br2(aq) at that moment?

Answer :

The question is incomplete, here is the complete question:

Consider the following reaction in aqueous solution: [tex]5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)[/tex]

If the rate of disappearance of [tex]Br^-(aq)[/tex] at a particular moment during the reaction is [tex]3.5\times 10^{-4}Ms^{-1}[/tex], what is the rate of appearance of Br₂(aq) at that moment?

Answer: The rate of appearance of [tex]Br_2(aq.)[/tex] for the reaction is [tex]2.1\times 10^{-4}[/tex]

Explanation:

We are given:

Rate of disappearance of [tex]Br^-(aq.)=-\frac{d[Br^-]}{dt}=3.5\times 10^{-4}Ms^{-1}[/tex]

For the given chemical equation:

[tex]5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)[/tex]

As we know, rate of the reaction remains the same.

So,

[tex]\text{Rate of appearance of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}[/tex]

From the reaction:

[tex]\frac{1}{5}\frac{d[Br^-]}{dt}=\frac{1}{3}\frac{d[Br_2]}{dt}[/tex]

Putting values in above equation, we get:

[tex]\frac{1}{5}\times (3.5\times 10^{-4})=\frac{1}{3}\frac{d[Br_2]}{dt}\\\\\frac{d[Br_2]}{dt}=\frac{3}{5}\times (3.5\times 10^{-4})=2.1\times 10^{-4}[/tex]

Hence, the rate of appearance of [tex]Br_2(aq.)[/tex] for the reaction is [tex]2.1\times 10^{-4}[/tex]