Question

Consider the following reaction in aqueous solution:

\(\)

\({\bf{5B}}{{\bf{r}}^ - }\left( {aq} \right) + BrO_3^ - \left( {aq} \right) + 6{H^ + }\left( {aq} \right) \to 3B{r_2}\left( {aq} \right) + 3{H_2}O\left( l \right)\)

If the rate of disappearance of Br^– (aq) at a particular moment during the reaction is 3.5 × 10⁻⁴ M s ⁻¹, what is the rate of appearance of\({\bf{B}}{{\bf{r}}_{\bf{2}}}\)(aq) at that moment?

Short answer

Rate of appearance of \({\bf{B}}{{\bf{r}}_{\bf{2}}}\)(aq) is \(2.1 \times {10^{ - 4}}M{s^{ - 1}}\)

Step-by-step solution

Rate of reaction 

\({\bf{5B}}{{\bf{r}}^ - }\left( {aq} \right) + BrO_3^ - \left( {aq} \right) + 6{H^ + }\left( {aq} \right) \to 3B{r_2}\left( {aq} \right) + 3{H_2}O\left( l \right)\)

Rate =\(\frac{{{\bf{ - \Delta (B}}{{\bf{r}}^{\bf{ - }}}{\bf{)}}}}{{{\bf{5\Delta t}}}}{\bf{ = }}\frac{{{\bf{ - \Delta (BrO}}_{\bf{3}}^{\bf{ - }}{\bf{)}}}}{{{\bf{\Delta t}}}}{\bf{ = }}\frac{{{\bf{ - \Delta (}}{{\bf{H}}^{\bf{ + }}}{\bf{)}}}}{{{\bf{6\Delta t}}}}{\bf{ = }}\frac{{{\bf{\Delta (B}}{{\bf{r}}_{\bf{2}}}{\bf{)}}}}{{{\bf{3\Delta t}}}}{\bf{ = }}\frac{{{{\bf{H}}_{\bf{2}}}{\bf{O}}}}{{{\bf{3\Delta t}}}}\)

Rate of appearance of Br2 (aq)

Given the rate of disappearance of Br^- (aq) is \({\bf{3}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M/s}}\)

Rate of appearance is equal to rate of disappearance

\(\frac{{{\bf{ - \Delta (B}}{{\bf{r}}^{\bf{ - }}}{\bf{)}}}}{{{\bf{5\Delta t}}}}{\bf{ = }}\frac{{{\bf{\Delta (B}}{{\bf{r}}_{\bf{2}}}{\bf{)}}}}{{{\bf{3\Delta t}}}}\)

\(\frac{{{\bf{ - \Delta (B}}{{\bf{r}}^{\bf{ - }}}{\bf{)}}}}{{{\bf{\Delta t}}}}{\bf{ = }}\frac{{{\bf{5\Delta (B}}{{\bf{r}}_{\bf{2}}}{\bf{)}}}}{{{\bf{3\Delta t}}}}\)

\(\begin{aligned}{}\frac{{{\bf{ - \Delta (B}}{{\bf{r}}^{\bf{ - }}}{\bf{)}}}}{{{\bf{\Delta t}}}}{\bf{ = - 3}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}{{\bf{s}}^{{\bf{ - 1}}}}\\\frac{{{\bf{5\Delta (B}}{{\bf{r}}_{\bf{2}}}{\bf{)}}}}{{{\bf{3\Delta t}}}}{\bf{ = - 3}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}{{\bf{s}}^{{\bf{ - 1}}}}\\\frac{{{\bf{\Delta (B}}{{\bf{r}}_{\bf{2}}}{\bf{)}}}}{{{\bf{\Delta t}}}}{\bf{ = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}{{\bf{s}}^{{\bf{ - 1}}}}\end{aligned}\)