For the reaction BrO₃^- + 5Br^-+ 6H⁺ $\rarr$ 3Br₂ + 3H₂O at a particular time, -$\Delta$[BrO₃^-]/$\Delta$t = 1.5 * 10^-2 M/s. What is -$\Delta$[Br^-]/$\Delta$t at the same instant

Question

Quizplus · Accepted Answer

According to the rate law for the reaction, the rate of disappearance of BrO₃^- is proportional to its concentration, and the proportionality constant is 1.5 F1F1F1S1F1F1F10 10^-2 M/s. Similarly, the rate of disappearance of Br^- is proportional to its concentration, and the proportionality constant can be found by dividing the overall rate law by the stoichiometric coefficient of Br^-, which gives: -F1F1F1S1F1F1F10[Br^-]/F1F1F1S1F1F1F10t = (1.5 F1F1F1S1F1F1F10 10^-2 M/s)/5 = 0.3 F1F1F1S1F1F1F10 10^-2 M/s. Therefore, the answer is B, 7.5 F1F1F1S1F1F1F10 10^-2 M/s.

For the Reaction BrO3- + 5Br-+ 6H+ →\rarr→ 3Br2 + 3H2O at a Particular Time, -

For the Reaction BrO₃^- + 5Br^-+ 6H⁺ $\rarr$ 3Br₂ + 3H₂O at a Particular Time, -