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

The rate law for the given reaction can be determined as:
Rate = k[BrOS1U1B13S1U1B0S1U1P1-S1S1P0][BrS1U1P1-S1S1P0][H+]
The overall order of this reaction is 1+1+1=3.
At a particular time, 
-F1F1F1S1F1F1F10[BrOS1U1B13S1U1B0S1U1P1-S1S1P0]/F1F1F1S1F1F1F10t = 1.5 × 10S1U1P1-2S1S1P0 M/s
Let's assume the initial concentration of BrOS1U1B13S1U1B0S1U1P1-S1S1P0 to be x.
Therefore, at the same instant, from the given rate law:
-F1F1F1S1F1F1F10[BrS1U1P1-S1S1P0]/F1F1F1S1F1F1F10t = k(x)(5x)(6[HS1U1P1])
-F1F1F1S1F1F1F10[BrS1U1P1-S1S1P0]/F1F1F1S1F1F1F10t = 30kx^2[HS1U1P1]
-F1F1F1S1F1F1F10[BrS1U1P1-S1S1P0]/F1F1F1S1F1F1F10t = 30 x (1.5 × 10S1U1P1-2S1S1P0) x (x) x (6)
-F1F1F1S1F1F1F10[BrS1U1P1-S1S1P0]/F1F1F1S1F1F1F10t = 2.7 × 10S1U1P1-3S1S1P0 x (x)
-F1F1F1S1F1F1F10[BrS1U1P1-S1S1P0]/F1F1F1S1F1F1F10t = 2.7 × 10S1U1P1-3S1S1P0 M/s
Therefore, the answer is (B) 7.5 × 10S1U1P1-2S1S1P0 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, -