Looking for classes? Ksquare Career Institute, Bengaluru →
In the reaction $BrO_3^-(aq) + 5Br^-(aq) + 6H^+ \rightarrow 3Br_2(l) + 3H_2O(l)$, the rate of appearance of bromine ($Br_2$) is related to rate of disappearance of bromide ions as following
A
$\frac{d[Br_2]}{dt} = \frac{3}{5}\frac{d[Br^-]}{dt}$
B
$\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}$
C
$\frac{d[Br_2]}{dt} = -\frac{5}{3}\frac{d[Br^-]}{dt}$
D
$\frac{d[Br_2]}{dt} = \frac{5}{3}\frac{d[Br^-]}{dt}$
Detailed Solution
For the given reaction: Rate $= -\frac{1}{5}\frac{d[Br^-]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt}$
The negative sign appears with $Br^-$ because its concentration decreases with time.
$\frac{1}{3}\frac{d[Br_2]}{dt} = -\frac{1}{5}\frac{d[Br^-]}{dt}$
$\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}$
The negative sign appears with $Br^-$ because its concentration decreases with time.
$\frac{1}{3}\frac{d[Br_2]}{dt} = -\frac{1}{5}\frac{d[Br^-]}{dt}$
$\frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt}$