For the reaction 5Br−+BrO3−+6H+→3Br2(aq)+3H2O(I) if the rate of disappearance of Br− is 2.5 x 10- 2 mol L-1 s-1the rate of appearance of Br2will be

A
$2.5 \times 10^{- 2} mol L^{- 1} s^{- 1}$
B
$1.5 \times 10^{- 2} mol L^{- 1} s^{- 1}$
C
$1.25 \times 10^{- 2} mol L^{- 1} s^{- 1}$
D
$3.0 mol L^{- 1} s^{- 1}$

Solution:

we have $- \frac{1}{5} \frac{d [{Br}^{-}]}{d t} = \frac{1}{3} \frac{d [{Br}_{2}]}{d t}$

Hence , $\frac{d [{Br}_{2}]}{d t} = - \frac{3}{5} \frac{d [{Br}^{-}]}{d t} = - \frac{3}{5} (- 2.5 \times 10^{- 2} {molL}^{- 1} s^{- 1}) = 1.5 \times 10^{- 2} {molL}^{- 1} s^{- 1}$

For the reaction $5 {Br}^{-} + {BrO}_{3}^{-} + 6 H^{+} \to 3 {Br}_{2} (aq) + 3 H_{2} O (I)$ if the rate of disappearance of ${Br}^{-}$ is $2.5 x 10^{- 2} mol L^{- 1} s^{- 1}$ the rate of appearance of ${Br}_{2}$ will be

Solution:

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