Question

In the reaction $BrO _{3}^{-}( aq )+5 Br ^{-}( aq )+6 H ^{+} \rightarrow 3 Br _{2}(l)+3 H _{2} O (l)$ The rate of appearance of bromine $\left( Br _{2}\right)$ 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. $\frac{d [Br_2]}{dt} = - \frac{5}{3} \frac{d [Br^-]}{dt}$
• C. $\frac{d [Br_2]}{dt} = \frac{5}{3} \frac{d [Br^-]}{dt}$
• D. $\frac{d [Br_2]}{dt} =\frac{3}{5} \frac{d [Br^-]}{dt}$

Answer 1

The Correct Option is A

The given reaction is:

BrO _{3}^{-}(aq) + 5 Br^{-}(aq) + 6 H^{+} \rightarrow 3 Br_{2}(l) + 3 H_{2}O(l)

In this reaction, bromide ions Br^{-} disappear, and bromine Br_{2} appears. We need to find the relationship between the rate of appearance of bromine and the rate of disappearance of bromide ions.

The stoichiometry of the reaction gives:

• For every 1 mole of BrO_3^{-} reacted, 5 moles of Br^{-} are consumed.
• For every 5 moles of Br^{-} consumed, 3 moles of Br_{2} are produced.

The rates of disappearance or appearance are related to their stoichiometric coefficients. Therefore, the rate formula can be written as:

\frac{1}{5} \frac{d[Br^{-}]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt}

Rewriting this equation, we determine:

\frac{d[Br_2]}{dt} = - \frac{3}{5} \frac{d[Br^{-}]}{dt}

The minus sign indicates that as bromide ions disappear, bromine is formed. Thus, the correct relationship is:

\frac{d[Br_2]}{dt} = - \frac{3}{5} \frac{d[Br^{-}]}{dt}