Question

Consider the following reaction in aqueous solution, \[ 5Br^- (aq) + BrO_3^- (aq) + 6H^+ (aq) \rightarrow 3Br_2 (aq) + 3H_2O (l) \] If the rate of appearance of Br\(_2\) at a particular moment during the reaction is \( 0.025 \, \mathrm{M \, s^{-1}} \), what is the rate of disappearance (in \( \mathrm{M \, s^{-1}} \)) of Br\(^- \) at that moment?

• A. 0.025 M s$^{-1}$
• B. 0.042 M s$^{-1}$
• C. 0.075 M s$^{-1}$
• D. 0.125 M s$^{-1}$

Hint:

Use coefficient ratio: multiply by $\frac{\text{reactant coeff}}{\text{product coeff}}$.

Answer 1

The Correct Option is B

Concept: Rate is proportional to stoichiometric coefficients: \[ \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} \]

Step 1: From reaction: 5Br$^-$ → 3Br$_2$.

Step 2: Relation: \[ \frac{1}{5}\left(-\frac{d[Br^-]}{dt}\right) = \frac{1}{3}\frac{d[Br_2]}{dt} \]

Step 3: Substitute given rate: \[ \frac{1}{5} \times \text{Rate of disappearance of Br}^- = \frac{1}{3} \times 0.025 \]

Step 4: Calculate: \[ \text{Rate of disappearance of Br}^- = \frac{5}{3} \times 0.025 = 0.0417 \approx 0.042 \] Conclusion:
Rate = 0.042 M s$^{-1}$