Question

For the reaction ${H^+ + BrO_3 ^- + 3Br^- -> 5Br+2 + H_2O}$ which of the following relation correctly represents the consumption & formation of reactants and products :

• A. $\frac{d [Br_2]}{dt} =-\frac{5}{3} \frac{d [Br^-]}{dt}$
• B. $\frac{d [Br_2]}{dt} =- \frac{d [Br^-]}{dt}$
• C. $\frac{d [Br_2]}{dt} =\frac{3}{5} \frac{d [Br^-]}{dt}$
• D. $\frac{d [Br_2]}{dt} =-\frac{3}{5} \frac{d [Br^-]}{dt}$

Answer 1

The Correct Option is D

To solve this problem, we need to apply the concept of reaction rates and stoichiometry in chemical reactions. The reaction given is:

{H^+ + BrO_3^- + 3Br^- \rightarrow 5Br_2 + H_2O}

The stoichiometric coefficients in a balanced chemical reaction represent the molar proportions of reactants and products. They help define the rates at which substances are consumed or produced.

The reaction we have is:

{H^+ + BrO_3^- + 3Br^- \rightarrow 5Br_2 + H_2O}

From the balanced equation, the stoichiometric coefficients are:

• For [Br^-]: Coefficient = 3
• For [Br_2]: Coefficient = 5

The rate of change in concentration of a substance is proportional to its stoichiometric coefficient. As per stoichiometry, for every 3 moles of Br^- consumed, 5 moles of Br_2 are produced.

Using the rate expression for the reaction, we express the rates of consumption and formation as:

• Change in concentration of Br^-: \frac{d [Br^-]}{dt}
• Change in concentration of Br_2: \frac{d [Br_2]}{dt}

According to the reaction stoichiometry:

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

However, this expression in terms of the rates should consider the direction (signs) of consumption and formation:

• Because Br^- is consumed: \frac{d [Br^-]}{dt}\ is negative.
• Because Br_2 is produced: \frac{d [Br_2]}{dt}\ is positive.

Thus, converting this to an expression relating their absolute values with correct signs:

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

Therefore, the correct option representing the relationship between the rates of change of Br^- and Br_2 is:

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

This option correctly accounts for the stoichiometric ratio as well as the signs indicating consumption and formation.