Question

In the following reaction, how is the rate of appear ance of the underlined product related to the rate of disappearance of the underlined reactant? \(\mathrm{BrO}_{3}^{-}(\mathrm{aq})+5 \mathrm{Br}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow 3 \mathrm{Br}_{2}\) (I) \(+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}\) (b) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (c) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (d) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\)

Short answer

The correct relation is option (d), \( \frac{\mathrm{d}[\mathrm{Br}_{2}]}{\mathrm{dt}} = -\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}} \).

Step-by-step solution

Understand the Stoichiometry

The balanced chemical equation is given as follows: \[ \mathrm{BrO}_{3}^{-} + 5 \mathrm{Br}^{-} + 6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_{2} + 3 \mathrm{H}_{2} \mathrm{O} \] This equation tells us that 5 moles of \( \mathrm{Br}^{-} \) are consumed for every 3 moles of \( \mathrm{Br}_{2} \) produced.

Relate Reaction Rates

In general, the rate of disappearance of a substance in a reaction is related to the rate of appearance of another by their stoichiometric coefficients. So, for every 5 moles of \( \mathrm{Br}^{-} \) that disappear, 3 moles of \( \mathrm{Br}_{2} \) appear. This implies the rates can be related by: \[ \frac{1}{3} \frac{\mathrm{d}[\mathrm{Br}_{2}]}{\mathrm{dt}} = \frac{1}{5} \left| \frac{\mathrm{d}[\mathrm{Br}^{-}]}{\mathrm{dt}} \right| \] Therefore, we get: \[ \frac{\mathrm{d}[\mathrm{Br}_{2}]}{\mathrm{dt}} = -\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}} \]

Verify Against Choices

By comparing our derived relationship to the given options, we find that option (d) is correct: \( \frac{\mathrm{d}[\mathrm{Br}_{2}]}{\mathrm{dt}} = -\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}} \). The negative sign indicates that \( \mathrm{Br}^{-} \) is being used up, hence its concentration decreases over time while \( \mathrm{Br}_{2} \) forms and its concentration increases.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry, focusing on the quantitative relationships between the reactants and products in a chemical reaction. These relationships are determined by the balanced chemical equation, which shows how many moles of each substance are involved.

• The balanced equation for the reaction between bromate ions \((\mathrm{BrO}_3^-)\) and bromide ions \((\mathrm{Br}^-)\) with protons \((\mathrm{H}^+)\) is used to derive these relationships.
• It tells us that 5 moles of bromide ions react with 1 mole of bromate to produce 3 moles of bromine \((\mathrm{Br}_2)\).

Stoichiometry allows us to predict the amount of products formed, such as \(\mathrm{Br}_2\), based on the amount of reactants consumed, like \(\mathrm{Br}^-\). This understanding is critical for calculating the reaction rates discussed in the original exercise.

Rate Law

Rate laws describe the relationship between the concentration of reactants and the speed of the reaction—how fast or slow a reaction proceeds.

• The rate law for a chemical reaction is determined experimentally and can differ from the stoichiometric coefficients of the balanced equation.
• It often takes the form of an expression: \(\text{Rate} = k [A]^m [B]^n\) where \(k\) is the rate constant, and \(m\) and \(n\) are the reaction orders with respect to each reactant.

In our reaction, we are specifically interested in how the disappearance rate of \(\mathrm{Br}^-\) compares with the appearance rate of \(\mathrm{Br}_2\). The rate law helps chemists understand these relationships, leading to insights on how changing reactant concentrations can speed up or slow down the reaction.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the steps involved in chemical reactions. It helps describe not only how quickly a reaction occurs but what factors affect this speed.

• Kinetics involves understanding the transition state, reaction intermediates, and the energy changes that occur during a reaction.
• This field also explores how conditions like concentration, temperature, and catalysts influence rates.

For our reaction, kinetics shows us that the absence or presence of \(\mathrm{Br}^-\) drastically changes the formation rate of bromine \((\mathrm{Br}_2)\). Understanding these concepts ensures better control and efficiency in chemical processes and reactions.

Balanced Chemical Equations

Balanced chemical equations are crucial as they lie at the heart of understanding chemical reactions.

• These equations ensure that the law of conservation of mass is respected, with equal numbers of each type of atom on both sides of an equation.
• In our exercise, this balance allows us to deduce that 5 moles of \(\mathrm{Br}^-\) are used to produce 3 moles of \(\mathrm{Br}_2\).

The coefficients derived from a balanced equation help in determining the stoichiometric ratios. This balance gives us a clear picture of the reactant-to-product conversion, vital when calculating rates and predicting reaction outcomes. These fundamental stoichiometric concepts underlie the entire study of chemical reactions, aiding in both theoretical and practical aspects of chemistry.