Question

For the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}$$ it was found that at a particular instant bromine was being formed at the rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). At that instant, at what rate is (a) water being formed? (b) bromide ion being oxidized? (c) \(\mathrm{H}^{+}\) being consumed?

Short answer

Question: At a particular moment in a reaction, the rate of bromine formation is 0.039 mol/L⋅s. Determine the rates of formation or consumption of water, bromide ion, and hydrogen ion at that instant. Answer: At that particular instant, water is being formed at a rate of 0.039 mol/L⋅s, bromide ions are being oxidized at a rate of 0.065 mol/L⋅s, and hydrogen ions are being consumed at a rate of 0.078 mol/L⋅s.

Step-by-step solution

a) Rate of water formation

To determine the rate of water formation, we need to compare the stoichiometric coefficients of water and bromine in the balanced equation. We can see that the coefficients are 3 and 3, respectively. Since the rate of bromine formation has been given as \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), the rate of water formation would be the same since their coefficients are equal. Therefore, the rate of water formation is: $$0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$

b) Rate of bromide ion oxidation

To find the rate at which bromide ions are being oxidized, we need to compare the stoichiometric coefficients of bromide ions and bromine in the balanced equation. We can see that they are 5 and 3, respectively. To find the rate of bromide ion oxidation, we can set up a proportion: $$\frac{\text{Rate of Br}^{-}}{\text{Rate of Br}_{2}} = \frac{5}{3}$$ Therefore, the rate of bromide ion oxidation is: $$\text{Rate of Br}^{-} = \frac{5}{3} \times 0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.065 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$

c) Rate of H+ ion consumption

To determine the rate at which H+ ions are being consumed, we need to compare the stoichiometric coefficients of H+ ions and bromine in the balanced equation. We can see that they are 6 and 3, respectively. To find the rate of H+ ion consumption, we can set up a proportion: $$\frac{\text{Rate of H}^{+}}{\text{Rate of Br}_{2}} = \frac{6}{3}$$ Therefore, the rate of H+ ion consumption is: $$\text{Rate of H}^{+} = 2 \times 0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.078 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$ In summary, at that particular instant, water is being formed at a rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), bromide ions are being oxidized at a rate of \(0.065 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), and H+ ions are being consumed at a rate of \(0.078 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\).

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation, which shows the proportionate relationship between reactants and products.
In the given reaction:

• 5 \( \text{Br}^- \)(aq) + \( \text{BrO}_3^- \)(aq) + 6 \( \text{H}^+ \)(aq) \( \rightarrow 3 \text{Br}_2 \)(aq) + 3 \( \text{H}_2\text{O} \)

The coefficients in front of each compound represent the stoichiometric ratios. For example, the 5:3 ratio between \( \text{Br}^- \) and \( \text{Br}_2 \) tells us that 5 moles of bromide ions are needed to produce 3 moles of bromine. Using stoichiometry allows us to calculate the reaction rates of different substances in the equation.
This concept is crucial for determining the rate of formation or consumption of each substance in a chemical reaction, as seen in how you calculate the rate of bromide ion oxidation and other compounds in the given exercise.

Rate Laws

Rate laws help us understand the speed of a chemical reaction. They provide a mathematically expressed relationship between the concentrations of reactants and the reaction rate.
In a typical rate law, the rate of a reaction is often expressed as:

• \( \text{Rate} = k[A]^m[B]^n \)

where \( k \) is the rate constant, and \([A]^{m}\) and \([B]^{n}\) represent the concentrations of the reactants raised to their respective powers, which are the orders of the reaction.
In the context of stoichiometry, while rate laws are determined experimentally and not directly from the stoichiometric coefficients, they help us validate our calculated rates of transformation, like those derived from the coefficients of the balanced equation in the exercise.
Such calculations ensure that our understanding of how quickly a substance is formed or consumed aligns with practical observations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They are described by balanced equations that show all reactants and the resulting products.
For example, in the given reaction:

• 5 \( \text{Br}^- \)(aq) + \( \text{BrO}_3^- \)(aq) + 6 \( \text{H}^+ \)(aq) \( \rightarrow 3 \text{Br}_2 \)(aq) + 3 \( \text{H}_2\text{O} \)

This balanced equation tells us how many units of each reactant are needed and how many units of product are formed. The process involves breaking and forming bonds, and the coefficients describe this quantitative transformation.
Understanding chemical reactions also involves recognizing the conditions under which the reaction occurs, such as in the solution (noted as \( \text{aq} \) for aqueous).
Furthermore, knowing the role of catalysts and the energy changes involved can deepen the understanding of reaction dynamics, although these are not directly explored in this exercise. Through these principles, chemical reactions reveal the fundamental processes that sustain both natural and synthetic chemical transformations.