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The oxidation of \(\mathrm{Br}^{-}\) by \(\mathrm{BrO}_{3}^{-}\) in acidic solution is described by the equation \(5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)\) The reaction is first order in \(\mathrm{Br}^{-}\), first order in \(\mathrm{BrO}_{3}^{-}\), and second order in \(\mathrm{H}^{+}\) (a) Write the rate law. (b) What is the overall reaction order? (c) How does the reaction rate change if the \(\mathrm{H}^{+}\) concentration is tripled? (d) What is the change in rate if the concentrations of both \(\mathrm{Br}^{-}\) and \(\mathrm{BrO}_{3}^{-}\) are halved?

Short Answer

Expert verified
(a) Rate = k[Br鈦籡[BrO鈧冣伝][H鈦篯虏; (b) Order is 4; (c) Rate increases by 9 times; (d) Rate decreases to 1/4.

Step by step solution

01

Write the rate law

To write the rate law for the reaction, use the information given about the order of reaction with respect to each reactant. The rate law expresses the reaction rate as a function of the concentration of reactants, each raised to the power of their respective reaction order. Based on the problem, the rate law is: \[\text{Rate} = k [\mathrm{Br}^{-}]^1 [\mathrm{BrO}_3^{-}]^1 [\mathrm{H}^{+}]^2\] Where \( k \) is the rate constant.
02

Determine Overall Reaction Order

The overall reaction order is the sum of the orders of the reaction with respect to each reactant. Here, the order is:- 1 for \( \mathrm{Br}^{-} \)- 1 for \( \mathrm{BrO}_3^{-} \)- 2 for \( \mathrm{H}^{+} \)Therefore, the overall reaction order is:\[1 + 1 + 2 = 4\]
03

Effect of Tripling [H鈦篯 Concentration

When the concentration of \( \mathrm{H}^{+} \) is tripled, the effect on the rate can be determined using the order of reaction with respect to \( \mathrm{H}^{+} \), which is 2. The rate is proportional to \( [\mathrm{H}^{+}]^2 \), therefore:If \( [\mathrm{H}^{+}] \) is tripled, \[\text{New rate} = 3^2 = 9 imes \text{original rate}\]So the reaction rate increases by a factor of 9.
04

Effect of Halving [Br鈦籡 and [BrO鈧冣伝] Concentrations

When the concentrations of both \( \mathrm{Br}^{-} \) and \( \mathrm{BrO}_3^{-} \) are halved, the effect on the rate can be determined from their respective orders (1 for both). If the concentration of each is halved, their contributions to the rate are:\[\text{New rate} = \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^1 = \frac{1}{4} \times \text{original rate}\]Thus, the reaction rate decreases to a quarter of the original rate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law
A rate law is an expression that describes how the rate of a reaction depends on the concentration of its reactants. It is derived from experimental data and tells us the relationship between the concentration and reaction rate. In our oxidation reaction example, the rate law can be written based on the reaction orders provided:
  • First order in \( \mathrm{Br}^{-} \)
  • First order in \( \mathrm{BrO}_3^{-} \)
  • Second order in \( \mathrm{H}^{+} \)
So, the rate law is: \[ \text{Rate} = k [\mathrm{Br}^{-}]^1 [\mathrm{BrO}_3^{-}]^1 [\mathrm{H}^{+}]^2 \]Here, \( k \) is the rate constant, a unique value for each reaction that changes with factors like temperature. Determining the exponents or orders is crucial for constructing the proper rate law. It allows chemists to predict how changes in concentration will affect the speed of the reaction.
Reaction Order
The reaction order provides us with the sum of the powers to which reactant concentrations are raised in the rate law. This value can influence how significantly a reaction's rate changes with varying concentrations of reactants. In our provided example, the reaction orders are:
  • First order for \( \mathrm{Br}^{-} \)
  • First order for \( \mathrm{BrO}_3^{-} \)
  • Second order for \( \mathrm{H}^{+} \)
The overall reaction order is calculated by summing these individual orders: 1 + 1 + 2 = 4. Knowing the overall reaction order helps in identifying the relationship between reactant concentrations and the reaction rate as a whole. If a reaction is of higher overall order, it typically means the reaction rate is more sensitive to concentration changes.
Concentration Effect
Concentration effect refers to how changes in reactant concentrations can impact the rate of a chemical reaction. According to the rate law, each reactant's contribution is determined by its order. Here's how different concentration changes affect the rate:
  • Tripling the concentration of \( \mathrm{H}^{+} \) which is second order, increases the rate by a factor of 9. This is calculated as \((3^2 = 9)\).
  • Halving the concentrations of both \( \mathrm{Br}^{-} \) and \( \mathrm{BrO}_3^{-} \) (both first order) leads to the rate decreasing by 75%. This is seen as \(\left( \frac{1}{2} \right)^1 \times \left( \frac{1}{2} \right)^1 = \frac{1}{4}\).
Changes in concentration give a clear picture of how dynamically the reaction rate can respond, allowing targeted adjustments to speed up or slow down the reaction in practical scenarios.
Oxidation Reaction
An oxidation reaction is a type of chemical reaction where an element or compound loses electrons. In our example equation, \( \mathrm{Br}^{-} \) ions are oxidized to \( \mathrm{Br}_2 \) as they lose electrons. This process takes place in acidic conditions, which is common because the presence of protons (\( \mathrm{H}^{+} \)) often facilitates the redox reaction. Key characteristics of oxidation reactions include:
  • Involvement of electron transfer.
  • Often accompanied by changes in oxidation states.
  • Frequently involve coordination with other reactions, such as reduction.
Learning about oxidation reactions is crucial due to their prevalence in industrial applications and biological processes. Understanding the behavior of reactants in these reactions can aid in designing better chemical processes and understanding biological systems.

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Most popular questions from this chapter

Beginning with the integrated rate law, derive a general equation for the half-life of a zeroth-order reaction of the type \(\mathrm{A} \rightarrow\) Products. How does the length of each half-life compare with the length of the previous one? Make the same comparison for firstorder and second-order reactions.

Consider the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g) .\) The reaction of a fixed amount of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) is studied in a cylinder fitted with a movable piston. Indicate the effect of each of the following changes on the rate of the reaction: (a) An increase in temperature at constant volume (b) An increase in volume at constant temperature (c) The addition of a catalyst (d) The addition of argon (an inert gas) at constant volume

Ammonia is manufactured in large amounts by the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ (a) How is the rate of consumption of \(H_{2}\) related to the rate of consumption of \(\mathrm{N}_{2}\) ? (b) How is the rate of formation of \(\mathrm{NH}_{3}\) related to the rate of consumption of \(\mathrm{N}_{2} ?\)

When the temperature of a gas is raised by \(10{ }^{\circ} \mathrm{C}\), the collision frequency increases by only about \(2 \%\), but the reaction rate increases by \(100 \%\) (a factor of 2 ) or more. Explain.

The reaction \(\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)\) occurs in one step. The activation energy is \(132 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta \mathrm{E}\) is \(-226 \mathrm{~kJ} / \mathrm{mol}\). (a) Is the reaction endothermic or exothermic? (b) What is the activation energy for the reverse reaction? (c) Does the reaction rate increase or decrease when temperature increases? Explain.

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