Give two reasons to measure initial rates in a kinetics study.

Short Answer

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1. Avoids complications from product accumulation.2. Determines reaction order with respect to reactants accurately.

Step by step solution

01

- Understand Initial Rates

Initial rates refer to the reaction rates measured at the very beginning of a reaction, where the concentrations of reactants have not significantly changed.
02

- Reason 1: Avoiding Complications

Measuring initial rates helps avoid complications that arise from product accumulation and possible reverse reactions, ensuring a more straightforward interpretation of the reaction kinetics.
03

- Reason 2: Determining Reaction Order

Initial rates allow for the determination of the reaction order with respect to each reactant, as they provide a clear relationship between reactant concentration and rate without interference from changing conditions or side reactions.

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Key Concepts

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

Reaction Kinetics
Reaction kinetics is the study of how fast chemical reactions occur and the various factors that influence these rates. It provides crucial insights into the mechanisms of reactions, helping chemists understand the step-by-step process that leads to products. At the heart of this study are reaction rates, which are measured as the change in concentration of reactants or products per unit time. By analyzing these rates, scientists can deduce important information about the energy barriers and molecular steps involved.
Factors influencing reaction rates include:
  • Concentration of reactants
  • Temperature
  • Presence of catalysts
  • Surface area of solid reactants
Understanding these factors allows chemists to control and optimize reactions, making reaction kinetics a fundamental area of study in chemistry.
Initial Reaction Rates
Initial reaction rates refer to the rates measured at the beginning of a reaction, generally just after the reactants are mixed. This early phase is critical because it provides a simplified view of the reaction without the complications that arise later.
Measuring initial reaction rates has several advantages:
  • It avoids complications from product accumulation, which can sometimes interfere with accurate measurements.
  • There is less chance for reverse reactions to occur, ensuring that the data reflects only the forward reaction.
  • Concentration changes are minimal, so the rate reflects the initial concentrations of the reactants more accurately.
Because of these reasons, initial rates are often used to simplify the study of reaction kinetics and to obtain clearer data.
Reaction Order Determination
Determining the reaction order is a crucial part of studying reaction kinetics. The reaction order with respect to a reactant is an exponent to which its concentration term in the rate equation is raised. It provides insight into how changes in concentration affect the rate of reaction.
Using initial rates to determine reaction order involves:
  • Measuring the initial rate of reaction at different concentrations of the reactants.
  • Analyzing how the rate changes as the concentration of each reactant is varied.
  • Using this data to deduce the reaction order for each reactant via mathematical relationships.
For example, if doubling the concentration of a reactant doubles the rate, the reaction is first-order with respect to that reactant. If it quadruples the rate, the reaction is second-order. Knowing the reaction order helps chemists to write the rate law and understand the mechanisms behind the reaction.

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Most popular questions from this chapter

If a slow step precedes a fast step in a two-step mechanism, do the substances in the fast step appear in the overall rate law? Explain.

16.89 A slightly bruised apple will rot extensively in about 4 days at room temperature \(\left(20^{\circ} \mathrm{C}\right)\). If it is kept in the refrigerator at \(0^{\circ} \mathrm{C}\), the same extent of rotting takes about 16 days. What is the activation energy for the rotting reaction?

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

Many drugs decompose in blood by a first-order process. (a) Two tablets of aspirin supply \(0.60 \mathrm{~g}\) of the active compound. After 30 min, this compound reaches a maximum concentration of \(2 \mathrm{mg} / 100 \mathrm{~mL}\) of blood. If the half-life for its breakdown is \(90 \mathrm{~min},\) what is its concentration (in \(\mathrm{mg} / 100 \mathrm{~mL}\) ) \(2.5 \mathrm{~h}\) after it reaches its maximum concentration? (b) For the decomposition of an antibiotic in a person with a normal temperature \(\left(98.6^{\circ} \mathrm{F}\right)\), \(k=3.1 \times 10^{-5} \mathrm{~s}^{-1} ;\) for a person with a fever (temperature of \(\left.101.9^{\circ} \mathrm{F}\right), k=3.9 \times 10^{-5} \mathrm{~s}^{-1}\). If the person with the fever must take another pill when \(\frac{2}{3}\) of the first pill has decomposed, how many hours should she wait to take a second pill? A third pill? (Assume that the pill is effective immediately.) (c) Calculate \(E_{\mathrm{a}}\) for decomposition of the antibiotic in part (b).

Reactions between certain haloalkanes (alkyl halides) and water produce alcohols. Consider the overall reaction for \(t\) -butyl bromide ( 2 -bromo- 2 -methylpropane): \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}(a q)+\mathrm{H}^{+}(a q)+\mathrm{Br}^{-}(a q)\) The experimental rate law is rate \(=k\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\right] .\) The accepted mechanism for the reaction is \((1)\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}(a q) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(a q)+\mathrm{Br}^{-}(a q) \quad[\mathrm{slow}]\) (2) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}_{2}^{+}(a q) \quad[\) fast\(]\) (3) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}_{2}^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}(a q) \quad[\) fast\(]\) (a) Why doesn't \(\mathrm{H}_{2} \mathrm{O}\) appear in the rate law? (b) Write rate laws for the elementary steps. (c) What reaction intermediates appear in the mechanism? (d) Show that the mechanism is consistent with the experimental rate law.

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