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Chapter 14: Problem 88

Consider two reactions. Reaction (1) has a constant halflife, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

Short Answer

Expert verified
From the given observations, we can conclude that Reaction (1) follows a first-order rate law with Rate = k[A], while Reaction (2) follows a second-order rate law with Rate = k[A]^2. This is based on the half-life characteristics of these reactions - a constant half-life for first-order reactions and an increasing half-life for second-order reactions.

Step by step solution

01

Understand half-life and rate law relationship

Half-life is defined as the time in which half of the reactant is consumed during a reaction. The relation between the rate law and half-life can provide insights into the reaction order. For example, if the half-life is constant, we can conclude that the reaction is of first order, while a varying half-life may indicate a second-order or higher-order reaction.
02

Analyzing Reaction (1)

The first reaction has a constant half-life. This behavior is characteristic of a first-order reaction. In first-order reactions, the rate of the reaction is directly proportional to the reactant's concentration, with a rate law in the form: Rate = k[A], where k is the rate constant, and [A] represents the concentration of the reactant A.
03

Analyzing Reaction (2)

The second reaction has a half-life that increases as the reaction proceeds. This behavior is not typical for a first-order reaction since they have a constant half-life. However, it is common for second-order reactions. In second-order reactions, the rate of the reaction depends on the concentration of the reactants squared, with a rate law in the form: Rate = k[A]^2.
04

Conclusions

From the given observations, we can conclude that Reaction (1) follows a first-order rate law, while Reaction (2) follows a second-order rate law. These conclusions are based on the half-life characteristics of these reactions - a constant half-life for first-order reactions and an increasing half-life for second-order reactions.

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Key Concepts

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

half-life
Half-life is a crucial concept in the study of chemical reactions, specifically relating to how quickly reactants are consumed. Imagine you have a certain quantity of a reactant. The half-life is the time it takes for exactly half of that original amount to react. It's a useful metric because it remains consistent for any sample size.

Understanding half-life helps in determining the order of a reaction. For example:
  • If a reaction has a constant half-life, it typically suggests a first-order reaction. This means no matter how much reactant you start with, the time it takes for half of it to remain will always be the same.
  • If the half-life of a reaction increases over time, it might indicate a second-order reaction. In this case, the half-life gets longer as the reactant concentration decreases because the reaction rate slows down more significantly as the reactants are used up.
Recognizing these patterns in half-life can be key to identifying reaction behaviors and understanding the underlying rate laws.
first-order reaction
A first-order reaction is unique because its reaction rate is directly proportional to the concentration of one reactant. This relationship results in several notable features:

  • The reaction rate decreases as the reactant is consumed, but the half-life stays constant, irrespective of the initial concentration.
  • The rate law for a first-order reaction is often expressed as: \[ ext{Rate} = k[A] \]where \( k \) is the rate constant and \( [A] \) represents the molar concentration of reactant A.
  • To determine the half-life for a first-order reaction, a simple formula is used: \[ t_{1/2} = \frac{0.693}{k} \]This tells us that the half-life is inversely related to the rate constant \( k \), meaning a higher rate constant results in a shorter half-life.

These characteristics make first-order reactions relatively predictable and are common in many natural and industrial processes.
second-order reaction
When dealing with second-order reactions, you encounter a situation where the rate of the reaction is dependent on the concentration of one or more reactants squared. Let's dive into the distinct features of these reactions:

  • The rate law for a second-order reaction can be represented as:\[ ext{Rate} = k[A]^2 \]for a reaction that only involves one reactant.
  • As the reaction progresses, the reactant concentration decreases more significantly, leading to a longer half-life. This is because, unlike first-order reactions, the rate slows down more notably as the reactants are consumed.
  • The half-life for a second-order reaction depends inversely on the initial concentration of reactants, expressed by:\[ t_{1/2} = \frac{1}{k[A_0]} \]where \( [A_0] \) is the initial concentration. Notably, as \( [A_0] \) decreases, the half-life increases, reflecting how reactants deplete slower over time.

These characteristics provide insights into the dynamic nature of second-order reactions, influencing how they are studied and applied.
rate law
The rate law is a mathematical expression that helps us understand the speed at which a chemical reaction occurs. It is determined experimentally and tells us how the concentration of reactants affects the reaction rate. Let's break it down:

  • The rate law takes the form: \[ ext{Rate} = k[A]^m[B]^n \]Here, \( k \) is the rate constant, \( [A] \) and \( [B] \) are the concentrations of reactants, and \( m \) and \( n \) are the reaction orders with respect to each reactant.
  • The overall reaction order is the sum of the exponents \( m \) and \( n \). This overall order highlights how various concentrations affect the reaction speed.
  • Rate laws vary for each reaction and must be determined from experimental data, as they provide deeper insights into the mechanism of the reaction.
Understanding the rate law is essential for predicting how reactions progress under different conditions, aiding in both theoretical studies and practical applications.

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Most popular questions from this chapter

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M}\), while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

Consider the reaction of peroxydisulfate ion $\left(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right)\( with iodide ion (I \)^{-}$ ) in aqueous solution: $$ \mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}(a q)+3 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{SO}_{4}{ }^{2-}(a q)+\mathrm{I}_{3}^{-}(a q) $$ At a particular temperature the rate of disappearance of $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}$ varies with reactant concentrations in the following manner: \begin{tabular}{llll} \hline & & & Initial Rate \\ Experiment & {\(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right](M)\)} & {\(\left[\mathrm{I}^{-}\right](M)\)} & \((M / s)\) \\ \hline 1 & \(0.018\) & \(0.036\) & \(2.6 \times 10^{-6}\) \\ 2 & \(0.027\) & \(0.036\) & \(3.9 \times 10^{-6}\) \\ 3 & \(0.036\) & \(0.054\) & \(7.8 \times 10^{-6}\) \\ 4 & \(0.050\) & \(0.072\) & \(1.4 \times 10^{-5}\) \\ \hline \end{tabular} (a) Determine the rate law for the reaction. (b) What is the average value of the rate constant for the disappearance of $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}$ based on the four sets of data? (c) How is the rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\) related to the rate of disappearance of \(I^{-} ?(\mathrm{~d})\) What is the rate of disappearance of I when \(\left[\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right]=0.025 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.050 \mathrm{M} ?\)

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00\) L of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{~h}\) if the gas is collected in a 10.0-L container. (Assume that the products do not dissolve in chloroform.)

Metals often form several cations with different charges. Cerium, for example, forms \(\mathrm{Ce}^{3+}\) and \(\mathrm{Ce}^{4+}\) ions, and thallium forms \(\mathrm{Tl}^{+}\) and \(\mathrm{Tl}^{3+}\) ions. Cerium and thallium ions react as follows: $$ 2 \mathrm{Ce}^{4+}(a q)+\mathrm{Tl}^{+}(a q) \longrightarrow 2 \mathrm{Ce}^{3+}(a q)+\mathrm{Tl}^{3+}(a q) $$ This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of \(\mathrm{Mn}^{2+}(a q)\), according to the following mechanism: $$ \begin{aligned} \mathrm{Ce}^{4+}(a q)+\mathrm{Mn}^{2+}(a q) & \longrightarrow \mathrm{Ce}^{3+}(a q)+\mathrm{Mn}^{3+}(a q) \\ \mathrm{Ce}^{4+}(a q)+\mathrm{Mn}^{3+}(a q) & \longrightarrow \mathrm{Ce}^{3+}(a q)+\mathrm{Mn}^{4+}(a q) \\ \mathrm{Mn}^{4+}(a q)+\mathrm{Tl}^{+}(a q) & \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Tl}^{3+}(a q) \end{aligned} $$ (a) Write the rate law for the uncatalyzed reaction. (b) What is unusual about the uncatalyzed reaction? Why might it be a slow reaction? (c) The rate for the catalyzed reaction is first order in \(\left[\mathrm{Ce}^{4+}\right]\) and first order in \(\left[\mathrm{Mn}^{2+}\right]\). Based on this rate law, which of the steps in the catalyzed mechanism is rate determining? (d) Use the available oxidation states of \(\mathrm{Mn}\) to comment on its special suitability to catalyze this reaction.

Consider the following reaction between mercury(II) chloride and oxalate ion: \(2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q)\) \(2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ? $$ \begin{array}{llll} \text { Experiment } & {\left[\mathrm{HgCl}_{2} \mathrm{~J}(\mathrm{M})\right.} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right](M)} & \text { Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2}\right)\) is \(0.25 \mathrm{M}\), if the temperature is the same as that used to obtain the data shown?
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