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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

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form NOF and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(A=6.0 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the structure for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

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 \(0.010 \mathrm{M}\) ?

The activation energy of a certain reaction is \(65.7 \mathrm{~kJ} / \mathrm{mol}\) How many times faster will the reaction occur at \(50^{\circ} \mathrm{C}\) than at \(0^{\circ} \mathrm{C}\) ?

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after \(5.12 \mathrm{~s}\) assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

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