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What are the units for the rate constants of first-order and second-order reactions?

Short Answer

Expert verified
First-order: \( \text{s}^{-1} \), Second-order: \( \text{L·mol}^{-1}\text{s}^{-1} \).

Step by step solution

01

Understand Reaction Order

A first-order reaction depends on the concentration of a single reactant. A second-order reaction can depend on the concentration of two reactants or the concentration of one reactant squared.
02

Rate Law for First-Order Reaction

For a first-order reaction, the rate law is given by \( \text{Rate} = k[A] \), where \([A]\) is the concentration of the reactant and has units of mol/L (molarity).
03

Determine Units for First-Order Rate Constant

Rearrange the rate law for the first-order reaction to solve for the rate constant \( k \): \( k = \frac{\text{Rate}}{[A]} \). The rate of reaction has units of mol/(L·s), so \( k \) must have units of \( \text{s}^{-1} \) to balance the units.
04

Rate Law for Second-Order Reaction

For a second-order reaction, the rate law is \( \text{Rate} = k[A]^2 \) if depending on one reactant, or \( \text{Rate} = k[A][B] \) if depending on two different reactants.
05

Determine Units for Second-Order Rate Constant

Using \( \text{Rate} = k[A]^2 \), rearrange to solve for rate constant: \( k = \frac{\text{Rate}}{[A]^2} \). The rate of reaction has units of mol/(L·s) and the concentration squared is mol²/L², so \( k \) for a second-order reaction is \( \text{L·mol}^{-1}\text{s}^{-1} \).

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Key Concepts

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

Rate Constants
In the world of chemical kinetics, the rate constant is a fundamental factor. It symbolizes the speed of a reaction at a given temperature and is represented by the letter "k."
  • A high rate constant indicates a fast reaction.
  • A low rate constant indicates a slow reaction.
In a reaction's rate equation, the rate constant connects the reaction rate with the reactant concentrations. However, the units of a rate constant vary depending on the order of the reaction. Understanding these units is vital for interpreting how quickly reactants turn into products under certain conditions. By examining units, chemists can also infer the reaction order and gain insights into reaction mechanisms.
First-Order Reactions
First-order reactions have a direct proportionality between the rate of the reaction and the concentration of a single reactant. This means if you double the concentration of the reactant, the rate of the reaction will also double.The rate law for a first-order reaction takes the form:\[ \text{Rate} = k[A] \]Here, "[A]" represents the concentration of the reactant, with units of molarity (M) or mol/L. Since the rate has units of mol/(L·s), the rate constant \( k \) for a first-order reaction must have units of \( \,s^{-1} \) to ensure dimensional consistency.
Second-Order Reactions
Second-order reactions can depend on either the concentration of two different reactants or on one reactant squared. This involves quite fascinating nuances in how concentration changes affect reaction rates.The general form of the rate law for a second-order reaction can be either:\[ \text{Rate} = k[A]^2 \]or\[ \text{Rate} = k[A][B] \]In both expressions, the rate refers to the product formed over time and shares the same unit of mol/(L·s). For second-order reactions, the rate constant \( k \) must balance the equation:- For a single reactant: \( \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1} \)- For two different reactants: The same units, as they multiply similar concentration terms.
Units of Measurement
Understanding units of measurement in chemical kinetics is crucial for any chemistry student.
  • For first-order reactions, the rate constant has units of \( s^{-1} \).
  • For second-order reactions, it's \( \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1} \).
Each unit tells us about how reactant concentration impacts the rate. The time element (seconds in these units) helps illustrate the temporal aspect of how fast reactants convert to products.
Rate Law
The rate law is a mathematical equation that expresses the connection between the rate of a chemical reaction and the concentration of its reactants. It is essential to deciphering reaction dynamics.In a typical rate law:\[ \text{Rate} = k[A]^m[B]^n \]- \( [A] \) and \( [B] \) are the molar concentrations of reactants.- \( m \) and \( n \) depict the reaction order concerning each reactant.By fitting experimental data to this form, chemists can uncover both the reaction order and rate constant. This forms the gateway to deeper insights into the mechanism driving the transformation of reactants to products. Understanding and deriving the rate law is central to predicting how changes in concentration affect the pace at which a reaction proceeds.

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Most popular questions from this chapter

The second-order rate constant for the dimerization of a protein (P) \(\mathrm{P}+\mathrm{P} \longrightarrow \mathrm{P}_{2}\) is \(6.2 \times 10^{-3} / M \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). If the concentration of the protein is \(2.7 \times 10^{-4} M,\) calculate the initial rate \((M / \mathrm{s})\) of formation of \(\mathrm{P}_{2}\). How long (in seconds) will it take to decrease the concentration of \(\mathrm{P}\) to \(2.7 \times 10^{-5} \mathrm{M}\) ?

(a) Consider two reactions, \(\mathrm{A}\) and \(\mathrm{B}\). If the rate constant for reaction B increases by a larger factor than that of reaction A when the temperature is increased from \(T_{1}\) to \(T_{2},\) what can you conclude about the relative values of the activation energies of the two reactions? (b) If a bimolecular reaction occurs every time an \(\mathrm{A}\) and a \(\mathrm{B}\) molecule collide, what can you say about the orientation factor and activation energy of the reaction?

When \(6 \mathrm{~g}\) of granulated \(\mathrm{Zn}\) is added to a solution of \(2 \mathrm{M}\) \(\mathrm{HCl}\) in a beaker at room temperature, hydrogen gas is generated. For each of the following changes (at constant volume of the acid) state whether the rate of hydrogen gas evolution will be increased, decreased, or unchanged: (a) \(6 \mathrm{~g}\) of powdered \(\mathrm{Zn}\) is used, (b) \(4 \mathrm{~g}\) of granulated \(\mathrm{Zn}\) is used, (c) \(2 M\) acetic acid is used instead of \(2 M \mathrm{HCl}\), (d) temperature is raised to \(40^{\circ} \mathrm{C}\).

The rate constant for the second-order reaction: $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after \(22 \mathrm{~s}\). (b) Calculate the half-lives when [NOBr] \(_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\)

The thermal decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) obeys first-order kinetics. At \(45^{\circ} \mathrm{C}\), a plot of \(\ln \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) versus \(t\) gives a slope of \(-6.18 \times 10^{-4} \mathrm{~min}^{-1}\). What is the half-life of the reaction?

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