Chapter 15: Problem 17
Explain the meaning of the orientation factor in the collision model.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 15: Problem 17
Explain the meaning of the orientation factor in the collision model.
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Why is the reaction rate for reactants defined as the negative of the change in reactant concentration with respect to time, whereas for products it is defined as the change in reactant concentration with re- spect to time (with a positive sign)?
Using the idea that reactions occur as a result of collisions between particles, explain why reaction rates depend on the temperature of the reaction mixture.
Explain the difference between the rate law for a reaction and the in- tegrated rate law for a reaction. What relationship does each kind of rate law express?
The first-order integrated rate law for a reaction \(\mathrm{A} \longrightarrow\) products is derived from the rate law using calculus. \begin{equation} \begin{aligned} \text { Rate } &=k[\mathrm{A}] \quad \text { (first-order rate law) } \\ \text { Rate } &=\frac{d[\mathrm{A}]}{d t} \\\ \frac{d[\mathrm{A}]}{d t} &=-k[\mathrm{A}] \end{aligned} \end{equation} The equation just given is a first-order, separable differential equa- tion that can be solved by separating the variables and integrating: \begin{equation} \begin{aligned} \frac{d[\mathrm{A}]}{[\mathrm{A}]} &=-k d t \\\ \int_{[\mathrm{A}]_{0}}^{[\mathrm{A]}} \frac{d[\mathrm{A}]}{[\mathrm{A}]} &=-\int_{0}^{t} k d t \end{aligned} \end{equation} In the integral just given, \([\mathrm{A}]_{0}\) is the initial concentration of \(\mathrm{A} . \mathrm{We}\) then evaluate the integral: \begin{equation} \begin{aligned}[\ln [\mathrm{A}]]_{[\mathrm{A}]_{0}}^{[\mathrm{Al}} &=-k[t]_{0}^{t} \\ \ln [\mathrm{A}]-\ln [\mathrm{A}]_{0} &=-k t \end{aligned} \end{equation} \begin{equation} \ln [\mathrm{A}]=-k t+\ln [\mathrm{A}]_{0}(\text { integrated rate law }) \end{equation} \begin{equation} \begin{array}{l}{\text { a. Use a procedure similar to the one just shown to derive an inte- }} \\ {\text { grated rate law for a reaction } A \longrightarrow \text { products, which is one-half- }} \\ {\text { order in the concentration of } A \text { (that is, Rate }=k[A]^{1 / 2} )}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Use the result from part a to derive an expression for the half-life }} \\ {\text { of a one-half-order reaction. }}\end{array} \end{equation}
\(\begin{aligned} \text { Consider the reaction. } \\ & 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned}\) \begin{equation} \begin{array}{l}{\text { a. Express the rate of the reaction in terms of the change in }} \\ {\text { concentration of each of the reactants and products. }} \\ {\text { b. In the first } 15.0 \text { s of the reaction, 0.015 mol of } \mathrm{O}_{2} \text { is produced in a }} \\ {\text { reaction vessel with a volume of } 0.500 \text { L. What is the average rate }} \\ {\text { of the reaction during this time interval? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Predict the rate of change in the concentration of } \mathrm{N}_{2} \mathrm{O} \text { during this }} \\ {\text { time interval. In other words, what is } \Delta\left[\mathrm{N}_{2} \mathrm{O}\right] / \Delta t?}\end{array} \end{equation}
What do you think about this solution?
We value your feedback to improve our textbook solutions.