Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant?

Short Answer

Expert verified
(a) A successful chemical reaction between two colliding molecules depends on two factors: 1) having enough activation energy (Ea) to overcome the energy barrier and break necessary bonds in reactant molecules, and 2) correct orientation during the collision, which allows the reactive parts of the molecules to come into direct contact (steric factor). (b) According to the collision model, temperature affects the rate constant by increasing both the kinetic energy of molecules and the collision frequency. This results in more molecules having the required activation energy and higher chances of collisions with the correct orientation, leading to an increased rate constant.

Step by step solution

01

Factors determining a successful chemical reaction from molecular collision

To have a successful chemical reaction following a collision between two molecules, the following two conditions must be met: 1. The molecules must have enough energy, called the activation energy (Ea), to overcome the energy barrier and react. Activation energy is the minimum amount of energy needed to break the required bonds in the reactant molecules. 2. The molecules must be oriented correctly when they collide, such that the reactive parts of the molecules come into direct contact. This is called the steric factor.
02

Collision Model

The collision model is a theory which states that a chemical reaction can occur only when reactant molecules come into direct contact with sufficient energy and the correct orientation. This theory primarily depends on two factors: collision frequency (how often the molecules collide) and the fraction of collisions possessing the minimum activation energy (Ea) and correct orientation.
03

Temperature effect on the value of the rate constant

According to the collision model, increasing the temperature leads to an increase in the rate constant because temperature has two main effects on the reaction: 1. Increased temperature leads to an increase in the kinetic energy of the molecules. This means that more molecules will possess the required activation energy (Ea) to overcome the energy barrier and successfully react. The higher the temperature, the greater the fraction of molecules with enough energy to react. 2. Increased temperature also increases the collision frequency between molecules. When molecules have higher energy, they move faster and collide more frequently. Hence, the chances of having successful collisions with the correct orientation increase. These two factors combine to make the overall rate of the reaction much faster, increasing the value of the rate constant at high temperatures.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) From your everyday experience, give two examples of the effects of temperature on the rates of reactions. (c) What is the difference between average rate and instantaneous rate?

The following data were collected for the rate of disappearance of \(\mathrm{NO}\) in the reaction $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow$ \(2 \mathrm{NO}_{2}(g)\) \begin{tabular}{llll} \hline & & Initial Rate \\ Experiment & {\([\mathrm{NO}](M)\)} & {\(\left[\mathrm{O}_{2}\right](M)\)} & $(M / s)$ \\ \hline 1 & \(0.0126\) & \(0.0125\) & \(1.41 \times 10^{-2}\) \\ 2 & \(0.0252\) & \(0.0125\) & \(5.64 \times 10^{-2}\) \\ 3 & \(0.0252\) & \(0.0250\) & \(1.13 \times 10^{-1}\) \\ \hline \end{tabular} (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of NO when \([\mathrm{NO}]=0.0750 \mathrm{M}\) and $\left[\mathrm{O}_{2}\right]=0.0100 \mathrm{M} ?(\mathrm{e})$ What is the rate of disappearance of \(\mathrm{O}_{2}\) at the concentrations given in part (d)?

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \rightarrow \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow & \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Indicate how the elementary reactions add to give the overall reaction. (Hint: You will need to multiply the coefficients of one of the equations by 2.) (b) Based on the rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

The following is a quote from an article in the August 18,1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(\left.59{ }^{\circ} \mathrm{F}\right]\) lowers the reaction rate six times; a 36-degree drop [from \(77^{\circ} \mathrm{F}\) to \(\left.41{ }^{\circ} \mathrm{F}\right]\) produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the 36 -degree drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of \(2.7\) years, calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\) (slow) \(\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \quad\) (fast) (a) Write the rate law for each of the elementary reactions of the mechanism. (b) Write the chemical equation for the overall process. (c) Identify the intermediate, if any, in the mechanism. (d) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free