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

Consider the reaction, $$\mathrm{CH}_{3} \mathrm{Cl}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+3 \mathrm{HCl}(g)$$ (a) Express the rate of the reaction with respect to each of the reactants and products. (b) If the instantaneous rate of the reaction with respect to \(\mathrm{HCl}\) is \(0.029 \mathrm{M} \mathrm{s}^{-1}\), what is the instantaneous rate of the reaction?

Short Answer

Expert verified
The rate of the reaction with respect to each of the reactants and products is: Rate = -d[CH3Cl]/dt = -(1/3)d[Cl2]/dt = (1/3)d[HCl]/dt = d[CCl4]/dt. The instantaneous rate of the reaction is 0.0097 M/s.

Step by step solution

01

Write the rate expressions for the reactants

To express the rate of the reaction with respect to reactants, you apply the general rate law, which in this case is: Rate = - (1/a) * (d[CH3Cl]/dt) = - (1/b) * (d[Cl2]/dt), where 'a' and 'b' are stoichiometric coefficients in the balanced equation, and [X] denotes the concentration of X. For this reaction, a = 1 and b = 3.
02

Write the rate expressions for the products

Similarly, express the rate of formation of the products as: Rate = (1/c) * (d[CCl4]/dt) = (1/d) * (d[HCl]/dt), where 'c' and 'd' are the stoichiometric coefficients for the products. For this reaction, c = 1 and d = 3.
03

Relate the rates of all substances

Using the coefficients from the balanced equation, relate the rates of consumption and production to each other. The rates of consumption of CH3Cl and Cl2 and the rates of production of CCl4 and HCl are related by their respective stoichiometric coefficients, 1:3:1:3.
04

Calculate the instantaneous rate of the reaction

Given that the rate of formation of HCl is 0.029 M/s, we can calculate the rate of the reaction. Since 3 mol of HCl is produced for every 1 mol of CH3Cl that reacts, the rate of the reaction is (1/3) times the rate of formation of HCl, which is 0.029 M/s / 3 = 0.0097 M/s.

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!

Key Concepts

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

Reaction Rate Expression
The reaction rate expression is a mathematical representation that defines how the rate of a chemical reaction is related to the concentration of the reactants. In the context of the exercise, the general form of the rate is given by a negative change in the concentration of reactants over time (for example, \( -\frac{1}{a} \frac{d[\mathrm{CH}_3\mathrm{Cl}]}{dt} \) for CH3Cl), which signifies that as the reaction progresses, the amount of reactants decreases.

In practice, you'll often find the rate expression for a reaction depends heavily on its stoichiometry. Hence, understanding the balanced chemical equation is crucial. For the given reaction, \( \mathrm{CH}_{3}\mathrm{Cl}(g) + 3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g) + 3 \mathrm{HCl}(g) \), it's apparent that the coefficients translate directly into the rate expressions, highlighting their importance.
Instantaneous Reaction Rate
The instantaneous reaction rate is the rate at which a reaction is proceeding at a specific moment in time. Unlike the average rate, which considers the change in concentration over an extended period, the instantaneous rate focuses on a very short interval, as to consider it almost at a specific point in time.

To determine the instantaneous rate of production of HCl in our example, we used the given rate (0.029 M/s) and recognized that, due to the stoichiometric relationship, this represents three times the rate of the overall reaction. Hence, by dividing by the stoichiometric coefficient of HCl (3), we arrived at the rate of the overall reaction (0.0097 M/s), a process critically tied to the concept of stoichiometry.
Stoichiometry
Stoichiometry is the section of chemistry that involves quantitative relationships between reactants and products in a chemical reaction. The stoichiometric coefficients (like the 1 for CH3Cl and 3 for Cl2) indicate the proportions in which substances react and are formed. These coefficients are essential when determining the reaction rate expressions, as they allow us to relate the rate of consumption of reactants to the rate of production of products.

As seen in the given exercise, the relationship establishes that for every 1 mole of CH3Cl consumed, 3 moles of Cl2 are also used up, and concurrently, 1 mole of CCl4 and 3 moles of HCl are produced, outlining a direct and quantitative connection between all substances involved.
Chemical Kinetics
Chemical kinetics is the study of the rates of chemical processes and the factors that affect these rates. It encompasses the analysis of how different conditions like temperature, concentration, and the presence of a catalyst can influence the speed of a reaction.

In our scenario, understanding chemical kinetics helps to explain why adjusting the concentration of HCl changes the instantaneous rate of the reaction. While the exercise didn't delve into these additional complexities, in real-world kinetic analysis, these factors are crucial for predicting how fast reactions will occur under various conditions, allowing chemists to optimize and control these processes effectively.

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

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