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

The rate of a reaction may be expressed as: \(+\frac{1}{2} \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{d} t}=-\frac{1}{3} \frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d} t}=+\frac{1}{4} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}\) \(=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t} .\) The reaction is (a) \(4 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}+3 \mathrm{D}\) (b) \(\mathrm{B}+3 \mathrm{D} \rightarrow 4 \mathrm{~A}+2 \mathrm{C}\) (c) \(4 \mathrm{~A}+2 \mathrm{C} \rightarrow \mathrm{B}+3 \mathrm{D}\) (d) \(2 \mathrm{~A}+3 \mathrm{~B} \rightarrow 4 \mathrm{C}+\mathrm{D}\)

Short Answer

Expert verified
Option (a) correctly represents the reaction because the coefficients match the inverses of the rate factors for each substance.

Step by step solution

01

Understand the rate of reaction

The given rate of reaction shows how the concentration of reactants and products change over time. The positive sign indicates the formation of a substance, while the negative sign indicates the consumption of a substance.
02

Compare the coefficients

The stoichiometric coefficients in the reaction equation will be inversely proportional to the rates of change of the concentrations. These coefficients are crucial for correctly balancing the reaction equation.
03

Match the coefficients with reaction options

Find the reaction option where the coefficients of substances A, B, C, and D match the given rate of reaction when multiplied by corresponding rate factors.
04

Verify each option

Evaluate each option (a) to (d) to check which one corresponds to the rate of reaction given.

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.

Chemical Kinetics
Chemical kinetics is a subdivision of physical chemistry that deals with understanding the speed or rate at which chemical reactions occur and the factors affecting these rates. It is a fundamental concept because it not only gives insight into the reaction mechanisms but also helps in the design of chemical processes and predicting product concentrations over time.

When studying chemical kinetics, we generally focus on the rate of a reaction, which is expressed as the change in concentration of reactants or products over time. For instance, if you notice that a substance's concentration increases or decreases rapidly, you can infer that the reaction is proceeding quickly. Conversely, if there's little change over an extended period, the reaction is considered slow.

By understanding how different conditions such as temperature, pressure, and concentration of reactants affect the reaction rate, chemists can optimize conditions to make reactions more efficient in industrial applications.
Stoichiometry
When it comes to the realm of chemical reactions, stoichiometry is the mathematical relationship between the quantities of reactants and products. It's essentially the recipe for a chemical reaction, telling you how much of each reactant you need to produce a certain amount of product.

The law of conservation of mass dictates that in a chemical reaction, the total mass of the reactants must equal the total mass of the products. To ensure this balance, we use stoichiometric coefficients, which are the numbers placed before compounds within a chemical equation to indicate the proportions in which they react or are produced.

Understanding stoichiometry is crucial in determining how much of a chemical you'll need for a reaction. Whether you're scaling a reaction up to produce sufficient product for commercial use or simply trying to predict the outcome of an experiment in a lab, stoichiometry provides the calculations necessary to keep a balanced and controlled reaction.
Differential Rate Expressions
Differential rate expressions are like the instantaneous snapshots of a reaction's rate at any given moment. They give us the instantaneous rate of change of reactant or product concentrations with respect to time, represented as a derivative. These expressions are based on calculus and offer a deep dive into the dynamics of chemical reactions.

The differential rate for a substance in a chemical reaction is noted by the symbol 'd', signifying a derivative with respect to time (t). For example, \(\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}\) represents the rate of change of concentration of substance A over time. Depending on the reaction and conditions, these rates can be positive or negative, indicating the formation or consumption of a substance, respectively.

Understanding differential rate expressions is essential for forming rate laws, predicting reaction kinetics, and even for monitoring the progress of a reaction in real time. In practical applications, they can be used to control reaction conditions to optimize product yield or to ensure safety in reactions that might otherwise happen too fast and potentially become hazardous.

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

For a first-order reaction, \(t_{0.75}=1386 \mathrm{~s}\). Its specific reaction rate is (a) \(10^{-3} \mathrm{~s}^{-1}\) (b) \(10^{-2} \mathrm{~s}^{-1}\) (c) \(10^{-4} \mathrm{~s}^{-1}\) (d) \(10^{-5} \mathrm{~s}^{-1}\)

In a first-order reaction, the activity of reactant drops from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times 10^{4} \mathrm{~s}\). The rate constant of the reaction, in \(\mathrm{s}^{-1}\), is (a) \(1.386 \times 10^{-4}\) (b) \(1.386 \times 10^{-3}\) (c) \(1.386 \times 10^{-5}\) (d) \(5.0 \times 10^{3}\)

For the reaction: \(\mathrm{aA}+\mathrm{bB} \rightarrow \mathrm{P}\), \(r=K[\mathrm{~A}]^{a} \cdot[\mathrm{B}]^{b} .\) If concentration of \(\mathrm{A}\) is doubled, the rate is doubled. If concentration of \(\mathrm{B}\) is doubled, the rate becomes four times. The correct relation is (a) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) (b) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+2 \cdot \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) (c) \(-2 \cdot \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\) (d) \(-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d} t}\)

Which is correct about first-order reaction? (a) \(t_{0.5}=50 \mathrm{~s}, t_{0.75}=100 \mathrm{~s}\) (b) \(t_{0.5}=50 \mathrm{~s}, t_{0,75}=75 \mathrm{~s}\) (c) \(t_{0.5}=50 \mathrm{~s}, t_{0,75}=50 \mathrm{~s}\) (d) \(t_{0.5}=50 \mathrm{~s}, t_{0,25}=25 \mathrm{~s}\)

For a reaction \(2 \mathrm{NH}_{3} \rightarrow \mathrm{N}_{2}+3 \mathrm{H}_{2}\), it is observed that \(-\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d} t}=K_{1}\left[\mathrm{NH}_{3}\right] ;+\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d} t}=K_{2}\left[\mathrm{NH}_{3}\right]\) and \(+\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t}=K_{3}\left[\mathrm{NH}_{3}\right]\) The correct relation between \(K_{1}, K_{2}\) and \(K_{3}\) is (a) \(K_{1}=K_{2}=K_{3}\) (b) \(2 K_{1}=3 K_{2}=6 K_{3}\) (c) \(3 K_{1}=6 K_{2}=2 K_{3}\) (d) \(6 K_{1}=3 K_{2}=2 K_{3}\)

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