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

For a reaction of order \(n\), the integrated form of the rate equation is: \((n-1) \cdot K \cdot t\) \(=\left(C_{0}\right)^{1-n}-(C)^{1-n}\), where \(C_{0}\) and \(C\) are the values of the reactant concentration at the start and after time ' \(t\) '. What is the relationship between \(t_{3 / 4}\) and \(t_{1 / 2}\), where \(t_{3 / 4}\) is the time required for \(C\) to become \(C_{0} / 4\). (a) \(t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]\) (b) \(t_{3 / 4}=t_{1 / 2}\left[2^{n-1}-1\right]\) (c) \(t_{3 / 4}=t_{1 / 2}\left[2^{n+1}-1\right]\) (d) \(t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n+1}+1\right]\)

Short Answer

Expert verified
The relationship between t_{3/4} and t_{1/2} is given by (b) t_{3 / 4}=t_{1 / 2}\left[2^{n-1}-1\right].

Step by step solution

01

Setting up the equations for t_{1/2} and t_{3/4}

First, set up the equation for the time required for the reactant concentration to drop to half its original value, t_{1/2}. When C = C_0/2, plug these values into the given equation to get an equation for t_{1/2}. Likewise, set up another equation for the time required to reduce the concentration to a quarter, t_{3/4}, by substituting C = C_0/4 into the original equation.
02

Simplify the t_{1/2} equation

Simplify the equation for t_{1/2} by reducing the term (C_0)^{1-n} and expressing the remaining terms in terms of C_0 and n.
03

Simplify the t_{3/4} equation

Similarly, simplify the equation for t_{3/4} by reducing the terms involving C_0 and expressing the result in terms of C_0 and n.
04

Find the relationship between t_{3/4} and t_{1/2}

Divide the equation for t_{3/4} by the equation for t_{1/2} to find a relationship between the two times. The resulting expression should match with one of the provided answer choices (a) through (d).

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 Order
Understanding the reaction order is crucial when exploring how chemical reactions proceed over time. The reaction order refers to the power to which the concentration of a reactant is raised in the rate law equation. It indicates how the rate of a reaction is affected by the concentration of that specific reactant.

In general, reaction orders can be zero, first, or second, but they can also adopt fractional or even negative values in some complex reactions. For a reaction of order zero, the rate is independent of the reactant's concentration. In a first-order reaction, the rate is directly proportional to the reactant's concentration. Meanwhile, a second-order reaction's rate is proportional to the square of the reactant's concentration. The concept of reaction order is fundamental in chemical kinetics as it helps predict how variations in concentration affect the reaction rate and thus the amount of time it takes for the reactants to transform into products.
Chemical Kinetics
Chemical kinetics is the field of chemistry that studies the rates at which reactions occur and the factors that influence these rates. The main objective is to understand the sequence of steps through which the reactants turn into products, known as the reaction mechanism.

Various factors can affect the reaction rate, including reactant concentrations, temperature, catalysts, and the physical state of the reactants. By studying kinetics, chemists are able to determine the reaction rate equation, which provides the relationship between the rate of a reaction and the concentrations of the reactants. This equation is fundamental for controlling chemical reactions in industrial processes, biological systems, and environmental changes. In the classroom, kinetic principles help students predict and control reactions effectively, playing a significant role in the design of chemical reactors and the synthesis of new materials.
Rate Equation
The rate equation, also known as the rate law, is a mathematical expression that relates the rate of a chemical reaction to the concentration of the reactants. It usually takes the form \( rate = k \cdot [A]^{m} \cdot [B]^{n} \), where \( k \) is the rate constant, \( [A] \) and \( [B] \) are the concentrations of reactants A and B, and \( m \) and \( n \) are the reaction orders with respect to A and B, respectively.

In the context of the integrated rate equation given in the exercise, we are looking at the equation after it has been integrated over time, which allows us to connect concentrations of reactants at different points in time with the elapsed time. This particular form of the rate equation is essential when we work with experimental data to determine reaction rates and orders. It also plays a central role in predicting how long a reaction will take to reach a certain level of completion, such as halfway or three quarters of the way to completion, which directly relates to the exercise at hand where we calculate \( t_{3/4} \) in relation to \( t_{1/2} \).

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

In \(80 \%\) ethanol at \(55^{\circ} \mathrm{C}\), isopropyl bromide reacts with hydroxide ion according to the following kinetics: $$ \begin{array}{l} -\frac{\mathrm{d}[\mathrm{RX}]}{\mathrm{d} t}=\left(4.8 \times 10^{-5} \mathrm{M}^{-1} \mathrm{~s}^{-1}\right) \\ {[\mathrm{RX}]\left[\mathrm{OH}^{-}\right]+2.4 \times 10^{-6} \mathrm{~s}^{-1}[\mathrm{RX}]} \end{array} $$ What percentage of isopropyl bromide reacts by the \(S_{\mathrm{N}_{2}}\) mechanism when \(\left[\mathrm{OH}^{-}\right]=0.01 \mathrm{M} ?\) (a) \(16.67 \%\) (b) \(83.33 \%\) (c) \(66.67 \%\) (d) \(33.33 \%\)

The rate of a reaction is approximately doubled for every \(10^{\circ} \mathrm{C}\) rise in temperature. If the temperature is raised by \(50^{\circ} \mathrm{C}\), the reaction rate increases by about (a) 10 times (b) 16 times (c) 32 times (d) 64 times

For a first-order reaction: \(\mathrm{A} \rightarrow\) Product, the initial concentration of \(\mathrm{A}\) is \(0.1 \mathrm{M}\) and after time \(40 \mathrm{~min}\), it becomes \(0.025 \mathrm{M}\). What is the rate of reaction at reactant concentration \(0.01 \mathrm{M} ?\) (a) \(3.465 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (b) \(3.465 \times 10^{-5} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (c) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (d) \(1.7325 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\)

Which of the following statements is incorrect? (a) For endothermic reactions, energy of activation is always greater than heat of reaction. (b) For exothermic reactions, energy of activation is always smaller than heat of reaction. (c) Rate of elementary reaction always increases with increase in temperature. (d) Temperature coefficient of a reaction is \(1.0\), when \(T \rightarrow \infty\).

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