Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2} ?\)

Short Answer

Expert verified
(a) The half-life for the reaction is \(t_{1/2} = \frac{0.693}{0.271\,s^{-1}} \approx 2.56\,s\). (b) The remaining concentration of \(\mathrm{I}_{2}\) after 5.12 s is \(A_t = 0.050\,M \cdot e^{-(0.271\,s^{-1})(5.12\,s)} \approx 0.0025\,M\).

Step by step solution

01

(a) Calculate the half-life for the reaction

We know that for a first-order reaction, the half-life is given by the formula \(t_{1/2} = \frac{0.693}{k}\), where \(t_{1/2}\) is the half-life and \(k\) is the rate constant. We have the rate constant, \(k = 0.271\,s^{-1}\). Now we can calculate the half-life: \(t_{1/2} = \frac{0.693}{0.271\,s^{-1}}\) Let's calculate it.
02

(b) Calculate the remaining concentration of I₂ after 5.12 s

For a first-order reaction, we can use the following equation to calculate the concentration of the reactant at a given time: \[A_t = A_0 \cdot e^{-kt}\] Where \(A_t\) is the concentration of the reactant at time \(t\), \(A_0\) is the initial concentration of the reactant, and \(k\) and \(t\) are the rate constant and the time, respectively. We have initial concentration \(A_0 = 0.050\,M\), rate constant \(k = 0.271\,s^{-1}\), and time \(t = 5.12\,s\). Now, we can calculate the remaining concentration of \(\mathrm{I}_{2}\): \[A_t = 0.050\,M \cdot e^{-(0.271\,s^{-1})(5.12\,s)}\] Let's calculate it.

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.

Rate Constant
In the realm of chemical kinetics, the rate constant, denoted by the symbol \( k \), is a critical parameter that influences the speed at which a chemical reaction proceeds. Think of it as the pace at which reactants are converted into products. In first-order reactions, the rate of the reaction is directly proportional to the concentration of one reactant.

For instance, if we look at the dissociation of molecular iodine (\( \mathrm{I}_2(g) \) into iodine atoms, the rate constant is given as \( 0.271 \, \mathrm{s}^{-1} \). This numeric value indicates that for every second that passes, a certain fraction of the \( \mathrm{I}_2 \) molecules present in the system will dissociate into iodine atoms. A higher rate constant implies a quicker reaction, meaning that reactants are used up or products are formed more swiftly. It’s essential to remember that the value of the rate constant is influenced by environmental factors, such as temperature and pressure, and is specific to each reaction.
Half-life of Reaction
The half-life of a chemical reaction, usually represented by \( t_{1/2} \), is the duration required for half of the reactant to be transformed or decomposed in a reaction. It's a straightforward way of gauging how fast a reaction occurs. The beauty of first-order reactions is that their half-lives are constant and independent of the starting concentration of the reactant. This characteristic offers the convenience of predicting how long it will take for half of any given amount of reactant to react, regardless of its initial quantity.

In our exercise with molecular iodine (\( \mathrm{I}_2 \)) at \( 625 \, \mathrm{K} \), we applied this first-order reaction principle to calculate the half-life using the formula \( t_{1/2} = \frac{0.693}{k} \). Knowing only the rate constant (\( k \)), we could determine that the half-life of the iodine dissociation reaction is a constant value, which is a critical piece of information in predicting the progress of the reaction over time.
Chemical Kinetics
Chemical kinetics is essentially the study of the rates at which chemical reactions occur and the factors that affect these rates. It’s the science that tells us how fast a chemical reaction takes place and what can be done to control the speed. Kinetics can involve complex math, but it boils down to some fundamental principles. These principles help scientists and engineers to design reactors, preserve food, manufacture drugs, and more – all by understanding and manipulating the rates of reactions.

In the case of the dissociation of \( \mathrm{I}_2 \) at high temperatures, we observe a first-order reaction where the rate depends only on the concentration of \( \mathrm{I}_2 \). Chemical kinetics can provide explanations for why the atoms do not readily recombine or why increasing the temperature can lead to a faster reaction by examining the microscopic interactions and the energy profiles involved in the reaction process.
Concentration Calculation
The art of concentration calculation lies at the heart of chemistry, allowing chemists to understand how much of a substance is present in a given volume and how its quantity changes over time during a reaction. It is vital when dealing with reactions which need specific reactant proportions. Using the equation \[A_t = A_0 \cdot e^{-kt}\], we can calculate the concentration of a reactant at any time during a first-order reaction.

In our exercise on molecular iodine, we solved for the remaining concentration of iodine after 5.12 seconds by plugging in the values into the exponential decay formula, which is a specific case of the more general concentration-time relationship in chemical kinetics. This formula encapsulates the exponential nature of first-order reaction decay and is instrumental for chemists when predicting how much reactant will remain at any point or when planning reactions to ensure complete reactant consumption.

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

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M},\) the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

(a) Explain the importance of enzymes in biological systems. (b) What chemical transformations are catalyzed (i) by the enzyme catalase, \((i i)\) by nitrogenase? (c) Many enzymes follow this generic reaction mechanism, where \(\mathrm{E}\) is enzyme, \(\mathrm{S}\) is substrate, ES is the enzyme-substrate complex (where the substrate is bound to the enzyme's active site), and \(\mathrm{P}\) is the product: 1\. \(\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES}\) 2\. \(\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P}\) What assumptions are made in this model with regard to the rate of the bound substrate being chemically transformed into bound product in the active site?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21{ }^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. (b) Assuming the collision factor is the same for both situations, estimate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

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