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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 Iâ‚‚ 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.

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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.

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Most popular questions from this chapter

A flask is charged with \(0.100 \mathrm{~mol}\) of \(\mathrm{A}\) and allowed to react to form \(\mathrm{B}\) according to the hypothetical gas-phase reaction \(\mathrm{A}(g) \longrightarrow \mathrm{B}(g)\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (s) } & 0 & 40 & 80 & 120 & 160 \\ \hline \text { Moles of A } & 0.100 & 0.067 & 0.045 & 0.030 & 0.020 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 -s interval in units of \(\mathrm{mol} / \mathrm{s}\). (c) What additional information would be needed to calculate the rate in units of concentration per time?

The temperature dependence of the rate constant for a reaction is tabulated as follows: $$ \begin{array}{lc} \hline \text { Temperature (K) } & k\left(M^{-1} \mathrm{~s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{array} $$ Calculate \(E_{a}\) and \(A\).

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?

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products? Explain.

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right)\). (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

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