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The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0\) min? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half- life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) 0.00822 mol of N2O5 will remain after 5 minutes, (b) it will take approximately 10.14 minutes for the quantity of N2O5 to drop to 0.010 mol, and (c) the half-life of N2O5 at 70°C is approximately 1.70 minutes.

Step by step solution

01

First-order Integrated Rate Law

For a first-order reaction, we use the Integrated Rate Law, which is given by: \(ln \frac{[A]_0}{[A]} = kt\), where [A]_0 and [A] are the initial and final concentrations, respectively, k is the rate constant, and t is time. We can rearrange this equation to find [A], the final concentration of N2O5 remaining: \([A] = \frac{[A]_0}{e^{kt}}\).
02

(a) Moles of N2O5 remaining after 5 minutes

We are given: [A]_0 = 0.0250 mol / 2.0 L = 0.0125 M, k = 6.82 x 10^{-3} s^{-1}, and t = 5.0 min = 300 s. Now we can plug these values into the Integrated Rate Law equation to find the final concentration of N2O5: [A] = 0.0125 M / e^(6.82 x 10^{-3} s^{-1} * 300 s) = 0.00411 M. To find the moles of N2O5 remaining, we multiply the concentration by the volume: Moles of N2O5 remaining = 0.00411 M * 2.0 L = 0.00822 mol.
03

(b) Time for N2O5 to drop to 0.010 mol

We are given: [A]_0 = 0.0125 M, [A] = 0.010 mol / 2.0 L = 0.005 M, and k = 6.82 x 10^{-3} s^{-1}. We can rearrange the Integrated Rate Law equation to find t: \(t = \frac{ln \frac{[A]_0}{[A]}}{k}\). Now we plug the values into the equation: t = ln(0.0125 / 0.005) / (6.82 x 10^{-3} s^{-1}) ≈ 608.4 s. Convert seconds to minutes: Time = 608.4 s / 60 s/min ≈ 10.14 min.
04

(c) Half-life of N2O5

We can use the half-life formula for a first-order reaction: \(t_{1/2} = \frac{ln 2}{k}\). We are given k = 6.82 x 10^{-3} s^{-1}. Plug this value into the equation: t_{1/2} = ln 2 / (6.82 x 10^{-3} s^{-1}) ≈ 101.7 s. Convert seconds to minutes: Half-life of N2O5 = 101.7 s / 60 s/min ≈ 1.70 min. In summary, (a) 0.00822 mol of N2O5 will remain after 5 minutes, (b) it will take approximately 10.14 minutes for the quantity of N2O5 to drop to 0.010 mol, and (c) the half-life of N2O5 at 70°C is approximately 1.70 minutes.

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Key Concepts

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

First-Order Reactions
First-order reactions are chemical processes where the rate at which a reactant converts into products is directly proportional to its concentration. In simple terms, the reaction rate depends only on the concentration of one reactant. This is a common type of reaction in both chemistry labs and the environment.

Characteristics of first-order reactions include:
  • The reaction rate decreases over time as the concentration of the reactant decreases.
  • A linear relationship between the logarithm of the reactant concentration and time.
  • Common in radioactive decay and various chemical reactions involving gases.
Understanding this concept is crucial for solving problems related to reaction kinetics, as it lays the foundation for calculating the remaining concentration of a reactant or predicting how long a reaction will take to reach a certain level of completion.
Rate Constant
The rate constant, often denoted by the symbol \(k\), is a crucial parameter in chemical kinetics that defines the speed of a reaction. For a first-order reaction, the rate constant has units of \;(s^{-1})\, indicating the reaction rate per second. It is called a constant because it remains the same under constant temperature, making it extremely useful for calculations.

Key points about the rate constant:
  • The rate constant determines how fast a reaction proceeds. A larger rate constant means a faster reaction.
  • The value of \(k\) is temperature dependent—a higher temperature often increases the rate constant.
  • In the given exercise, \(k = 6.82 \times 10^{-3} \; s^{-1}\) at \(70^{\circ}C\), helps us calculate the time needed for the reaction to progress to a specific point.
Understanding the rate constant is essential when working with the integrated rate law and calculating reaction times or remaining concentrations.
Integrated Rate Law
The integrated rate law for a first-order reaction provides a mathematical relationship between the concentration of the reactant and time. This formula is useful to determine the concentration of reactants at any given time during the reaction.

The integrated rate law is expressed as:\[ln \left(\frac{[A]_0}{[A]}\right) = kt\]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant. This equation can be rearranged to solve for different variables based on the available data.

Benefits of using the integrated rate law include:
  • Finding the remaining concentration of a reactant after a specified period.
  • Determining the time required to reach a certain concentration.
  • Calculating the rate constant if concentrations and time are known.
In essence, the integrated rate law acts like a guide for predicting the future or investigating past states of a reaction, making it a powerful tool in kinetic studies.
Half-Life
Half-life is a crucial concept in understanding first-order reactions. It refers to the time required for the concentration of a reactant to decrease to half its initial value. Unlike zero-order or second-order reactions, the half-life for a first-order reaction is constant regardless of the initial concentration.

To calculate the half-life \(t_{1/2}\) for first-order reactions, you can use the formula:\[t_{1/2} = \frac{ln 2}{k}\]where \(ln 2\) is a natural logarithm constant approximately equal to 0.693 and \(k\) is the rate constant. This formula highlights that the half-life is inversely proportional to the rate constant.

A few important notes regarding half-life:
  • For first-order reactions, the half-life remains unchanged regardless of how much reactant is present initially.
  • It provides a convenient measure for comparing how fast different reactions are, as a smaller half-life indicates a faster reaction.
  • In the presented exercise, the calculated half-life of \(N_2O_5\) at \(70^{\circ}C\) is approximately 1.70 minutes, showing a relatively rapid decomposition.
Understanding half-life helps in predicting the dynamics of a reaction and is particularly useful in fields like pharmacology, environmental science, and nuclear physics.

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Most popular questions from this chapter

(a) For a second-order reaction, what quantity, when graphed versus time, will yield a straight line? (b) How do the half-lives of first-order and second- order reactions differ?

Consider the following reaction between mercury(II) chloride and oxalate ion: \(2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q)\) \(2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ? $$ \begin{array}{llll} \text { Experiment } & {\left[\mathrm{HgCl}_{2} \mathrm{~J}(\mathrm{M})\right.} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right](M)} & \text { Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2}\right)\) is \(0.25 \mathrm{M}\), if the temperature is the same as that used to obtain the data shown?

A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if (a) \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ;\) (b) \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ?

The following mechanism has been proposed for the gas-phase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{aligned} &\mathrm{H}_{2}(\mathrm{~g})+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) Write rate laws for each elementary reaction in the mechanism. (d) If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} &\mathrm{NO}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ &\mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

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