/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 120 The decomposition of \(\mathrm{N... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 120

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is a first-order reaction. At \(25{ }^{\circ} \mathrm{C}\), it takes \(5.2 \mathrm{~h}\) for the concentration to drop from \(0.120 \mathrm{M}\) to \(0.060 \mathrm{M}\).How many hours does it take for the concentration to drop from \(0.030 \mathrm{M}\) to \(0.015 \mathrm{M} ?\) From \(0.480 \mathrm{M}\) to \(0.015 \mathrm{M} ?\)

Short Answer

Expert verified
It takes 5.2 hours for 0.030 M to 0.015 M and 26.0 hours for 0.480 M to 0.015 M.

Step by step solution

01

Understanding first-order reactions

In a first-order reaction, the rate of reaction depends on the concentration of one reactant raised to the first power. The integrated rate law for a first-order reaction is \( [A]=[A]_0 e^{-kt} \). Alternatively, we can use the relation \( \ln \left(\frac{[A]_0}{[A]}\right) = kt \), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.
02

Calculate the rate constant

Using the data provided for the initial concentration \(0.120 \mathrm{M}\) and final concentration \(0.060 \mathrm{M}\) over time \(5.2 \mathrm{~h}\), we can find \(k\):\\[\ln \left(\frac{0.120}{0.060}\right) = k \cdot 5.2\] Simplify to find \(\ln(2) = 5.2k\). Solve for \(k\): \(k = \frac{\ln(2)}{5.2}\approx 0.1332 \mathrm{h}^{-1}\).
03

Determine time for concentration drop from 0.030 M to 0.015 M

Use the relation \( \ln \left(\frac{[A]_0}{[A]}\right) = kt \) with \([A]_0 = 0.030 \mathrm{M}\) and \([A] = 0.015 \mathrm{M}\), and \(k = 0.1332 \mathrm{h}^{-1}\): \\[ \ln(\frac{0.030}{0.015}) = 0.1332 \cdot t \] This simplifies to \( \ln(2) = 0.1332 \cdot t\). Solve for \(t\): \( t = \frac{\ln(2)}{0.1332} \approx 5.2 \mathrm{~h}\).
04

Determine time for concentration drop from 0.480 M to 0.015 M

Same method applies. Use \([A]_0 = 0.480 \mathrm{M}\) and \([A] = 0.015 \mathrm{M}\): \\[ \ln\left(\frac{0.480}{0.015}\right) = 0.1332 \cdot t \] Simplify the left side: \( \ln(32) = 0.1332t \). Solve for \(t\): \( t = \frac{\ln(32)}{0.1332} \approx 26.0 \mathrm{~h}\).

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Key Concepts

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

Integrated Rate Law
Integrated rate laws are crucial in understanding how the concentration of reactants changes over time during a chemical reaction. For first-order reactions, the integrated rate law is expressed as \[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \].Here,
  • \([A]_0\) represents the initial concentration of the reactant,
  • \([A]\) is the concentration at time \(t\),
  • and \(k\) is the rate constant.
This equation allows us to calculate either the rate constant, time elapsed, or the concentration at a specific time, given the other variables. It's derived from the basic idea that, in a first-order reaction, the rate of the reaction is directly proportional to the concentration of a single reactant.
Rate Constant
The rate constant, denoted by \(k\), is a key feature in reaction kinetics and provides essential information about the speed of a reaction. In the case of first-order reactions, \(k\) has units of \(\mathrm{h}^{-1}\) (when time is in hours).We can calculate \(k\) using the integrated rate law, knowing the initial and final concentrations as well as the time taken for that change:\[ k = \frac{\ln \left(\frac{[A]_0}{[A]}\right)}{t} \].In our specific case, knowing that the concentration of \(\mathrm{N}_2\mathrm{O}_5\) decreases from 0.120 M to 0.060 M in 5.2 hours, the value of \(k\) is found to be approximately 0.1332 \(\mathrm{h}^{-1}\). This constant is a representation of how quickly the reactant is consumed in the reaction, influencing how rapidly equilibrium is reached.
Logarithmic Calculations
Logarithmic calculations are often used in kinetics to simplify the mathematical handling of exponential decay or growth processes. When dealing with first-order reactions, the natural logarithm \(\ln\) helps us linearize the relationship between reactant concentration and time.For example, given the relation \( \ln \left(\frac{[A]_0}{[A]}\right) = kt \), you can solve for different variables. If we're looking to find how long it will take for a concentration shift, rearranging gives us \[ t = \frac{\ln \left(\frac{[A]_0}{[A]}\right)}{k} \].These calculations are a handy tool because they transform the multiplicative decaying nature of first-order processes into a much simpler additive relationship, making it easier to analyze and predict reaction behavior.
Reaction Kinetics
Reaction kinetics delves into the rates and mechanisms of chemical reactions. By studying these dynamics, we can predict how substances change over time. In first-order reactions, the rate is directly proportional to only one reactant's concentration, making it relatively straightforward to model and predict. Understanding the kinetics of a reaction helps in various practical applications like industrial synthesis, where optimizing reaction time and yield are crucial. By applying the integrated rate law and logarithmic calculations, chemists can optimize reactions to run efficiently under specific conditions, often tweaking variables to harness a desired outcome. Embracing these concepts allows for a deeper comprehension of how individual molecules behave on their journey through a chemical reaction.

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Most popular questions from this chapter

Give the molecularity and the rate law for each of the following elementary reactions: (a) \(\mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g)\) (b) \(\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)\) (c) \(\mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{Cl}(g)+\mathrm{Cl}(g)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{N}_{2}(g)\)

The following reaction has a first-order rate law:$$ \begin{aligned} &\mathrm{Co}(\mathrm{CN})_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)^{2-}(a q)+\mathrm{I}(a q) \longrightarrow \mathrm{Co}(\mathrm{CN})_{5} \mathrm{I}^{3-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\ &\text { Rate }=k\left[\mathrm{Co}(\mathrm{CN})_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)^{2-}\right] \end{aligned} $$Suggest a possible reaction mechanism, and show that your mechanism agrees with the observed rate law.

Comment on the following statement: "A catalyst increases the rate of a reaction, but it is not consumed because it does not participate in the reaction."

Bromomethane is converted to methanol in an alkaline solution. The reaction is first order in each reactant. $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$(a) Write the rate law. (b) How does the reaction rate change if the OH concentration is decreased by a factor of \(5 ?\) (c) What is the change in rate if the concentrations of both reactants are doubled?

The gas-phase reaction of hydrogen and iodine monochloride is first order in \(\mathrm{H}_{2}\) and first order in ICl. What is the rate law, and what are the units of the rate constant? $$ \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{ICl}(g) \longrightarrow 2 \mathrm{HCl}(g)+\mathrm{I}_{2}(g) $$
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