/*! 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 43 As described in Exercise 14.41 ,... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 43

As described in Exercise 14.41 , the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of \(60 \mathrm{kPa}\), what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

Short Answer

Expert verified
(a) The partial pressure of the sulfuryl chloride after 60 seconds is approximately 16.16 kPa. (b) The time it takes for the partial pressure to decrease to one-tenth of its initial value is approximately 50.67 seconds.

Step by step solution

01

(a) Calculate the partial pressure of SO鈧侰l鈧� after 60 seconds

Using the first-order decay equation: \(P(t) = P_0 e^{-kt}\) Plug in the given values: \(P(60) = 60~kPa \times e^{- (4.5 \times 10^{-2}~s^{-1})(60~s)}\) Calculate the result: \(P(60) \approx 16.16 ~kPa\) The partial pressure of the sulfuryl chloride after 60 seconds is approximately 16.16 kPa.
02

(b) Calculate the time for the partial pressure to decrease to one-tenth of its initial value

Since we want to find the time when the partial pressure of SO鈧侰l鈧� is one-tenth of its initial value (60 kPa), we can set up the equation as follows: \(\frac{1}{10}(60~kPa) = 60~kPa \times e^{-(4.5 \times 10^{-2}~s^{-1})t}\) Divide both sides by the initial pressure (60 kPa): \(\frac{1}{10} = e^{-(4.5 \times 10^{-2}~s^{-1})t}\) Take the natural logarithm of both sides: \(ln(\frac{1}{10}) = -(4.5 \times 10^{-2}~s^{-1})t\) Divide by the rate constant: \(\frac{ln(\frac{1}{10})}{-(4.5 \times 10^{-2}~s^{-1})} = t\) Calculate the result: \(t \approx 50.67~s\) The time it takes for the partial pressure to decrease to one-tenth of its initial value is approximately 50.67 seconds.

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Key Concepts

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

First-order reaction
In chemistry, a first-order reaction refers to processes where the rate is directly proportional to a single reactant's concentration. This means as the concentration of the reactant decreases, the reaction rate decreases as well. The mathematical expression for a first-order reaction is given by:\[Rate = k[A]\]Here, \(k\) is the rate constant, and \([A]\) is the concentration of the reactant at any time. Notably, first-order reactions exhibit a constant half-life, which exclusively depends on the rate constant and not on the initial concentration. This makes calculations straightforward. For example, in the decomposition of sulfuryl chloride, the formulation:\[P(t) = P_0 e^{-kt}\]expresses how the pressure, representing concentration in gaseous reactions, decreases exponentially over time.
Rate constant
The rate constant, often symbolized as \(k\), is a crucial factor in kinetic studies. It provides insight into how fast a reaction proceeds. For first-order reactions, the rate constant holds the dimensions of inverse time, such as \(\mathrm{s}^{-1}\). It is independent of the concentration, depending only on the nature of the reactants and the temperature at which the reaction occurs.

For example, the rate constant for the decomposition of sulfuryl chloride at 660 K is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). This value signifies that at this specific temperature, the reaction rate progresses at a measurable and calculable pace. Changes in temperature would lead to variations in the rate constant, hence affecting the speed of the reaction.
Decomposition reaction
Decomposition reactions involve a single compound breaking down into two or more simpler substances. They are essential in various natural and industrial processes. Often requiring external energy like heat, light, or electricity to initiate, these reactions illustrate how complex molecules can degrade over time.

In our exercise, the decomposition reaction of sulfuryl chloride shows its breakdown into simpler products, described generally by:\[\text{AB} \rightarrow \text{A} + \text{B}\]During this process, sulfuryl chloride separates into sulfur dioxide and chlorine gas under certain conditions. The kinetics, particularly in first-order reactions, provide a clear understanding of the time-dependent nature of these processes. This reveals not just the speed but also the mechanism by which the decomposition occurs.

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Most popular questions from this chapter

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .(\mathbf{a})\) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .(\mathbf{b})\) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(280 \mathrm{pm}\) (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are \(154 \mathrm{~kJ} / \mathrm{mol}\) and \(136 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?
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