/*! 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 45 The reaction $\mathrm{SO}_{2} \m... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 45

The reaction $\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\( is first order in \)\mathrm{SO}_{2} \mathrm{Cl}_{2}$. Using the following kinetic data, determine the magnitude and units of the first-order rate constant: $$ \begin{array}{cc} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{kPa}) \\ \hline 0 & 101.3 \mathrm{kPa} \\ 2500 & 95.95 \mathrm{kPa} \\ 5000 & 90.69 \mathrm{kPa} \\ 7500 & 85.92 \mathrm{kPa} \\ 10,000 & 81.36 \mathrm{kPa} \\ \hline \end{array} $$

Short Answer

Expert verified
The magnitude of the first-order rate constant for the reaction \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) + \mathrm{Cl}_{2}(g)\) is approximately \(8.78 \times 10^{-4}\) and the units are \(\mathrm{s^{-1}}\).

Step by step solution

01

Write down the integrated rate law for a first-order reaction

The integrated rate law for a first-order reaction is: \[ln\frac{[A]_0}{[A]_t} = kt\] Where: - \([A]_0\) is the initial concentration (pressure in this case) of the reactant at time \(t = 0\) - \([A]_t\) is the concentration (pressure) of the reactant at time \(t\) - \(k\) is the rate constant - \(t\) is the time elapsed
02

Choose two data points from the table

We will choose two data points from the given table to make a comparison: - Time (s) = 0, Pressure (kPa) = 101.3 - Time (s) = 2500, Pressure (kPa) = 95.95
03

Apply the integrated rate law

Plugging the selected data points into the integrated rate law equation, we get: \[ln\frac{101.3}{95.95} = k \cdot 2500\]
04

Solve for the rate constant k

Solve the equation for \(k\): \[k = \frac{ln\frac{101.3}{95.95}}{2500}\] Calculate \(k\): \[k = \frac{ln\frac{101.3}{95.95}}{2500} \approx 8.78 \times 10^{-4} \mathrm{s^{-1}}\] The magnitude of the first-order rate constant is \(8.78 \times 10^{-4}\) and the units are \(\mathrm{s^{-1}}\).

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Most popular questions from this chapter

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\), and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?

(a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism? (d) What is meant by the term rate-determining step?

(a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is \(160 \mathrm{~kJ} / \mathrm{mol}\). Calculate the fraction of methyl isonitrile molecules that has an energy of \(160.0 \mathrm{~kJ}\) or greater at \(500 \mathrm{~K}\). (b) Calculate this fraction for a temperature of \(520 \mathrm{~K}\). What is the ratio of the fraction at \(520 \mathrm{~K}\) to that at \(500 \mathrm{~K} ?\)

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the halflife for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?
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