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Chapter 11: Problem 56

The decomposition of nitrosyl chloride $$\mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ is a second-order reaction. If it takes \(0.20 \mathrm{~min}\) to decompose \(15 \%\) of a \(0.300 \mathrm{M}\) solution of nitrosyl chloride, what is \(k\) for the reaction?

Short Answer

Expert verified
Answer: The reaction rate constant (k) for the decomposition of nitrosyl chloride is approximately 0.0536 M⁻¹s⁻¹.

Step by step solution

01

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

The integrated rate law for a second-order reaction is given by: $$\frac{1}{[\mathrm{A}]_{t}}=\frac{1}{[\mathrm{A}]_{0}}+kt$$ where \([\mathrm{A}]_{t}\) is the concentration of \(\mathrm{A}\) at time \(t\), \([\mathrm{A}]_{0}\) is the initial concentration of \(\mathrm{A}\), \(k\) is the reaction rate constant, and \(t\) is the time elapsed.
02

Determine the initial concentration and the concentration at the given time

We are told the initial concentration of nitrosyl chloride is \([\mathrm{NOCl}]_{0}=0.300\,\mathrm{M}\). Since we need to find the concentration after decomposing \(15\%\) of the initial concentration, we have: $$[\mathrm{NOCl}]_{t}=0.85\times[\mathrm{NOCl}]_{0}=0.85\times0.300\,\mathrm{M}=0.255\,\mathrm{M}$$
03

Determine the given time and convert it to the appropriate unit

The given time for the decomposition is \(0.20\,\mathrm{minutes}\), which needs to be converted to seconds: $$t=0.20\,\mathrm{minutes}\times\frac{60\,\mathrm{s}}{1\,\mathrm{min}}=12\,\mathrm{s}$$
04

Plug the values into the integrated rate law and solve for k

Now, we can plug in the values for \([\mathrm{NOCl}]_{0}\), \([\mathrm{NOCl}]_{t}\), and \(t\) into the integrated rate law and solve for \(k\): $$\frac{1}{0.255\,\mathrm{M}}=\frac{1}{0.300\,\mathrm{M}}+k(12\,\mathrm{s})$$ Solving for \(k\) yields: $$k=\frac{\frac{1}{0.255\,\mathrm{M}}-\frac{1}{0.300\,\mathrm{M}}}{12\,\mathrm{s}}\approx0.0536\,\mathrm{M^{-1}s^{-1}}$$ So, the reaction rate constant for the decomposition of nitrosyl chloride is approximately \(k=0.0536\,\mathrm{M^{-1}s^{-1}}\).

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Most popular questions from this chapter

Experimental data are listed for the hypothetical reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\\{[\mathrm{~A}]} & 0.32 & 0.24 & 0.20 & 0.16 & 0.14 & 0.12 \\ \hline\end{array}$$ (a) Plot these data as in Figure \(11.2\). (b) Draw a tangent to the curve to find the instantaneous rate at \(30 \mathrm{~s}\). (c) Find the average rate over the 10 to \(40 \mathrm{~s}\) interval. (d) Compare the instantaneous rate at \(30 \mathrm{~s}\) with the average rate over the thirty-second interval.

Consider the reaction $$\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{SCN}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}$$ The following data were obtained: $$\begin{array}{ccc}\hline\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\right] & {\left[\mathrm{SCN}^{-}\right]} & \text {Initial Rate }(\mathrm{mol} /\mathrm{L} \cdot \mathrm{min}) \\ \hline 0.025 & 0.060 & 6.5 \times 10^{-4} \\\0.025 & 0.077 & 8.4 \times 10^{-4} \\\0.042 & 0.077 & 1.4 \times 10^{-3} \\ 0.042 & 0.100 & 1.8 \times 10^{-3} \\\\\hline\end{array}$$ (a) Write the rate expression for the reaction. (b) Calculate \(k\). (c) What is the rate of the reaction when \(15 \mathrm{mg}\) of \(\mathrm{KSCN}\) is added to \(1.50 \mathrm{~L}\) of a solution \(0.0500 \mathrm{M}\) in \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ?

For the reaction between hydrogen and iodine, $$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ the experimental rate expression is rate \(=k\left[\mathrm{H}_{2}\right] \times\left[\mathrm{I}_{2}\right] .\) Show that this expression is consistent with the mechanism $$\begin{array}{cc}\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) & \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) & \text { (slow) }\end{array}$$

Consider the following hypothetical reaction: $$\mathrm{X}(g) \longrightarrow \mathrm{Y}(g)$$ A 200.0-mL flask is filled with \(0.120\) moles of \(\mathrm{X}\). The disappearance of \(\mathrm{X}\) is monitored at timed intervals. Assume that temperature and volume are kept constant. The data obtained are shown in the table below. $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{X} & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline\end{array}$$ (a) Make a similar table for the appearance of Y. (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two twenty-minute intervals. (c) What is the average rate of appearance of \(\mathrm{Y}\) between the 20 - and 60-minute intervals?

Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)
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