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Chapter 14: Problem 17

The isomerization of methyl isonitrile (CH \(_{3} \mathrm{NC}\) ) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C}\), and the following data were obtained: \begin{tabular}{ll} \hline Time (s) & {\(\left[\mathrm{CH}_{3} \mathrm{NCl}(M)\right.\)} \\ \hline 0 & \(0.0165\) \\ 2,000 & \(0.0110\) \\ 5,000 & \(0.00591\) \\ 8,000 & \(0.00314\) \\ 12,000 & \(0.00137\) \\ 15,000 & \(0.00074\) \\ \hline \end{tabular} (a) Calculate the average rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Graph \(\left[\mathrm{CH}_{3} \mathrm{NC}\right]\) versus time, and determine the instantaneous rates in \(M / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).

Short Answer

Expert verified
The average rate of reaction between each time interval is as follows: 1. Time interval 0 - 2000 s: \(2.75\times10^{-6} M/s\) 2. Time interval 2000 - 5000 s: \(1.70\times10^{-6} M/s\) 3. Time interval 5000 - 8000 s: \(9.23\times10^{-7} M/s\) 4. Time interval 8000 - 12000 s: \(4.43\times10^{-7} M/s\) 5. Time interval 12000 - 15000 s: \(2.10\times10^{-7} M/s\) To find the instantaneous rate at t=5000 s and t=8000 s, plot the concentration of CH3NC versus time, draw tangent lines to the curve at t=5000 s and t=8000 s, and calculate the slope of each tangent line (change in concentration over change in time for a small interval around each time point). The slopes of the tangent lines will give you the instantaneous rates in M/s at t = 5000s and t = 8000s.

Step by step solution

01

Calculate average rate of reaction between each time interval

To calculate the average rate of reaction between each time interval, we will use the formula: Average rate of reaction = \(\frac{Change \ in \ concentration}{Change \ in \ time}\) Let's calculate the average rate for each time interval: 1. Time interval 0 - 2000 s: \(\frac{0.0165-0.0110}{2000-0} = \frac{0.0055}{2000} = 2.75\times10^{-6} M/s\) 2. Time interval 2000 - 5000 s: \(\frac{0.0110-0.00591}{5000-2000} = \frac{0.00509}{3000} = 1.70\times10^{-6} M/s\) 3. Time interval 5000 - 8000 s: \(\frac{0.00591-0.00314}{8000-5000} = \frac{0.00277}{3000} = 9.23\times10^{-7} M/s\) 4. Time interval 8000 - 12000 s: \(\frac{0.00314-0.00137}{12000-8000} = \frac{0.00177}{4000} = 4.43\times10^{-7} M/s\) 5. Time interval 12000 - 15000 s: \(\frac{0.00137-0.00074}{15000-12000} = \frac{0.00063}{3000} = 2.10\times10^{-7} M/s\)
02

Graph the concentration of CH3NC versus time

Now, we need to plot the concentration of CH3NC over time using the given data points. For this, you can use any graphing tool like Microsoft Excel, Desmos, or create a hand-drawn plot, whichever is more comfortable for you.
03

Determine the instantaneous rate at t=5000 s and t=8000 s

To find the instantaneous rate of reaction at a specific time, we need to determine the slope of the tangent line at that point on the graph. For finding the instantaneous rate at t = 5000 s and t = 8000 s, follow these steps: 1. Draw the tangent lines to the curve at t = 5000 s and t = 8000 s. 2. Calculate the slope of each tangent line by determining the change in concentration over the change in time for a small interval around each time point. 3. The slope of each tangent line will give you the instantaneous rate of reaction at those time points. Please note that the last step may require interpolation. After completing these steps, you will get the instantaneous rates in M/s at t = 5000s and t = 8000s.

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Key Concepts

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

Isomerization Reaction
An isomerization reaction is a process where a molecule is transformed into another molecule which has the same molecular formula but a different structural arrangement. These types of reactions are widespread and important in both biological and chemical systems, often leading to changes in chemical and physical properties. For instance, in the given exercise, methyl isonitrile ((CH_3NC)) undergoes isomerization to become acetonitrile ((CH_3CN)), a structural isomer with the same atoms but different connectivity.

Understanding isomerization is crucial because it underpins numerous biochemical pathways and industrial processes. For example, in the biological context, vitamin A's vision-aiding properties rely on an isomerization reaction in the eye. In industry, the transformation of straight-chain hydrocarbons to branched ones improves fuel quality in the cracking process.
Average Rate of Reaction
The average rate of a chemical reaction is a measure of how quickly the concentration of a reactant or product changes over a particular time interval. In the context of our textbook problem, the average rate is given by the change in concentration of methyl isonitrile ((CH_3NC)) divided by the time span over which this change occurs.

The formula to calculate this rate is: \[ \text{Average rate of reaction} = \frac{\text{Change in concentration}}{\text{Change in time}} \]
By determining the average rate of reaction at different intervals, we see how the speed of the reaction changes over time. This information is vital for understanding reaction kinetics and for designing processes at a commercial scale where controlling the rate of reaction is key to efficiency and safety.
Instantaneous Rate of Reaction
While the average rate gives us information over a period, the instantaneous rate of reaction tells us about the rate at a specific moment. It's akin to looking at the speedometer of a car at a precise instant as opposed to calculating the average speed over a long trip. Mathematically, it is the slope of the tangent to the concentration vs. time curve at a point of interest.

To find the instantaneous rate, we often graph the concentration of the reactant over time and draw a tangent line at the time of interest, then calculate its slope. This process can be interpreted as zooming in on the curve until a small portion becomes nearly straight. The formula used is similar to that of average rate but for a very tiny change in time - essentially approaching zero. It's crucial in understanding how the reaction proceeds at any given time and can impact decisions in controlling chemical processes that have strict time-dependent processes.

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Most popular questions from this chapter

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

The following mechanism has been proposed for the gas-phase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (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 decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64{ }^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\)

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Zinc metal dissolves in hydrochloric acid according to the reaction $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(a q)--\rightarrow \operatorname{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Suppose you are asked to study the kinetics of this reaction by monitoring the rate of production of \(\mathrm{H}_{2}(g)\). (a) By using a reaction flask, a manometer, and any other common laboratory equipment, design an experimental apparatus that would allow you to monitor the partial pressure of \(\mathrm{H}_{2}(g)\) produced as a function of time. (b) Explain how you would use the apparatus to determine the rate law of the reaction. (c) Explain how you would use the apparatus to determine the reaction order for \(\left[\mathrm{H}^{+}\right]\) for the reaction. (d) How could you use the apparatus to determine the activation energy of the reaction? (e) Explain how you would use the apparatus to determine the effects of changing the form of \(\mathrm{Zn}(s)\) from metal strips to granules.
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