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Chapter 14: Problem 21

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) 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: $$ \begin{array}{rl} \hline \text { Time (s) } & {\left[\mathrm{CH}_{3} \mathrm{NC}\right](\boldsymbol{M})} \\ \hline 0 & 0.0165 \\ 2,000 & 0.0110 \\ 5,000 & 0.00591 \\ 8,000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate the average rate of reaction, in \(M / s\), for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s}\). (c) Graph [CH \(\left._{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M /\) s at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).

Short Answer

Expert verified
(a) For the given time intervals, the calculated average rates of reaction are as follows: 1. \(-2.75 \times 10^{-6} M/s\) 2. \(-1.7 \times 10^{-6} M/s\) 3. \(-9.23 \times 10^{-7} M/s\) 4. \(-4.42 \times 10^{-7} M/s\) 5. \(-2.1 \times 10^{-7} M/s\) (b) The overall average rate of reaction for the entire time of observation (\(t=0s\) to \(t=15000s\)) is \(-1.05 \times 10^{-6} M/s\). (c) To find the instantaneous rates at \(t=5000s\) and \(t=8000s\), plot the given data with concentration of \(CH_3NC\) on the y-axis and time on the x-axis, draw a tangent line to the curve at those points, and determine the slope of these tangent lines. The slope represents the instantaneous rates at those specific times.

Step by step solution

01

Time intervals and average rates

For each time interval, the average rate of reaction is given by the change in the concentration of the reactant divided by the change in time. That is: \[ \text{Average rate} = \frac{[\text{A}]_{\text{final}}-[\text{A}]_{\text{initial}}}{t_{\text{final}}-t_{\text{initial}}} \] We need to apply this formula for all the given time intervals: 1. \(t = 0s\) to \(t = 2000s\) 2. \(t = 2000s\) to \(t = 5000s\) 3. \(t = 5000s\) to \(t = 8000s\) 4. \(t = 8000s\) to \(t = 12000s\) 5. \(t = 12000s\) to \(t = 15000s\) **Step 2: Calculate average rates for each interval**
02

Average rates calculation

For each interval, we can use the formula from step 1 and the concentrations given in the table to find the average rates: 1. \(t = 0s\) to \(t = 2000s\) \[ \text{Average rate} = \frac{0.0110 - 0.0165}{2000 - 0} = -2.75 \times 10^{-6} M/s \] 2. \(t = 2000s\) to \(t = 5000s\) \[ \text{Average rate} = \frac{0.00591 - 0.0110}{5000 - 2000} = -1.7 \times 10^{-6} M/s \] 3. \(t = 5000s\) to \(t = 8000s\) \[ \text{Average rate} = \frac{0.00314 - 0.00591}{8000 - 5000} = -9.23 \times 10^{-7} M/s \] 4. \(t = 8000s\) to \(t = 12000s\) \[ \text{Average rate} = \frac{0.00137 - 0.00314}{12000 - 8000} = -4.42 \times 10^{-7} M/s \] 5. \(t = 12000s\) to \(t = 15000s\) \[ \text{Average rate} = \frac{0.00074 - 0.00137}{15000 - 12000} = -2.1 \times 10^{-7} M/s \] **Step 3: Calculate the average rate of reaction over the entire time of observation**
03

Average rate over entire time

To find the overall average rate of reaction from \(t = 0s\) to \(t = 15000s\), we can use the initial and final concentrations, along with the total time period: \[ \text{Overall average rate} = \frac{0.00074 - 0.0165}{15000 - 0} = -1.05 \times 10^{-6} M/s \] **Step 4: Graphing and finding instantaneous rates**
04

Graph and instantaneous rates

To graph the concentration of \(CH_3NC\) versus time and find the instantaneous rates at \(t=5000s\) and \(t=8000s\), we need to plot the given data and find the slope of the tangent line at those two points. In this answer, we're not going to provide the actual graphical plot. However, we will give a general idea of the process: 1. Plot the concentration of \(CH_3NC\) on the y-axis and time on the x-axis. 2. Draw a smooth curve through the data points. 3. For \(t=5000s\) and \(t=8000s\), draw a tangent line to the curve at those points. 4. Determine the slope of these tangent lines, which represent the instantaneous rates at those specific times.

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Key Concepts

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

Isomerization
Isomerization is a chemical process by which a molecule is transformed into another molecule with the same atoms, but in a different arrangement, hence resulting in a molecule with different chemical properties. In the context of our example, methyl isonitrile (\( \text{CH}_3\text{NC} \)) is converted into acetonitrile (\( \text{CH}_3\text{CN} \) ). This reaction is vital in both industrial applications and biological systems, notably in the interconversion of sugars in biochemical pathways.

Isomerization reactions can occur in various phases, but in gas phase reactions, the molecules are more spread out and collisions between them can lead to different dynamics compared to reactions in liquid solutions. Understanding the rate at which isomerization occurs is key to controlling these reactions in synthetic and natural environments.

Importance of Reaction Rate in Isomerization

The speed, or rate, of isomerization is crucial because it can affect the yield and purity of the desired product. In industrial processes, optimizing this rate can lead to improved efficiency and cost savings. For students, grasping the underlying principles governing these rates is essential for predictive and analytical purposes in chemical kinetics.
Chemical Kinetics
Chemical kinetics deals with the speed at which chemical reactions occur and the factors affecting this speed. It is an important field in chemistry because it not only helps in understanding how fast reactions will proceed, but it also provides insights into the reaction mechanism and transition states. The reaction rate is often measured in terms of the change in concentration of reactants (or products) over time.

When calculating reaction rates, we usually differentiate between average and instantaneous rates. The average rate measures the speed of a reaction over a certain time interval, offering a broader view of the reaction's progress. Contrastingly, the instantaneous rate is the rate at a specific moment in time and is analogous to taking a 'snapshot' of the reaction rate. This can be visualized by the slope of a tangent line on a concentration vs. time graph.

Exercise Improvement Advice

To better understand kinetics, students should practice by calculating both average and instantaneous rates, as shown in the textbook exercise. They could graph the concentrations to comprehend how the rate changes over time, and this learning process will significantly help with their comprehension of reaction mechanisms.
Gas Phase Reactions
Gas phase reactions are chemical reactions that take place in the gaseous state. These types of reactions can behave quite differently from those in liquid or solid states due to the higher energy and freedom of movement of the molecules involved. In the gas phase, molecules collide more frequently and energetically, which can lead to faster reaction rates and unique reaction pathways.

The study of gas phase reactions is vital in fields ranging from environmental science to astrophysics. For instance, understanding these reactions is essential for controlling air pollution and for grasping the chemical processes that occur in the atmosphere of Earth and other planets.

Characteristics of Gas Phase Reactions

Temperature is a key factor in influencing the rate of gas phase reactions. As demonstrated in the textbook exercise with the isomerization of methyl isonitrile, conducting the reaction at a high temperature (\( 215^\text{o}C \) ) significantly increases the molecules' kinetic energy, thus enhancing the rate of reaction. Students should note that under different conditions, such as pressure and temperature, the behavior and rate of gas phase reactions can alter dramatically.

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Most popular questions from this chapter

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathrm{c})\) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g),\) is studied at \(383{ }^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{rl} \hline \text { Time }(\mathbf{s}) & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the rate constant? (c) If you used the method of initial rates to obtain the order for \(\mathrm{NO}_{2},\) predict what reaction rates you would measure in the beginning of the reaction for initial concentrations of \(0.200 \mathrm{M}, 0.100 \mathrm{M},\) and \(0.050 \mathrm{M} \mathrm{NO}_{2}\)

Sketch a graph for the generic first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) that has concentration of \(\mathrm{A}\) on the vertical axis and time on the horizontal axis. (a) Is this graph linear? Explain. (b) Indicate on your graph the half-life for the reaction.

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & \text { Rate Constant }\left(\mathbf{s}^{-1}\right) \\ \hline 300 & 3.2 \times 10^{-11} \\ 320 & 1.0 \times 10^{-9} \\ 340 & 3.0 \times 10^{-8} \\ 355 & 2.4 \times 10^{-7} \\ \hline \end{array} $$ From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\)
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