Problem 21 - Chemistry The Central Science In Si Units

Question

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{tabular}{rr} \hline Time (s) & {\(\left[\mathrm{CH}_{3} \mathrm{NC}\right](M)\)} \\ \hline 0 & 0.0165 \\ 2000 & 0.0110 \\ 5000 & 0.00591 \\ 8000 & 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) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s}\). \((\mathbf{c})\) Which is greater, the average rate between \(t=2000\) and \(t=12,000 \mathrm{~s}\), or between \(t=8000\) and \(t=15,000 \mathrm{~s} ?\) (d) Graph [CH \(\left._{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).

Step-by-Step Solution

Answer

The overall average rate from 0 to 15000 s is \(1.15 \times 10^{-6} \, M/s\). The average rate from 2000 to 12000 s is greater than from 8000 to 15000 s. Instantaneous rates at 5000 s and 8000 s are approximately \(1.13 \times 10^{-6} \, M/s\) and \(5.30 \times 10^{-7} \, M/s\), respectively.

Step 1: Calculate Average Rate Intervals (a)

To find the average rate of reaction for a given interval, we use the formula: \[ \text{Average Rate} = -\frac{\Delta[\mathrm{CH}_{3}\mathrm{NC}]}{\Delta t} \]Compute \(\Delta[\mathrm{CH}_{3}\mathrm{NC}]\) and \(\Delta t\) for each interval given in the data and calculate:1. For \(t=0\) to \(t=2000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.0110 - 0.0165}{2000 - 0} = 2.75 \times 10^{-6} \ M/s \]2. For \(t=2000\) to \(t=5000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.00591 - 0.0110}{5000 - 2000} = 1.70 \times 10^{-6} \ M/s \]3. For \(t=5000\) to \(t=8000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.00314 - 0.00591}{8000 - 5000} = 9.23 \times 10^{-7} \ M/s \]4. For \(t=8000\) to \(t=12000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.00137 - 0.00314}{12000 - 8000} = 4.43 \times 10^{-7} \ M/s \]5. For \(t=12000\) to \(t=15000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.00074 - 0.00137}{15000 - 12000} = 2.10 \times 10^{-7} \ M/s \]

Step 2: Calculate Average Rate Over Entire Time (b)

The average rate over the entire period from \(t=0\) to \(t=15000\, \mathrm{s}\) is given by:\[ \text{Average Rate} = -\frac{0.00074 - 0.0165}{15000 - 0} = 1.15 \times 10^{-6} \ M/s \]

Step 3: Compare Two Specific Average Rates (c)

Calculate the average rates for two specific intervals:1. Between \(t=2000\) and \(t=12000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.00137 - 0.0110}{12000 - 2000} = 9.63 \times 10^{-7} \ M/s \]2. Between \(t=8000\) and \(t=15000\, \mathrm{s}\):\[ \text{Average Rate} = -\frac{0.00074 - 0.00314}{15000 - 8000} = 3.43 \times 10^{-7} \ M/s \]The average rate from \(t=2000\) to \(t=12000\, \mathrm{s}\) is greater.

Step 4: Graph Concentration vs. Time (d)

Plot \([\mathrm{CH}_{3}\mathrm{NC}]\) on the y-axis and time on the x-axis. Use the points provided in the table to create the graph. This will help in estimating the instantaneous rates at specific times.

Step 5: Determine Instantaneous Rates at Specific Times (d)

From the graph, draw tangents at \(t=5000\, \mathrm{s}\) and \(t=8000\, \mathrm{s}\). Calculate the slope of these tangents to determine the instantaneous rate.1. At \(t=5000\, \mathrm{s}\), estimate the slope of the tangent: approximately \(1.13 \times 10^{-6}\, M/s\).2. At \(t=8000\, \mathrm{s}\), estimate the slope: approximately \(5.30 \times 10^{-7}\, M/s\).

Key Concepts

Rate of Reaction

The rate of reaction is a key concept in reaction kinetics, describing how quickly a reaction occurs. It is essentially the speed at which reactants are converted into products. This can be measured by observing the change in concentration of reactants or products over a specific time period. In the given exercise, the rate of reaction is calculated by monitoring how the concentration of methyl isonitrile egin{itemize}

Changes as it transforms into acetonitrile.

The formula used to calculate the average rate is

\( \text{Average Rate} = -\frac{\Delta [\mathrm{CH}_{3}\mathrm{NC}]}{\Delta t} \)

where \( \Delta [\mathrm{CH}_{3}\mathrm{NC}] \) is the change in concentration and \( \Delta t \) is the change in time.

The negative sign indicates that the concentration of methyl isonitrile decreases over time. Calculating the average rate over different intervals can demonstrate how the reaction speed changes as the reaction progresses.
By fitting the data into this equation, you can get detailed insights into the behavior of the reaction over any given period.

Chemical Concentration Over Time

Chemical concentration over time helps in understanding how substances involved in a reaction vary as the reaction advances. In the exercise data, concentrations of methyl isonitrile were recorded at different times:

• These measurements show a declining trend of concentration over time, typical for reactions.
• The data helps plot a graph of concentration against time, giving a visual representation of the reaction progress.

By plotting egin{itemize}

\([\mathrm{CH}_{3}\mathrm{NC}]\) on the y-axis and time on the x-axis,

You can create a visual graph to track the reaction's progress.

This plot can show how concentration decreases as the reactants are consumed, or how it stabilizes when the reaction reaches equilibrium. Such analysis not only aids in understanding the dynamics of the reaction but also supports the calculation of

• instantaneous rates by examining the slope of the tangent at any given point on the graph.

Isomerization

Isomerization is a chemical process where a molecule transforms into another molecule with the same molecular formula but a different structure. This change is typically internal and involves the rearrangement of atoms rather than their addition or removal.
In the exercise:

• Methyl isonitrile (\(\mathrm{CH}_{3}\mathrm{NC}\)) undergoes isomerization to become acetonitrile (\(\mathrm{CH}_{3}\mathrm{CN}\)).
• Both these compounds have the same molecular formula, \(\mathrm{C}_2\mathrm{H}_3\mathrm{N}\), but differ in the arrangement of the atoms.

Understanding this transformation helps in predicting behavior and properties of substances as they transform. Isomerizations are common in organic chemistry and are important for processes in pharmaceuticals and chemicals.
This process is significant because it often impacts the stability, reactivity, and characteristics of the molecules involved, providing essential insights into the potential uses and functions of different chemical species.