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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: $$ \begin{array}{rc} \hline \text { Time (s) } & {\left[\mathrm{CH}_{3} \mathrm{NC}\right](M)} \\ \hline 0 & 0.0165 \\ 2000 & 0.0110 \\ 5000 & 0.00591 \\ 8000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (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 \(t=15,000 \mathrm{~s} .(\mathbf{c})\) Which is greater, the average rate between \(t=2000\) and \(t=12,000 \mathrm{~s}\), or between \(t=8000\) and \(t=15,000 \mathrm{~s} ?(\mathbf{d})\) 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
In summary, the average rates of reaction for various time intervals are: - Between \(t=0\) and \(t=2000\) s: \(2.75 \times 10^{-6} \frac{M}{s}\) - Between \(t=2000\) and \(t=5000\) s: \(1.70 \times 10^{-6} \frac{M}{s}\) - Between \(t=5000\) and \(t=8000\) s: \(9.23 \times 10^{-7} \frac{M}{s}\) - Between \(t=8000\) and \(t=12000\) s: \(4.43 \times 10^{-7} \frac{M}{s}\) - Between \(t=12000\) and \(t=15000\) s: \(2.10 \times 10^{-7} \frac{M}{s}\) The overall average rate of reaction is \(1.05 \times 10^{-6} \frac{M}{s}\). The average rate between \(t=2000\) and \(t=12000\) s is greater than the average rate between \(t=8000\) and \(t=15000\) s. To find the instantaneous rates of reaction at \(t=5000\) s and \(t=8000\) s, plot the concentration data as a function of time and calculate the slope of the tangent line at these points.

Step by step solution

01

Calculate the average rate of reaction

The average rate of reaction is given by the formula: \[ \text{Average rate} = \frac{\Delta [\text{concentration}]}{\Delta \text{time}} \] We will use this formula to calculate the average rate of reaction between each measurement. 1. Between \(t=0\) and \(t=2000\) s: \[ \text{Average rate} = \frac{0.0165 - 0.0110}{2000 - 0} = \frac{0.0055}{2000} = 2.75 \times 10^{-6} \frac{M}{s} \] 2. Between \(t=2000\) and \(t=5000\) s: \[ \text{Average rate} = \frac{0.0110 - 0.00591}{5000 - 2000} = \frac{0.00509}{3000} = 1.70 \times 10^{-6} \frac{M}{s} \] 3. Between \(t=5000\) and \(t=8000\) s: \[ \text{Average rate} = \frac{0.00591 - 0.00314}{8000 - 5000} = \frac{0.00277}{3000} = 9.23 \times 10^{-7} \frac{M}{s} \] 4. Between \(t=8000\) and \(t=12000\) s: \[ \text{Average rate} = \frac{0.00314 - 0.00137}{12000 - 8000} = \frac{0.00177}{4000} = 4.43 \times 10^{-7} \frac{M}/{s} \] 5. Between \(t=12000\) and \(t=15000\) s: \[ \text{Average rate} = \frac{0.00137 - 0.00074}{15000 - 12000} = \frac{0.00063}{3000} = 2.10 \times 10^{-7} \frac{M}{s} \] #b) Calculate the average rate of reaction over the entire time of the data#
02

Calculate the overall average rate of reaction

To find the average rate of reaction for the entire time, we will use the formula for the average rate between \(t=0\) and \(t=15000\) s: \[ \text{Average rate} = \frac{0.0165 - 0.00074}{15000 - 0} = \frac{0.01576}{15000} = 1.05 \times 10^{-6} \frac{M}{s} \] #c) Compare the average rate between t=2000 and t=12000 s, and between t=8000 and t=15000 s#
03

Compare the average rates

We have previously calculated the average rates between these time intervals: - Between \(t=2000\) and \(t=12000\) s, the average rate is equal to the sum of the average rates between \(t=2000\), \(t=5000\), \(t=8000\) and \(t=12000\) which is: \(2.75 \times 10^{-6} + 1.70 \times 10^{-6} + 9.23 \times 10^{-7} + 4.43 \times 10^{-7} = 5.66 \times 10^{-6} \frac{M}{s}\) - Between \(t=8000\) and \(t=15000\) s, the average rate is equal to the sum of the average rates between \(t=8000\), \(t=12000\) and \(t=15000\) which is: \(9.23 \times 10^{-7} + 4.43 \times 10^{-7} + 2.10 \times 10^{-7} = 1.58 \times 10^{-6} \frac{M}{s}\) The average rate between \(t=2000\) and \(t=12000\) s is greater than the average rate between \(t=8000\) and \(t=15000\) s. #d) Graph CH3NC vs. time and determine the instantaneous rates at t=5000 s and t=8000 s#
04

Determine the instantaneous rates

We'll graph the concentration of CH3NC as a function of time. Using a graph, we can calculate the slope of the tangent line to the curve at specific time points (like \(t=5000\) and \(t=8000\) s) to find the instantaneous rate. Unfortunately, we cannot draw the graph in this format, so we suggest using any graph plotting tool or software to create a graph with the data provided and then determining the tangent's slopes at \(t=5000\) s and \(t=8000\) s.

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Key Concepts

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

Isomerization
Isomerization is a fascinating chemical process that involves the transformation of a molecule into another molecule with the same atoms, but in a different arrangement. In the context of this exercise, isomerization is demonstrated through the conversion of methyl isonitrile ( CH_3NC ) to acetonitrile ( CH_3CN ) in the gas phase. This rearrangement involves breaking and forming bonds, and it highlights an essential aspect of chemical kinetics, where the molecular structure is altered without changing the chemical formula.
  • It's a vital process in synthetic chemistry.
  • Plays a role in many biological pathways.
  • It can affect physical properties like boiling point and solubility.
Understanding isomerization helps us grasp how molecular changes can impact reactions and processes.
Average Reaction Rate
The average reaction rate gives us an overview of how fast a reaction is occurring over a specific time interval. It is calculated by measuring how much the concentration of a reactant or product changes over a given time. In this exercise, we use the formula:\[\text{Average rate} = \frac{\Delta [\text{concentration}]}{\Delta \text{time}}\]This allows us to see the reaction's speed between each measurement.
  • Between \(t=0\) to \(t=2000\) s, we saw a rate of \(2.75 \times 10^{-6} M/s\).
  • The rate decreases over time, indicating the reaction slows as it proceeds.
  • This approach is great for comparing reaction rates over different time spans.
Using average rates can help scientists determine the efficiency of chemical processes.
Instantaneous Reaction Rate
The instantaneous reaction rate offers a snapshot of how quickly a reaction is happening at a specific moment. Unlike the average rate, which covers a time span, the instantaneous rate is the slope of the tangent line at a particular point on a concentration vs. time graph.
  • Provides precise information on reaction speed at specific times.
  • Useful for understanding dynamic changes in reaction mechanisms.
  • Calculated by finding the derivative of the concentration concerning time.
You can visualize this by plotting the concentration vs. time data and finding the tangent's slope at the point of interest, such as \(t=5000\) or \(t=8000\) seconds.
Concentration vs. Time Graph
A concentration vs. time graph is a powerful tool in chemical kinetics, illustrating how the concentration of a reactant or product changes over time. By plotting the concentration of CH_3NC against time from the data given, we can analyze both average and instantaneous rates.
  • Helps in visualizing the progress of a reaction.
  • Allows the determination of reaction order.
  • Essential for understanding the reaction’s kinetics.
This graph shows the downward trend in concentration as the reaction proceeds. The steeper the curve, the faster the reaction at that point. Software tools and graphing calculators can be used to create these graphs, enhancing analysis and understanding of chemical kinetics.

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Most popular questions from this chapter

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O}\). At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1} .\) Calculate the half- life at this temperature. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

The dimerization of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) to \(\mathrm{C}_{4} \mathrm{~F}_{8}\) has a rate constant \(k=0.045 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(450 \mathrm{~K} .\) (a) Based on the unit of \(k\) what is the reaction order in \(\mathrm{C}_{2} \mathrm{~F}_{4} ?(\mathbf{b})\) If the initial concentration of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) is \(0.100 \mathrm{M}\), how long would it take for the concentration to decrease to \(0.020 \mathrm{M}\) at \(450 \mathrm{~K}\) ?

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(\mathrm{Cl}_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l} \text { Step } 1: \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X} \\\ \text { Step } 2: \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D} \end{array} $$ \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

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