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 $t=8000 \mathrm{~s}$.

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#

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#

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#

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.