Question

From the following data for the first-order gas-phase isomerization of \(\mathrm{CH}_{3} \mathrm{NC}\) at \(215^{\circ} \mathrm{C}\), calculate the firstorder rate constant and half-life for the reaction: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure CH }_{3} \text { NC (torr) } \\ \hline 0 & 502 \\ 2,000 & 335 \\ 5,000 & 180 \\ 8,000 & 95.5 \\ 12,000 & 41.7 \\ 15,000 & 22.4 \end{array} $$

Short answer

The average rate constant for the first-order gas-phase isomerization of CH3NC is \(5.80 \times 10^{-4} s^{-1}\), and the half-life of the reaction is approximately 1194.8 seconds.

Step-by-step solution

Write the integrated rate law for a first-order reaction

For a first-order reaction, the integrated rate law is expressed as follows: \[k = \frac{2.303}{t} \log{\frac{P_0}{P_t}}\] Where \(k\) is the rate constant, \(t\) is the time, \(P_0\) is the initial pressure, and \(P_t\) is the pressure at time \(t\).

Calculate the rate constant for each data point

Using the given data, we will calculate the rate constant for each data point using the integrated rate law. Since the initial pressure (\(P_0\)) is the same (502 torr) for each data point, we can reuse this value in our calculations: - At t = 2,000 s: \(k = \frac{2.303}{2000} \log{\frac{502}{335}}\) - At t = 5,000 s: \(k = \frac{2.303}{5000} \log{\frac{502}{180}}\) - At t = 8,000 s: \(k = \frac{2.303}{8000} \log{\frac{502}{95.5}}\) - At t = 12,000 s: \(k = \frac{2.303}{12000} \log{\frac{502}{41.7}}\) - At t = 15,000 s: \(k = \frac{2.303}{15000} \log{\frac{502}{22.4}}\) Calculate the numerical values for each rate constant to get: - \(k_1 = 5.27 \times 10^{-4} s^{-1}\) - \(k_2 = 5.62 \times 10^{-4} s^{-1}\) - \(k_3 = 6.07 \times 10^{-4} s^{-1}\) - \(k_4 = 5.97 \times 10^{-4} s^{-1}\) - \(k_5 = 6.08 \times 10^{-4} s^{-1}\)

Determine the average rate constant for the reaction

Take the average of the calculated rate constants from Step 2: \[k_{avg} = \frac{k_1 + k_2 + k_3 + k_4 + k_5}{5}\] \[k_{avg} = \frac{5.27 \times 10^{-4}+5.62 \times 10^{-4} + 6.07 \times 10^{-4} + 5.97 \times 10^{-4} + 6.08 \times 10^{-4}}{5}\] \[k_{avg} = 5.80 \times 10^{-4} s^{-1}\] The average rate constant for the first-order gas-phase isomerization of CH3NC is \(5.80 \times 10^{-4} s^{-1}\).

Calculate the half-life of the reaction

For a first-order reaction, the half-life (\(t_{1/2}\)) is calculated using the following formula: \[t_{1/2} = \frac{0.693}{k}\] Thus, plug in our average rate constant: \[t_{1/2} = \frac{0.693}{5.80 \times 10^{-4} s^{-1}}\] \[t_{1/2} = 1194.8 s\] The half-life of the first-order gas-phase isomerization of CH3NC is approximately 1194.8 seconds.

Key concepts

Rate Constant Calculation

In the study of chemical kinetics, understanding how to calculate the rate constant for a reaction is pivotal. For first-order reactions, the calculation uses the integrated rate law formula:\[ k = \frac{2.303}{t} \log\left(\frac{P_0}{P_t}\right) \]Here, \( k \) is the rate constant, \( t \) is the time elapsed, \( P_0 \) is the initial pressure, and \( P_t \) is the pressure of the reactant at time \( t \). This formula arises from the natural logarithmic relationship between the concentration (or pressure, in gaseous reactions) of the reactant and time in first-order reactions.
To find \( k \), you input the known values from your data—initial and current pressures, and the time elapsed—and solve for the logarithmic relationship.
For multiple data points, as is common in experiments, calculating \( k \) for each will yield individual rate constants. An average of these values gives a reliable measurement, considering slight experimental variations.
Consistently finding a similar rate constant confirms the reaction is indeed first-order.

Half-life of Reaction

The half-life \((t_{1/2})\) of a reaction is the time taken for half of the reactant to be consumed. For a first-order reaction like the isomerization of \( \mathrm{CH}_3 \mathrm{NC} \), it remains consistent regardless of the starting concentration or pressure.The formula to calculate the half-life in a first-order reaction is:\[ t_{1/2} = \frac{0.693}{k} \]This constant nature of half-life in first-order reactions is particularly useful. It allows predictions of reaction progress over time without needing constant recalculation of decay times.
Using the average rate constant \( k \) obtained from observations, \( t_{1/2} \) can be easily found. For instance, if the average rate constant \( k \) calculated is \( 5.80 \times 10^{-4} \) s\(^{-1}\), the half-life is calculated as \( 1194.8 \) seconds.
This simplification is advantageous during kinetic studies and practical applications, such as estimating when a certain percentage of reactants will have transformed.

Gas-phase Isomerization

Gas-phase isomerization reactions involve the transformation of one isomer into another in the gaseous state. These reactions are significant for understanding chemical kinetics and mechanisms. In our context, the isomerization of \( \mathrm{CH}_3 \mathrm{NC} \) into another species demonstrates a classic first-order reaction.First-order gas-phase isomerizations are crucial because:

• They provide insights into molecular stability and reactivity.
• They are often simple to follow experimentally, providing clear kinetic data.
• They establish the foundational principles used to understand more complex reaction systems.

In these reactions, the process entails breaking and reforming of bonds, often leading to different spatial configurations of atoms.
Understanding the rate at which isomerization occurs (via rate constants and half-lives) informs chemists about reaction dynamics, necessary conditions for control, and how to optimize reactions for desired outcomes.
Gas-phase reactions, given their homogenous nature, allow for more accurate studies of reaction kinetics, which are crucial in both fundamental and applied chemistry.