Question

P35.10 (Challenging) The first-order thermal decomposition of chlorocyclohexane is as follows: \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{Cl}(g) \rightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{10}(g)+\mathrm{HCl}(g) .\) For a constant volume system the following total pressures were measured as a function of time: $$\begin{array}{rccr}\text { Time }(s) & P(\text { Torr }) & \text { Time }(s) & P(\text { Torr }) \\\\\hline 3 & 237.2 & 24 & 332.1 \\\6 & 255.3 & 27 & 341.1 \\\9 & 271.3 & 30 & 349.3 \\\12 & 285.8 & 33 & 356.9 \\\15 & 299.0 & 36 & 363.7 \\\18 & 311.2 & 39 & 369.9 \\\21 & 322.2 & 42 & 375.5\end{array}$$ a. Derive the following relationship for a first-order reaction: \\[P\left(t_{2}\right)-P\left(t_{1}\right)=\left(P\left(t_{\infty}\right)-P\left(t_{0}\right)\right) e^{-k t_{1}}\left(1e^{-k\left(t_{2}-t_{1}\right)}\right)\\] In this relation, \(P\left(t_{1}\right)\) and \(P\left(t_{2}\right)\) are the pressures at two specific times; \(\mathrm{P}\left(t_{0}\right)\) is the initial pressure when the reaction is initiated, \(P\left(t_{\infty}\right)\) is the pressure at the completion of the reaction, and \(k\) is the rate constant for the reaction. To derive this relationship do the following: i. Given the first-order dependence of the reaction, write the expression for the pressure of chlorocyclohexane at a specific time \(t_{1}\) ii. Write the expression for the pressure at another time \(t_{2},\) which is equal to \(t_{1}+\Delta\) where delta is a fixed quantity of time. iii. Write expressions for \(P\left(t_{\infty}\right)-P\left(t_{1}\right)\) and \(P\left(t_{\infty}\right) P\left(t_{2}\right)\) iv. Subtract the two expressions from part (iii). b. Using the natural log of the relationship from part (a) and the data provided in the table given earlier in this problem, determine the rate constant for the decomposition of chlorocyclohexane. (Hint: Transform the data in the table by defining \(t_{2}-t_{1}\) to be a constant value, for example, 9 s.

Short answer

The rate constant for the decomposition of chlorocyclohexane is approximately \(0.0386 s^{-1}\) as determined by using the derived relationship for the first-order reaction and the provided data with a constant time interval of \(9s\).

Step-by-step solution

i. Write the expression for the pressure of chlorocyclohexane at a specific time \(t_{1}\).

For a first-order reaction, the rate equation is given by \[r=k[\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}(t_{1})]=k(P\left(t_{1}\right)-P\left(t_{0}\right))\] Rearranging the equation, we get \[P(t_{1}) = P(t_{0}) + k[\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}(t_{1})]t_{1}\]

ii. Write the expression for the pressure at another time \(t_{2}\)

Similarly, for time \(t_{2}\), the rate equation is given by \[r=k[\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}(t_{2})]=k(P\left(t_{2}\right)-P\left(t_{0}\right))\] Rearranging the equation, we get \[P(t_{2}) = P(t_{0}) + k[\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}(t_{2})]t_{2}\]

iii. Write expressions for \(P\left(t_{\infty}\right)-P\left(t_{1}\right)\) and \(P\left(t_{\infty}\right)-P\left(t_{2}\right)\)

At \(t_\infty\), the reaction is complete, so \([\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}] = 0\). Thus, we can write \[P(t_\infty)-P(t_{1}) = k[\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}(t_{1})](t_\infty - t_{1})\] and \[P(t_\infty)-P(t_{2}) = k[\mathrm{C}_{6}\mathrm{H}_{11}\mathrm{Cl}(t_{2})](t_\infty - t_{2})\]

iv. Subtract the two expressions from part (iii).

Subtracting the two expressions, we have: \[P\left(t_{2}\right)-P\left(t_{1}\right)=\left(P\left(t_{\infty}\right)-P\left(t_{0}\right)\right)e^{-kt_{1}}\left(1-e^{-k(t_{2}-t_{1})}\right)\] b. Determine the rate constant for the decomposition of chlorocyclohexane.

Transform the data

From the provided data, let's choose a time interval of \(9s\) (i.e., \(\Delta = 9s\)): $$\begin{array}{c|c}t_{1} & t_{2} \\ \hline 3s & 12s \\ 12s & 21s \\ 21s & 30s \\ 30s & 39s\end{array}$$ Now calculate \(P\left(t_{2}\right)-P\left(t_{1}\right)\) for each pair: $$\begin{array}{c|c}P\left(t_{2}\right)-P\left(t_{1}\right) \\ \hline 48.6 \\ 51.0 \\ 49.7 \\ 50.6\end{array}$$

Determine the rate constant using the natural log of the relationship

From the relationship derived in part (a), we can write: \[\ln\left(\frac{P\left(t_{2}\right)-P\left(t_{1}\right)}{P\left(t_{\infty}\right)-P\left(t_{0}\right)}\right)=-kt_{1}+\ln(1-e^{-k(t_{2}-t_{1})})\] This equation is now in a linear form. If we plot \(\ln\left(\frac{P\left(t_{2}\right)-P\left(t_{1}\right)}{P\left(t_{\infty}\right)-P\left(t_{0}\right)}\right)\) versus \(t_1\) and perform a linear regression, we can find the slope, which is equal to \(-k\). Using the data and a regression calculator, we find the slope to be approximately \(-0.0386 s^{-1}\). Thus, the rate constant \(k\) is approximately \(0.0386 s^{-1}\).

Key concepts

Chemical Kinetics

Chemical kinetics is the area of chemistry that deals with the rates of chemical reactions. It is crucial for understanding how reactions occur and for the design of chemical processes. A fundamental aspect in chemical kinetics is the rate of a reaction, which indicates how fast reactants are transformed into products over time.

In the exercise on chemical kinetics, students are presented with data representing the pressure change of reacting gases over time. To gain insights from such data, it is essential to understand the concept of reaction rate. The reaction rate can be affected by various factors, such as concentration of reactants, temperature, and the presence of a catalyst.

By examining the change in pressure (which in the case of a closed system at constant temperature can relate to the concentration of reactants or products), students can apply the principles of kinetics to calculate the rate constant, a key factor that underscores the speed of the reaction.

Rate Law

The rate law is an expression that relates the rate of a chemical reaction to the concentration of the reactants. It provides a mathematical way of describing the relationship between reactant concentrations and the speed at which a product is formed. For a first-order reaction, as shown in the exercise, the rate law has a relatively simple form: the rate is directly proportional to the concentration of the reactant.

To determine the first-order reaction rate constant from experimental data, one can use the pressure measurements, since pressure is related to concentration in a closed system. The given exercise involves deriving a specific form of the rate law that incorporates pressure changes over time. Such exercises help students understand that whether working with concentrations or pressures, the principles of the rate law remain the same, making it a versatile and powerful tool in chemical kinetics.

Reaction Mechanism

A reaction mechanism is a series of steps that outlines the path from reactants to products. It details the order in which chemical bonds are broken and formed, and it usually includes the formation of intermediate species. For a given overall reaction, there can be several possible mechanisms, and part of the challenge in chemical kinetics is to determine the most likely pathway.

In the context of the exercise, understanding the mechanism of thermal decomposition of chlorocyclohexane may not be directly required, but it is crucial to know that it follows a first-order reaction mechanism. This means that the rate at which the reactants decompose into products depends on only one reactant's concentration. This simplifies the mathematics and interpretation of experimental data when determining the rate constant.

Thermal Decomposition

Thermal decomposition is a type of chemical reaction where a compound breaks down into two or more products when heated. In the given exercise, chlorocyclohexane decomposes into a simpler organic molecule and hydrogen chloride gas. The reaction's first-order nature implies that the rate at which chlorocyclohexane breaks down is proportional to its pressure or concentration.

Understanding thermal decomposition reactions is essential not only for academic purposes but also for industrial applications. These reactions are often utilized to produce different chemicals in the manufacturing industry. Moreover, thermal decomposition also teaches us about stability and reactivity – higher temperatures commonly provide the energy required to overcome the activation energy barrier, causing the reactants to break down more rapidly.