Problem 75 - Chemistry: An Atoms-Focused Approach

Question

The dimerization of ClO, $$2 \mathrm{ClO}(g) \rightarrow \mathrm{Cl}_{2} \mathrm{O}_{2}(g)$$ is second order in ClO. a. Use the following data to determine the value of \(k\) at \(298 \mathrm{K}\) $$\begin{array}{cc} \text { Time (s) } & \text { [ClO] (molecules/cm }^{3} \text { ) } \\ 0 & 2.60 \times 10^{11} \\ \hline 1.00 & 1.08 \times 10^{11} \\ \hline 2.00 & 6.83 \times 10^{10} \\ \hline 3.00 & 4.99 \times 10^{10} \\ \hline 4.00 & 3.93 \times 10^{10} \\ \hline \end{array}$$ b. Determine the half-life for the dimerization of C1O.

Step-by-Step Solution

Answer

Question: Determine the value of the rate constant and the half-life of the dimerization of ClO using the provided data. Answer: a. The rate constant, k, at 298K is approximately 2.54 × 10^(-11) s^(-1). b. The half-life of the dimerization of ClO is approximately 1.54 seconds.

Step 1: Write the equation for a second-order reaction

For a second-order reaction, we can use the following equation: $$\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$$ Here, \([A]_t\) : concentration of ClO at time t, \([A]_0\) : initial concentration of ClO, \(k\) : rate constant and \(t\) : time.

Step 2: Use the given data to determine the value of k

Let's choose two datapoints from the table provided; for example, the first and the second row: $$[A]_0 = 2.60 \times 10^{11} \mathrm{molecules/cm^3}$$ $$[A]_1 = 1.08 \times 10^{11} \mathrm{molecules/cm^3} \text{ at } t = 1.00\,s$$ Substitute these values into the second-order equation and solve for k. $$\frac{1}{1.08 \times 10^{11}} - \frac{1}{2.60 \times 10^{11}} = k(1.00)$$ Solve for k: $$k = \frac{1}{1.00}\left(\frac{1}{1.08 \times 10^{11}} - \frac{1}{2.60 \times 10^{11}}\right)$$ $$k \approx 2.54 \times 10^{-11} \mathrm{s^{-1}}$$

Step 3: Determine the half-life of the reaction

For a second-order reaction, the half-life, \(t_\frac{1}{2}\), can be calculated via: $$t_\frac{1}{2} = \frac{1}{k[A]_0}$$ Substitute for k and \([A]_0\) using values obtained in Step 2: $$t_\frac{1}{2} = \frac{1}{(2.54 \times 10^{-11}\,\mathrm{s^{-1}})(2.60 \times 10^{11}\,\mathrm{molecules/cm^3})}$$ Calculate the half-life: $$t_\frac{1}{2} \approx 1.54 \,s$$ To summarize the solution: a. The value of the rate constant, \(k\), at \(298\,\mathrm{K}\) is approximately \(2.54 \times 10^{-11}\,\mathrm{s^{-1}}\). b. The half-life for the dimerization of ClO is approximately \(1.54\,s\).

Key Concepts

Dimerization of ClO

Dimerization is a process where two identical molecules combine to form a single unit called a dimer. In the case of the ClO molecule, it undergoes dimerization as follows: \[2 \mathrm{ClO}(g) \rightarrow \mathrm{Cl}_{2} \mathrm{O}_{2}(g)\]This reaction is classified as a second-order reaction based on ClO. This means that the rate at which the reaction occurs is proportional to the square of the concentration of ClO. The study of such reactions is important for understanding how reaction rates change with concentration.In a second-order reaction, the formula used is essential: \[\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\]Here,

• \([A]_t\) represents the concentration of the reactant at time \(t\).
• \([A]_0\) is the initial concentration.
• \(k\) is the rate constant, a crucial value that reflects how fast the reaction happens at a given temperature.

Understanding the dimerization of ClO is critical not just for chemical kinetics but also in atmospheric chemistry, where such reactions can influence processes like ozone depletion.

Rate Constant Calculation

The rate constant, \(k\), is pivotal in defining the speed of a second-order reaction. It's specific to each reaction and influenced by temperature.To calculate \(k\) for the ClO dimerization at 298 K, we use the second-order reaction formula: \[\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\]Selecting data points from the given time-concentration table, such as the initial state (time = 0 seconds) and one second later, helps in determining \(k\). Let's see the steps:

• Initial concentration, \([A]_0 = 2.60 \times 10^{11}\) molecules/cm³.
• Concentration at \(t = 1.00\) s, \([A]_1 = 1.08 \times 10^{11}\) molecules/cm³.

By plugging these into the equation:\[\frac{1}{1.08 \times 10^{11}} - \frac{1}{2.60 \times 10^{11}} = k(1.00)\]Solving gives \(k \approx 2.54 \times 10^{-11} \mathrm{s^{-1}}\). This rate constant tells us how quickly ClO molecules dimerize in the given conditions. Such calculations are fundamental in predictive modeling, where knowing how fast reactions proceed aids in simulating various chemical systems.

Half-Life Determination

The half-life of a reaction tells us how long it takes for half of a substance to react or decay. In the context of second-order reactions, this concept is even more intriguing.For a second-order reaction like the dimerization of ClO, the equation for half-life \(t_{1/2}\) differs from first-order reactions.The formula is given by:\[t_\frac{1}{2} = \frac{1}{k[A]_0}\]Plug in the values obtained earlier:

• The rate constant \(k = 2.54 \times 10^{-11} \mathrm{s^{-1}}\).
• The initial concentration \([A]_0 = 2.60 \times 10^{11}\) molecules/cm³.

By substituting these into the formula, the half-life comes out to:\[t_{1/2} \approx 1.54 \mathrm{s}\]This means that in approximately 1.54 seconds, half the initial concentration of ClO would have dimerized to form \(\mathrm{Cl}_2\mathrm{O}_2\). This kind of calculation is useful not only in chemical laboratories but also in industries where reaction completion times are crucial for processes.