Question

The reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude of the first-order rate constant: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { (atm) } \\ \hline 0 & 1.000 \\ 2,500 & 0.947 \\ 5,000 & 0.895 \\ 7,500 & 0.848 \\ 10,000 & 0.803 \\ \hline \end{array} $$

Short answer

The magnitude of the first-order rate constant for the given reaction is approximately \(2.25 \times 10^{-5} \ \mathrm{s}^{-1}\).

Step-by-step solution

Understand the rate law for first-order reactions

For a first-order reaction, we can express the rate law as: $$ - \frac{d[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}{dt} = k [\mathrm{SO}_{2} \mathrm{Cl}_{2}] $$ Where \(k\) is the rate constant that we are looking for, and [\(\mathrm{SO}_{2}\mathrm{Cl}_{2}\)] represents the concentration of SO2Cl2 at any given time. Now we need to integrate this equation to find its relationship with time.

Integrate the rate equation

Integrate the rate equation with respect to time, and then separate the variables: $$ -\int\frac{d[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]} = \int k dt $$ This gives us: $$ -ln\ \frac{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]_{0}} = kt $$ Where \([\mathrm{SO}_{2}\mathrm{Cl}_{2}]_{0}\) is the initial concentration of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\).

Convert pressure data to concentration data

Since we are given pressure data, we can use the ideal gas formula to convert it into concentration data: $$ P = \frac{n}{V}RT = \frac{[\mathrm{SO}_{2} \mathrm{Cl}_{2}]}{M}RT $$ Where \(P\) is the partial pressure of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\), \(n\) is the number of moles, \(V\) is the volume, \(R\) is the ideal gas constant, \(T\) is the temperature, and \(M\) is the molar mass of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\). Since the volume, temperature, and molar mass remain constant during the reaction, the concentration of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) is proportional to its partial pressure: $$ [\mathrm{SO}_{2} \mathrm{Cl}_{2}] = k_{p}\times P $$ Where \(k_{p}\) is a proportionality constant.

Use the given data to calculate the rate constant

Since we have found that the concentration of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) (\(k_p\)×\(P\)) is proportional to its partial pressure, we can replace the original equation term with pressure and the proportionality constant. Then, we can plug in the data from any two time and pressure values to find the value of the rate constant: $$ -ln\ \frac{k_p \times P_{t}}{k_p \times P_{0}} = kt $$ Using data from time 0 and 5000, \( -ln\ \frac{0.895}{1} = k\times 5,000 \)

Solve for the rate constant

Rearrange the equation to solve for \(k\): $$ k = \frac{-ln\ \frac{0.895}{1}}{5,000} $$ Now, calculate the value of \(k\): $$ k \approx 2.25 \times 10^{-5} \ \mathrm{s}^{-1} $$ Therefore, the magnitude of the first-order rate constant for the given reaction is approximately \(2.25 \times 10^{-5} \ \mathrm{s}^{-1}\).