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 and units of the first-order rate constant: $$ \begin{array}{cc} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{kPa}) \\ \hline 0 & 101.3 \mathrm{kPa} \\ 2500 & 95.95 \mathrm{kPa} \\ 5000 & 90.69 \mathrm{kPa} \\ 7500 & 85.92 \mathrm{kPa} \\ 10,000 & 81.36 \mathrm{kPa} \\ \hline \end{array} $$

Short answer

The magnitude of the first-order rate constant for the reaction \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) + \mathrm{Cl}_{2}(g)\) is approximately \(8.78 \times 10^{-4}\) and the units are \(\mathrm{s^{-1}}\).

Step-by-step solution

Write down the integrated rate law for a first-order reaction

The integrated rate law for a first-order reaction is: \[ln\frac{[A]_0}{[A]_t} = kt\] Where: - \([A]_0\) is the initial concentration (pressure in this case) of the reactant at time \(t = 0\) - \([A]_t\) is the concentration (pressure) of the reactant at time \(t\) - \(k\) is the rate constant - \(t\) is the time elapsed

Choose two data points from the table

We will choose two data points from the given table to make a comparison: - Time (s) = 0, Pressure (kPa) = 101.3 - Time (s) = 2500, Pressure (kPa) = 95.95

Apply the integrated rate law

Plugging the selected data points into the integrated rate law equation, we get: \[ln\frac{101.3}{95.95} = k \cdot 2500\]

Solve for the rate constant k

Solve the equation for \(k\): \[k = \frac{ln\frac{101.3}{95.95}}{2500}\] Calculate \(k\): \[k = \frac{ln\frac{101.3}{95.95}}{2500} \approx 8.78 \times 10^{-4} \mathrm{s^{-1}}\] The magnitude of the first-order rate constant is \(8.78 \times 10^{-4}\) and the units are \(\mathrm{s^{-1}}\).