Question

As described in Exercise 14.41, the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 450 torr, what is the partial pressure of this substance after \(60 \mathrm{~s}\) ? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

Short answer

(a) The partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after 60 seconds is approximately 30.15 torr. (b) It takes approximately 51.18 seconds for the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decline to one-tenth its initial value.

Step-by-step solution

Identify the first-order reaction formula

A first-order reaction usually follows the equation: \[ ln(\frac{P_{t}}{P_{0}}) = -kt \] where: - \(P_{0}\) is the initial partial pressure of the substance, - \(P_{t}\) is the partial pressure of the substance at time \(t\), - \(t\) is the time (in seconds), - \(k\) is the rate constant which is given as \(4.5 \times 10^{-2} s^{-1}\), and - \(ln()\) denotes the natural logarithm.

Calculate the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after 60 seconds

We are given that the initial partial pressure \(P_{0} = 450\) torr and have to find the partial pressure after \(t = 60\) seconds. Plug these values into the first-order reaction formula: \[ ln(\frac{P_{60}}{450}) = -4.5 \times 10^{-2} \times 60 \] Now, we need to calculate the value of \(P_{60}\) (partial pressure after 60 seconds).

Solve for the partial pressure after 60 seconds

First, let's calculate the product of the rate constant and the time: \[ -4.5 \times 10^{-2} \times 60 = -2.7 \] So the equation becomes: \[ ln(\frac{P_{60}}{450}) = -2.7 \] Now, we can find the value of the fraction inside the logarithm by raising the constant to e: \[ \frac{P_{60}}{450} = e^{-2.7} \] Calculate \(e^{-2.7}\): \[ e^{-2.7} ≈ 0.067 \] Now, multiply both sides by 450 to find the value of \(P_{60}\): \[ P_{60} ≈ 450 \times 0.067 \approx 30.15\, torr \] So the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after 60 seconds is approximately 30.15 torr.

Calculate the time for the partial pressure to decline to one-tenth the initial value

We are supposed to find the time when the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) becomes one-tenth of its initial value. In other words, we need to find the time when: \[ P_{t} = \frac{1}{10} \times P_{0} \] So, let's plug the values into the first-order reaction formula: \[ ln(\frac{P_{t}}{P_{0}}) = ln(\frac{\frac{1}{10} \times P_{0}}{P_{0}}) = -kt \]

Solve for the time

First, we can simplify the equation: \[ ln(\frac{\frac{1}{10} \times 450}{450}) = -4.5 \times 10^{-2} \times t \] \[ ln(\frac{1}{10}) = -4.5 \times 10^{-2} \times t \] Now, calculate the natural logarithm of \(\frac{1}{10}\): \[ ln(\frac{1}{10}) = -2.303 \] So the equation becomes: \[ -2.303 = -4.5 \times 10^{-2} \times t \] Now, let's find the time \(t\): \[ t \approx \frac{-2.303}{-4.5 \times 10^{-2}} \approx 51.18\, seconds \] Thus, it takes approximately 51.18 seconds for the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decline to one-tenth its initial value.

Key concepts

Rate Constant Calculation

In chemical kinetics, the rate constant is a crucial value that helps us understand the speed at which a reaction occurs. For first-order reactions, the rate constant, denoted by \( k \), has units of \( ext{time}^{-1} \), often expressed as \( ext{s}^{-1} \) in reactions involving gases.

The rate constant can be found using experimental data that track the concentration or pressure changes of reactants over time. The general expression for a first-order reaction is given by the equation:

\[ ln(\frac{P_{t}}{P_{0}}) = -kt \]

where:

• \( P_{t} \) is the partial pressure at time \( t \),
• \( P_{0} \) is the initial partial pressure,
• \( k \) is the rate constant,
• \( t \) is time.

To calculate the rate constant, you generally need to rearrange this formula and use data from controlled laboratory experiments. Knowing \( k \) allows predictions about how fast the reaction progresses under the same conditions.

Partial Pressure Calculations

Partial pressure is essential in the study of reactions involving gases, as it tells us the pressure each gas in a mixture contributes to the total pressure. In the context of first-order reactions, partial pressures can determine how a gas's reactant quantity changes over time.

Given an initial partial pressure \( P_{0} \) and wanting to find the partial pressure \( P_{t} \) at any time \( t \), the first-order kinetic equation can be utilized:

Using the equation \[ ln(\frac{P_{t}}{P_{0}}) = -kt \], we can solve for \( P_{t} \):

• Rearrange to find \( P_{t} = P_{0} \times e^{-kt} \).
• This formula demonstrates how the partial pressure decreases exponentially with time, characteristic of first-order kinetics.

If you know \( k \) and the initial pressure, you can determine \( P_{t} \) for any \( t \). This is essential for understanding how quickly a reactant diminishes in a reaction.

Natural Logarithm in Kinetics

The natural logarithm (\( ln \)) is integral to first-order reactions. It appears in the main kinetics equation and provides a way to linearize exponential decay in processes involving pressure or concentration changes over time. Generally, the reason we use the natural logarithm is that many decay processes naturally follow exponential functions.

In the equation \( ln(\frac{P_{t}}{P_{0}}) = -kt \), the \( ln \) helps transform the ratios of pressures into a form that’s easily interpretable and mathematically manageable. Because of \( ln \)’s property of turning multiplication into addition and simplifying complex exponential relationships, it’s used to derive rates of change.

• The natural logarithm effectively simplifies complex decay dynamics into a linear function.
• In experiments, plotting \( ln(P_{t}/P_{0}) \) against time provides a straight line with a slope of \( -k \), confirming the reaction is first-order.

Understanding \( ln \) in kinetics helps students better grasp how scientists model and interpret reaction dynamics.