Question

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?

Short answer

The rate constant at 600 K is approximately \(3.01 \times 10^{-6} \mathrm{~s}^{-1}\). The half-life at 320°C is approximately \(3.14 \times 10^{4} \mathrm{~s}\).

Step-by-step solution

(Part a) Find the rate constant at 600 K

Given the half-life \(t_{1/2} = 2.3 \times 10^{5}\mathrm{~s}\) and the formula for the half-life of a first-order reaction \(t_{1/2} = \frac{0.693}{k}\). Now let's solve for the rate constant \(k\): 1. Use the half-life formula: \(t_{1/2} = \frac{0.693}{k}\). 2. Input the given value of \(t_{1/2} = 2.3 \times 10^{5}\mathrm{~s}\) to the formula: \(2.3 \times 10^{5}\mathrm{~s} = \frac{0.693}{k}\). 3. To find the value of \(k\), multiply both sides by \(k\): \(k \times (2.3 \times 10^{5}\mathrm{~s}) = 0.693\). 4. Finally, divide both sides by \((2.3 \times 10^{5}\mathrm{~s})\): \(k = \frac{0.693}{2.3 \times 10^{5}\mathrm{~s}}\). 5. Calculate the value of \(k\): \(k \approx 3.01 \times 10^{-6} \mathrm{~s}^{-1}\). The rate constant at 600 K is approximately \(3.01 \times 10^{-6} \mathrm{~s}^{-1}\).

(Part b) Find the half-life at 320°C

Given the rate constant \(k = 2.2 \times 10^{-5} \mathrm{~s}^{-1}\) at 320°C, let's find the half-life \(t_{1/2}\) using the formula for the half-life of a first-order reaction \(t_{1/2} = \frac{0.693}{k}\): 1. Use the half-life formula: \(t_{1/2} = \frac{0.693}{k}\). 2. Input the given value of \(k = 2.2 \times 10^{-5} \mathrm{~s}^{-1}\) to the formula: \(t_{1/2} = \frac{0.693}{2.2 \times 10^{-5} \mathrm{~s}^{-1}}\). 3. Calculate the value of \(t_{1/2}\): \(t_{1/2} \approx 3.14 \times 10^{4} \mathrm{~s}\). The half-life at 320°C is approximately \(3.14 \times 10^{4} \mathrm{~s}\).

Key concepts

Rate constant

Understanding the rate constant is essential when dealing with first-order reactions. The rate constant, denoted as \( k \), offers a way to quantify the speed of a chemical reaction. In first-order reactions, the rate is directly proportional to the concentration of one reactant. Ideal conditions, temperature, and the nature of reactants all affect the value of the rate constant.
For a first-order reaction, the formula \( t_{1/2} = \frac{0.693}{k} \) relates the half-life of the reaction to the rate constant. This expression shows that the half-life is inversely proportional to \( k \). If the rate constant is known, predicting how quick a chemical undergoes change becomes easier.
In the exercise provided, we determined the rate constant at a certain temperature (600 K) using this formula. By knowing the half-life, we plugged it into the equation, rearranged, and solved for \( k \). This calculation helps us understand the inherent speed of the decomposition process at that temperature.

Half-life calculation

Calculating the half-life of a reaction provides critical insights into how long it takes for a substance to reduce to half its original amount. This is particularly important in first-order reactions, like radioactive decay or the decomposition reaction described in the exercise.
The formula \( t_{1/2} = \frac{0.693}{k} \) allows one to calculate the half-life if the rate constant is known. When the half-life is constant, as in first-order reactions, the calculations are straightforward. For example, if you know \( k \), you simply substitute it into the formula to find the half-life. This provides a predictable measure of how fast a reaction proceeds over time.
In part b of the exercise, we applied this concept by using a given rate constant of \( 2.2 \times 10^{-5} \text{s}^{-1} \) at 320°C. Plugging this into the formula gave us a resulting half-life, indicating how quickly half of the original substance decomposes at that condition.

Decomposition reaction

Decomposition reactions are where a single compound breaks down into two or more simpler substances. These reactions are fundamental in chemistry due to their common occurrence and the insight they provide into reaction mechanisms. They are often endothermic, requiring energy input to occur, such as heat, electricity, or light.
In the specific decomposition of \( \text{SO}_2\text{Cl}_2 \) mentioned in the exercise, the compound decomposes into \( \text{SO}_2 \) and \( \text{Cl}_2 \). This is a first-order reaction, meaning the rate depends solely on the concentration of \( \text{SO}_2\text{Cl}_2 \).
Understanding decomposition reactions in terms of kinetics allows chemists to control and predict the reaction conditions and the time scales involved. This is extremely useful in industrial and laboratory settings where precise control of chemical processes is crucial.