Question

(a) The gas-phase decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right), \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(300^{\circ} \mathrm{C}\) the half-life for this process is two and a half days. What is the rate constant at this temperature? (b) At \(400^{\circ} \mathrm{C}\) the rate constant is \(0.19 \mathrm{~min}^{-1}\). What is the half-life at this temperature?

Short answer

(a) The rate constant at \(300^{\circ} \mathrm{C}\) is \(0.277\thinspace\text{day}^{-1}\). (b) The half-life at \(400^{\circ} \mathrm{C}\) is approximately 3.65 minutes.

Step-by-step solution

Recall the formula for half-life of a first-order reaction

For a first-order reaction, the half-life (t₁/₂) and the rate constant (k) are related by the following formula: \[ t_{1/2} = \frac{0.693}{k}\]

Plug in the given half-life and solve for k

We're given the half-life as 2.5 days. Let's plug this value into the formula and solve for the rate constant (k): \[ 2.5 \thinspace \text{days} = \frac{0.693}{k} \] Now we can solve for k: \[ k = \frac{0.693}{2.5\thinspace \text{days}} \approx 0.277\thinspace\text{day}^{-1} \] So, the rate constant at 300°C is \(0.277\thinspace\text{day}^{-1}\). b) Finding the half-life at 400°C

Recall the formula for half-life of a first-order reaction

As previously mentioned, for a first-order reaction, the half-life (t₁/₂) and the rate constant (k) are related by the following formula: \[ t_{1/2} = \frac{0.693}{k} \]

Plug in the given rate constant and solve for half-life

We're given the rate constant as \(0.19\thinspace\text{min}^{-1}\) at 400°C. Let's plug this value into the formula and solve for the half-life (t₁/₂): \[ t_{1/2} = \frac{0.693}{0.19\thinspace\text{min}^{-1}} \approx 3.65\thinspace\text{min} \] So, the half-life at 400°C is approximately 3.65 minutes.

Key concepts

Rate constant calculation

Understanding how to calculate the rate constant for a reaction is a crucial concept in chemistry. The rate constant (\(k\) ) is a number that helps us understand how fast or slow a reaction occurs. For first-order reactions, the rate constant can be directly tied to the half-life of the reaction.
First-order reactions are unique because their rate only depends on the concentration of one reactant. This means even if you have twice the amount of reactant, the change in rate will be proportional. It makes calculations a bit simpler.
For a first-order reaction, we use the formula:

• \( t_{1/2} = \frac{0.693}{k} \)

This formula shows that the half-life, \(t_{1/2}\), is inversely proportional to the rate constant, \(k\). When you know one, you can easily find the other. To find \(k\), you rearrange this formula to:

• \( k = \frac{0.693}{t_{1/2}} \)

This simple relationship is useful in predicting how long a reactant will last or how fast a reaction proceeds under certain conditions. In our example, once you plug in the half-life of 2.5 days, you can calculate the rate constant at 300°C as approximately \(0.277\thinspace\text{day}^{-1}\).

Half-life formula

In chemical kinetics, the half-life formula is particularly useful for first-order reactions. The concept of half-life refers to the time required for half of the reactant to be consumed in the reaction. For first-order reactions, this time period remains constant, which means that regardless of the starting concentration of the reactant, it will always take the same amount of time for half of it to disappear.

For first-order reactions, the formula is simply:

• \( t_{1/2} = \frac{0.693}{k} \)

This formula emerges from the natural logarithmic evolution of first-order reactions, linking the rate constant \(k\) to half-life \(t_{1/2}\). The 0.693 factor is derived from the natural logarithm of 2, which accounts for the time it takes for a substance to reduce to half its amount.
Let's consider an example at 400°C. Given the rate constant \(k\) of \(0.19\thinspace\text{min}^{-1}\), you can rearrange the formula to solve for the half-life:

• \( t_{1/2} = \frac{0.693}{0.19\thinspace\text{min}^{-1}} \)
• This equals approximately 3.65 minutes

This formula allows us to determine how quickly a reaction proceeds based on conditions such as temperature or catalysis.

Sulfuryl chloride decomposition

The decomposition of sulfuryl chloride, \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\), is a textbook example of a first-order reaction. In this process, sulfuryl chloride gas decomposes into sulfur dioxide gas (\(\text{SO}_2\)) and chlorine gas (\(\text{Cl}_2\)). What makes it interesting is its simplicity in focusing on how the concentration of a single reactant determines the rate of reaction.
In practical terms, understanding the decomposition of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) helps illustrate how reaction conditions, like temperature, can significantly impact reaction rates. At different temperatures, sulfuryl chloride decomposes at different rates. For example:

• At 300°C, using a half-life of 2.5 days, the rate constant is found to be \(0.277\text{day}^{-1}\).
• At 400°C, with the rate constant \(0.19\text{min}^{-1}\), the half-life is approximately 3.65 minutes.

This temperature dependence is generally due to the increase in kinetic energy, which promotes faster molecular collisions and thus quicker decompositions. Understanding these reactions under various conditions is crucial for applications like industrial chemistry and environmental science. By knowing the kinetics, chemists can better control and optimize reactions for desired outcomes.