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 \(4.5 \times 10^{-2} \mathrm{s}^{-1}\) .half-life for this reaction? (b) If you start with 0.050\(M \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Short answer

The half-life of the decomposition of sulfuryl chloride (SO2Cl2) at 660 K is 15.4 seconds. After 5.12 seconds, the concentration of I2 remaining is 0.0249 M.

Step-by-step solution

Determine the half-life of the first-order reaction

We will use the half-life formula for a first-order reaction: \[t_{1/2} = \frac{0.693}{k}\] Where \(t_{1/2}\) is the half-life, and k is the rate constant. Given the rate constant, k = \(4.5 \times 10^{-2}\, s^{-1}\), we can find the half-life: \[t_{1/2} = \frac{0.693}{4.5 \times 10^{-2} \,s^{-1}}\]

Calculate the half-life

Plug in the rate constant value into the equation and solve for the half-life: \[t_{1/2} = \frac{0.693}{4.5 \times 10^{-2}\, s^{-1}} = 15.4\, s\] So, the half-life of the reaction is 15.4 seconds.

Use the integrated rate law for a first-order reaction

To determine the amount of I2 remaining after 5.12 s, we will use the integrated rate law for a first-order reaction: \[ ln[\frac{A}{A_0}] = -kt\] Where A is the final concentration of I2, \(A_0\) is the initial concentration of I2, k is the rate constant, and t is the time. We are given the initial concentration of I2: \(A_0 = 0.050\, M\), the rate constant k, and the time: t = 5.12 s. Now we can solve for the final concentration, A.

Calculate the final concentration of I2

Rearrange the integrated rate law equation to solve for A: \[A = A_0 e^{-kt}\] Plug in the given values: \[A = 0.050\, M \times e^{-4.5 \times 10^{-2}\, s^{-1} \times 5.12\, s}\]

Find the remaining amount of I2

Evaluate the expression to find A: \[A = 0.050\, M \times e^{-(4.5 \times 10^{-2}\, s^{-1})(5.12\, s)} = 0.0249\, M\] So, after 5.12 seconds, there will be 0.0249 M of I2 remaining.

Key concepts

Rate Constant

The rate constant, often denoted by the symbol \( k \), is a fundamental value in chemical kinetics that quantifies the rate of a reaction. It serves as a unique identifier for the speed of a reaction under specific conditions. In first-order reactions, the rate constant has units of inverse time, often \( s^{-1} \), which shows how concentration changes as time progresses.

• A higher rate constant signifies a faster reaction.
  
• The magnitude of \( k \) can be influenced by factors such as temperature and presence of catalysts.
  
• For our reaction, the given rate constant is \( 4.5 \times 10^{-2} \, s^{-1} \). This indicates that, at the temperature of 660 K, the reaction is relatively rapid.

Integrated Rate Law

The integrated rate law is a mathematical expression that connects reactant concentrations over time, specifically designed for first-order reactions. It helps us determine concentrations in a reaction using time and the rate constant. For a first-order reaction, the relationship is expressed as:\[ln\left(\frac{[A]}{[A_0]}\right) = -kt\]- \([A]\) is the concentration of the reactant at time \( t \)
- \([A_0]\) is the initial concentration
- \( k \) is the rate constant
- \( t \) represents time.By rearranging this equation, we can solve for \([A]\), which allows us to know how much of the reactant remains after a certain period. In the provided solution, this was used to determine the amount of \( I_2 \) left after 5.12 seconds, resulting in a concentration of \( 0.0249 \, M \). This shows the powerful predictive nature of rate laws.

Half-Life of Reactions

Half-life in chemical kinetics is the time required for half of the reactant to be depleted. It is particularly straightforward to calculate for first-order reactions using the formula:\[t_{1/2} = \frac{0.693}{k}\]This formula directly arises from the natural log base, since the reaction's concentration diminishes exponentially.
Here are some important points about half-life:

• In first-order reactions, the half-life is constant and does not depend on initial concentration.
  
• This makes the first-order reactions unique because the half-life remains the same regardless of how much substance is present initially.
  
• In the exercise, with a rate constant \( k = 4.5 \times 10^{-2} \, s^{-1} \), the half-life was calculated to be 15.4 seconds, implying that half of the \( SO_2Cl_2 \) decomposes in this time span.

Understanding half-life is crucial for predicting how reactant concentrations evolve and gives insights into the duration required for significant reaction progress.