Problem 55 - Chemistry A Molecular Approach

Question

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and has a rate constant of \(1.42 \times 10^{-4} \mathrm{~s}^{-1}\) at a certain temperature. a. What is the half-life for this reaction? b. How long will it take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(25 \%\) of its initial concentration? c. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.00 \mathrm{M}\), how long will it take for the concentration to decrease to \(0.78 \mathrm{M} ?\) d. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(0.150 \mathrm{M},\) what is the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after \(2.00 \times 10^{2} \mathrm{~s}\) ? After \(5.00 \times 10^{2} \mathrm{~s} ?\)

Step-by-Step Solution

Answer

a. The half-life for this reaction is 4,887 s. b. It takes 9,774 s for the concentration to decrease to 25%. c. It takes 1,881 s for the concentration to decrease to 0.78 M. d. After 2.00 x 10^2 s, the concentration is 0.147 M, and after 5.00 x 10^2 s, it is 0.142 M.

Step 1: Calculating the half-life (t1/2) for a first-order reaction

The formula for the half-life of a first-order reaction is given by: t1/2 = ln(2) / k, where k is the rate constant. Here, k = 1.42 x 10^-4 s^-1. Plugging these values into the formula gives t1/2 = ln(2) / (1.42 x 10^-4 s^-1).

Step 2: Determining time to decrease to 25% concentration

To decrease to 25% of the initial concentration, the compound needs to go through two half-lives (since 100% -> 50% -> 25%). Therefore, the time taken will be 2 times the half-life calculated in Step 1.

Step 3: Time for concentration to decrease from 1.00 M to 0.78 M

For a first-order reaction, the relationship between time (t), initial concentration ([A]0), final concentration ([A]), and the rate constant (k) is given by: ln([A]0/[A]) = kt. Solve for t using [A]0 = 1.00 M, [A] = 0.78 M, and k = 1.42 x 10^-4 s^-1.

Step 4: Concentration after 2.00 x 10^2 s for 0.150 M initial concentration

Using the first-order relationship, solve for [A] after t = 2.00 x 10^2 s with [A]0 = 0.150 M and k = 1.42 x 10^-4 s^-1. This requires rearranging the first-order rate equation to solve for [A].

Step 5: Concentration after 5.00 x 10^2 s for 0.150 M initial concentration

Repeat the process in Step 4 for t = 5.00 x 10^2 s to find the concentration of SO2Cl2 after this time period.

Key Concepts

Chemical Kinetics

Chemical kinetics is the study of the rates of chemical processes and the factors that affect them. It helps us understand how quickly a reaction proceeds and what can be done to control the speed of the reaction. A key concept within chemical kinetics is the order of a reaction, which refers to how the rate is affected by the concentration of the reactants. For a first-order reaction, the rate is directly proportional to the reactant concentration. The rate constant, represented by 'k', is a crucial factor and determines the speed of the reaction at a given temperature; it is unique to each reaction and its conditions.

Understanding the rate constant and reaction order allows chemists to predict how concentrations will change over time, which is essential for industries such as pharmaceuticals, where reaction timing is critical.

Half-Life of a Reaction

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half its initial value. In the context of a first-order reaction, the half-life is independent of the initial concentration. This means that no matter how much reactant you start with, the time it takes for half of it to decompose or react will always be the same. The half-life (\( t_{1/2} \)) can be calculated using the rate constant (\( k \) using the formula: \( t_{1/2} = \frac{\ln(2)}{k} \), where \(\ln(2)\) is the natural logarithm of 2. This concept is extremely useful when determining how long a drug remains effective in the body or how quickly a pollutant will decrease to a safe level in the environment.

Rate Constant

The rate constant (\( k \) is a numerical value that represents the speed of a given chemical reaction. For a first-order reaction, it remains constant at a given temperature but can vary with different temperatures or the presence of a catalyst. The value of the rate constant can tell us a lot about a reaction; a larger value indicates a faster reaction while a smaller value suggests a slower one. In our exercise, the rate constant for the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2} \) is \(1.42 \times 10^{-4} \mathrm{s}^{-1} \), implying the reaction proceeds at a moderate pace at the given temperature. Knowing the rate constant allows us to predict the future concentration of reactants or products at any given time using mathematical equations.

Reaction Concentration Over Time

As a reaction progresses, the concentration of reactants decreases over time. In a first-order reaction, this decrease in concentration is exponential and not linear, meaning that the rate decreases as the reactants are used up. The relationship between time (\( t \) and concentration of reactants can be described by the first-order rate law: \( \ln \left(\frac{[A]_{0}}{[A]}\right) = kt \), where \( [A]_{0} \) is the initial concentration, \( [A] \)is the concentration at time \( t \) and \( k \) is the rate constant. This equation is pivotal in solving problems like those in the exercise, where we need to determine how long it will take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2} \) to decrease to a certain level. By manipulating the formula, we can solve for any variable, given the others are known.