Problem 32
Question
The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) at high temperature is a first-order reaction with a half-life of \(2.5 \times 10^{3}\) min. What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will remain after 750 min?
Step-by-Step Solution
Verified Answer
Approximately 81.2% of \(\mathrm{SO}_2 \mathrm{Cl}_2\) remains after 750 minutes.
1Step 1: Understanding Half-Life
For a first-order reaction, the half-life is the time it takes for half of the reactant to decompose. We are given a half-life of \(2.5 \times 10^3\) minutes for the decomposition of \(\mathrm{SO}_2 \mathrm{Cl}_2\). This means every \(2.5 \times 10^3\) minutes, half of the initial amount of \(\mathrm{SO}_2 \mathrm{Cl}_2\) remains.
2Step 2: Using the First-Order Kinetics Equation
The fraction of reactant remaining after time \(t\) can be calculated using the equation: \[ N_t = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \] where \(N_t\) is the remaining amount after time \(t\), \(N_0\) is the initial amount, \(t\) is the time elapsed, and \(t_{1/2}\) is the half-life.
3Step 3: Substitute Values into the Equation
Given \(t = 750\) min and \(t_{1/2} = 2.5 \times 10^3\) min, substitute into the equation: \[ N_t = N_0 \left(\frac{1}{2}\right)^{\frac{750}{2.5 \times 10^3}} \]
4Step 4: Calculate the Exponent
Calculate \(\frac{750}{2.5 \times 10^3} = 0.3\). This value is the exponent in the equation \(\left(\frac{1}{2}\right)^{0.3}\).
5Step 5: Compute the Fraction Remaining
Calculate \(\left(\frac{1}{2}\right)^{0.3} \approx 0.812\). This means approximately 81.2% of \(\mathrm{SO}_2 \mathrm{Cl}_2\) will remain after 750 minutes.
Key Concepts
Chemical KineticsHalf-Life CalculationDecomposition Reaction
Chemical Kinetics
Chemical kinetics is the study of how chemical reactions occur and how fast they happen. It focuses on understanding the rate of reactions and the factors affecting these rates. In chemical kinetics, reactions can be classified by their order, which indicates how the rate is affected by the concentration of reactants. For example, in a first-order reaction, the rate is directly proportional to the concentration of one reactant.
First-order reactions, like the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), have a characteristic feature: a constant half-life. This means that the time taken for half of the reactant to convert into product remains constant throughout the entire reaction. This is a key difference from higher-order reactions, where half-lives can vary depending on concentrations.
First-order reactions, like the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), have a characteristic feature: a constant half-life. This means that the time taken for half of the reactant to convert into product remains constant throughout the entire reaction. This is a key difference from higher-order reactions, where half-lives can vary depending on concentrations.
Half-Life Calculation
The concept of half-life is crucial in understanding first-order reactions. For any given first-order reaction, the half-life \( (t_{1/2}) \) is the time required for half of the initial amount of reactant to decompose. This can be specifically valuable in predicting how long a substance will last and its decomposition rate over time.
In our case with \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \), the half-life is provided as \(2.5 \times 10^{3}\) minutes. This means that after every \(2.5 \times 10^{3}\) minutes, half of the \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) has decomposed. By using the half-life equation for a first-order reaction:
In our case with \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \), the half-life is provided as \(2.5 \times 10^{3}\) minutes. This means that after every \(2.5 \times 10^{3}\) minutes, half of the \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) has decomposed. By using the half-life equation for a first-order reaction:
- \( N_t = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \)
Decomposition Reaction
Decomposition reactions involve a single compound breaking down into two or more simpler substances. These reactions are a key part of many natural and industrial chemical processes. In the example of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) decomposition, at high temperatures, it breaks into \( \mathrm{SO}_2 \) and \( \mathrm{Cl}_2 \).
This process is a perfect example of a first-order reaction, where the rate at which the decomposition occurs depends on the concentration of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \). As decomposition progresses, the reactant concentration decreases, but the time for half of the \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) to decompose remains the same, illustrating the characteristic constant half-life of first-order reactions.
Studying decomposition reactions helps chemists understand reaction mechanisms, improve processes such as mass production, and predict the behavior of materials under various conditions, offering essential insights into chemical dynamics.
This process is a perfect example of a first-order reaction, where the rate at which the decomposition occurs depends on the concentration of \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \). As decomposition progresses, the reactant concentration decreases, but the time for half of the \( \mathrm{SO}_{2} \mathrm{Cl}_{2} \) to decompose remains the same, illustrating the characteristic constant half-life of first-order reactions.
Studying decomposition reactions helps chemists understand reaction mechanisms, improve processes such as mass production, and predict the behavior of materials under various conditions, offering essential insights into chemical dynamics.
Other exercises in this chapter
Problem 30
Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 5.6-mg sample of the isotope, ho
View solution Problem 31
Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$\mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm
View solution Problem 33
Common sugar, sucrose, breaks down in dilute acid solution to form glucose and fructose. Both products have the same formula, \(\mathrm{C}_{6} \mathrm{H}_{12} \
View solution Problem 35
Data for the decomposition of dinitrogen oxide $$2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{N}_{2} (\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}
View solution