Question

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), decomposes by first- order kinetics, and \(k=2.81 \times 10^{-3} \mathrm{~min}^{-1}\) at a certain temperature. (a) Determine the half-life for the reaction. (b) Determine the time needed for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(10 \%\) of its initial concentration. (c) If \(14.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is sealed in a \(2500 .-\mathrm{L}\) reaction vessel and heated to the specified temperature, what mass will remain after \(1.5 \mathrm{~h}\) ?

Short answer

The half-life is approximately 246.9 minutes. It takes about 544.6 minutes for the concentration to decrease to 10% of its initial concentration. After 1.5 hours, approximately 3.51 g of sulfur dioxide chloride remains.

Step-by-step solution

Determine the half-life for the reaction

The half-life for a first-order reaction can be calculated using the formula: \( t_{1/2} = \frac{\ln(2)}{k} \).

Calculate the half-life using the given rate constant

Plug the given rate constant into the half-life equation: \( t_{1/2} = \frac{\ln(2)}{2.81 \times 10^{-3} \text{ min}^{-1}} \). Calculate the value to find the half-life.

Determine the time for concentration to reach 10% of the initial concentration

For a first-order reaction, the time required for the concentration to change to a percentage of the initial concentration is given by \( t = \frac{\ln(C_0/C)}{k} \), where \( C_0 \) is the initial concentration, and \( C \) is the final concentration. For 10%, \( \ln(C_0/C) = \ln(10) \).

Calculate the time to reach 10% concentration

Use the rate constant \( k \) and the natural logarithm of 10 to find the time: \( t = \frac{\ln(10)}{2.81 \times 10^{-3} \text{ min}^{-1}} \).

Calculate the initial moles of \(\mathrm{SO}_{2} \(\mathrm{Cl}_{2}\)\)

Using the molar mass of \(\mathrm{SO}_{2} \(\mathrm{Cl}_{2}\)\), find the initial moles in 14.0 g of the substance.

Use the integrated first-order rate law to find remaining moles

Apply the first-order integrated rate law, \( \ln(\frac{[A]_t}{[A]_0}) = -kt \), to calculate the remaining moles after 1.5 hours. Convert hours to minutes to match the minute units in \( k \).

Calculate the final mass of \(\mathrm{SO}_{2} \(\mathrm{Cl}_{2}\)\) remaining after 1.5 h

Convert the remaining moles back to grams using the molar mass to find the remaining mass.

Key concepts

Understanding Chemical Kinetics

Chemical kinetics is the study of the speed or rate at which chemical reactions occur and the factors that affect these rates. It's an essential area of chemistry because it helps us understand reaction mechanisms and predict the behavior of reactions under various conditions.

In kinetics, we often deal with the reaction rate, which is the change in concentration of a reactant or product over a specific period. For a reaction where a substance A transforms into products, the rate can be expressed as the negative change in concentration of A over time, usually denoted by \( -\frac{d[A]}{dt} \).

One crucial aspect of chemical kinetics is the order of the reaction, which relates to how the rate responds to changes in concentration. A first-order reaction implies that the rate is directly proportional to the concentration of the reactant. In simpler terms, if you double the amount of reactant, the reaction rate will also double.

Half-Life Calculation in Chemical Reactions

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 first-order reactions, half-life is a constant, meaning it does not depend on the initial concentration of the reactant.

The formula to calculate half-life for a first-order reaction is \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the reaction rate constant. This equation is derived from the integrated first-order rate law, and it reflects the exponential nature of the decay in concentration over time.

Understanding half-life is extremely useful as it provides a clear and measurable understanding of how quickly a reaction proceeds. It's a common concept not just in chemistry, but also in physics, biology, and even pharmacology.

Determining the Reaction Rate Constant

The reaction rate constant, denoted as \( k \) in equations, is a proportionality factor that relates the reaction rate to the reactant concentrations. In a first-order reaction, the rate constant can tell us how quickly the reaction occurs. A higher value of \( k \) indicates a faster reaction.

We obtain \( k \) experimentally by measuring the concentration of reactants or products over time and fitting this data to a kinetic model. The units of \( k \) vary depending on the order of the reaction, and for first-order reactions, it has units of reciprocal time (e.g., \( s^{-1} \) or \( min^{-1} \)).

Knowing the rate constant is fundamental for calculating other kinetics parameters, such as half-life, and for predicting how long it will take for a reaction to reach a certain conversion level, as in the case of the textbook exercise provided.

The Integrated Rate Law for First-Order Reactions

The integrated rate law is an expression that describes the concentration of a reactant in a reaction over time. For first-order reactions, this law takes the form \( \ln(\frac{[A]_t}{[A]_0}) = -kt \), where \( [A]_0 \) is the initial concentration of the reactant A, \( [A]_t \) is the concentration of A at time \( t \) and \( k \) is the reaction rate constant. This mathematical relationship is derived from the differential rate law by integrating it over the time the reaction has been proceeding.

By using this equation, we can calculate the concentration of a reactant at any point in time during a first-order reaction. It's a powerful tool for chemists because it connects the measurable quantities of concentration and time to the underlying kinetics of a reaction. In practical terms, this lets us determine, among other things, how much reactant remains after a certain period, just as we might calculate in the given textbook exercise.