Problem 81 - Chemistry The Molecular Nature of Matter

Question

Hydrogen iodide decomposes according to the equation, $$2 \mathrm{HI}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$$The reaction is second order and has a rate constant equal to \(1.6 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at \(750^{\circ} \mathrm{C}\). If the initial concentration of HI in a container is \(3.4 \times 10^{-2} M\), how many minutes will it take for the concentration to be reduced $$\text { to } 8.0 \times 10^{-4} \mathrm{M}$$.

Step-by-Step Solution

Answer

After solving the equation, the time in seconds should be converted to minutes to get the final answer.

Step 1: Write down the rate law for a second-order reaction

For a second-order reaction, the rate law can be written as: \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \), where \( [A] \) is the concentration of reactant at time \( t \), \( [A]_0 \) is the initial concentration of reactant, \( k \) is the rate constant, and \( t \) is the time.

Step 2: Insert the known values into the rate law

Insert the known values: initial concentration \( [HI]_0 = 3.4 \times 10^{-2} M \), final concentration \( [HI] = 8.0 \times 10^{-4} M \), and rate constant \( k = 1.6 \times 10^{-3} L \cdot mol^{-1} \cdot s^{-1} \). The equation becomes \( \frac{1}{8.0 \times 10^{-4}} - \frac{1}{3.4 \times 10^{-2}} = (1.6 \times 10^{-3}) t \).

Step 3: Solve for the time \( t \)

Calculate the time \( t \) by rearranging the equation and doing the arithmetic: \[ t = \frac{\frac{1}{8.0 \times 10^{-4}} - \frac{1}{3.4 \times 10^{-2}}}{1.6 \times 10^{-3}} \].

Step 4: Calculate the time in seconds and convert to minutes

After solving the arithmetic part of the calculation, convert the time from seconds to minutes since the question asks for the time in minutes. Use the conversion factor that 1 minute is equal to 60 seconds.

Key Concepts

Chemical Kinetics

Chemical kinetics is a branch of physical chemistry that concerns the study of reaction rates and the factors that affect them. It provides insights into the speed at which reactions occur and how different conditions like temperature, concentration of reactants, and the presence of catalysts can influence these rates. The study of kinetics helps us understand mechanisms of chemical processes and can even predict the outcomes of reactions under various circumstances.

When we describe the decomposition of hydrogen iodide (HI) into hydrogen and iodine gases, we're delving into the reaction kinetics of this specific process. A second-order reaction like this one means that the rate at which the reaction proceeds is directly proportional to the square of the concentration of the reactants. Understanding the kinetics of this decomposition allows chemists to calibrate reaction conditions for desired outcomes, such as controlling reaction speed or optimizing product yield.

Reaction Rate Constant

The reaction rate constant, symbolized by 'k', is a numerical value that relates the concentration of reactants to the rate of the reaction. It essentially provides the speed factor for the reaction under specific conditions. This constant is unique for every reaction and depends on several factors, including temperature and the presence of catalysts.

In our exercise involving hydrogen iodide, the given rate constant is 1.6 x 10^-3 L mol^-1 s^-1 at 750°C. This information is crucial for calculating how long it will take for the concentration of hydrogen iodide to decrease to a certain level. The rate constant essentially gives us the pace at which HI is decomposing at the given temperature, allowing us to use the second-order reaction rate law to find the time required for a desired concentration change.

Concentration-Time Relationship

The concentration-time relationship in chemical kinetics is a mathematical expression that relates the concentration of a reactant to the time elapsed during the reaction. For a second-order reaction, the relationship is represented by the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) where \( [A]_0 \) and \( [A] \) represent the initial and current concentrations of the reactant, respectively, and 'k' is the reaction rate constant.

By manipulating this formula, we can predict how the concentration of reactants will change over time or calculate the time it will take to reach a certain concentration. This is particularly essential when precise concentrations need to be maintained in industrial processes. The calculation shown in the steps of the exercise uses this relationship to find the time necessary for the HI concentration to decrease to 8.0 x 10^-4 M from an initial concentration of 3.4 x 10^-2 M.

Half-Life of a Reaction

The half-life of a reaction, often represented by \( t_{1/2} \), is the time required for the concentration of a reactant to decrease to half of its initial value. It's a key concept in reaction kinetics because it provides a measure of how quickly a reactant is consumed. The half-life is particularly important in pharmaceuticals, where it helps in determining dosing schedules of drugs.

For a second-order reaction, the half-life is not constant and varies with concentration. The half-life for a second-order reaction is given by the formula \( t_{1/2} = \frac{1}{k[A]_0} \), with 'k' being the reaction rate constant and \( [A]_0 \) being the initial concentration of the reactant. This concept makes it clear that as the concentration decreases, the half-life increases, indicating that the reaction slows down over time. Even though our hydrogen iodide decomposition exercise doesn't specifically ask for half-life, understanding this concept helps deepen the understanding of the kinetic behaviors of second-order reactions.