Question

From the following data for the second-order gas-phase decomposition of HI at \(430^{\circ} \mathrm{C}\), calculate the second-order rate constant and half- life for the reaction: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { [HI]/mol } \mathrm{dm}^{-3} \\ \hline 0 & 1 \\ 100 & 0.89 \\ \hline 200 & 0.8 \\ \hline 300 & 0.72 \\ \hline 400 & 0.66 \\ \hline \end{array} $$

Short answer

The second-order rate constant for the gas-phase decomposition of HI is approximately \(1.25 \times 10^{-3} \ \text{mol}^{-1}\text{dm}^3\text{s}^{-1}\), and the half-life of the reaction is 800 s.

Step-by-step solution

Write the second-order rate equation

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

Calculate the rate constant (k) using the given data

We can choose any of the given data points (other than the initial point) to calculate the rate constant (k). Let's choose the data point at t = 200 s, where [HI] = 0.8 mol/dm³. Using the second-order rate equation: $$ \frac{1}{0.8} - \frac{1}{1} = k(200) $$ Solving for k, we get: $$ k = \frac{\frac{1}{0.8} - \frac{1}{1}}{200} = \frac{(1 - 0.8) / 0.8}{200} = 1.25 \times 10^{-3} \ \text{mol}^{-1} \text{dm}^3 \text{s}^{-1} $$ So, the second-order rate constant (k) is approximately \(1.25 \times 10^{-3}\) mol⁻¹ dm³ s⁻¹.

Calculate the half-life (t1/2) using the rate constant (k) and initial concentration [HI]₀

The half-life of a second-order reaction can be calculated as: $$ t_{1/2} = \frac{1}{k[A]_0} $$ where: - \(t_{1/2}\) is the half-life - \(k\) is the second-order rate constant - \([A]_0\) is the initial concentration of the reactant Using the calculated rate constant (k) and the initial concentration [HI] = 1 mol/dm³: $$ t_{1/2} = \frac{1}{(1.25 \times 10^{-3})(1)} = \frac{1}{1.25 \times 10^{-3}} = 800 \ \text{s} $$ So, the half-life (t1/2) of the reaction is 800 s. In conclusion, the second-order rate constant for the gas-phase decomposition of HI is approximately \(1.25 \times 10^{-3} \ \text{mol}^{-1}\text{dm}^3\text{s}^{-1}\), and the half-life of the reaction is 800 s.

Key concepts

Rate Equation

For reactions, especially those of second-order kinetics, understanding the rate equation is crucial. An equation in chemical kinetics used to express the rate of a reaction is known as the rate equation. It shows the relationship between the concentration of the reactant and the rate at which it turns into products.
In the case of a second-order reaction, the rate equation takes the form:

• \( \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \)

This equation helps us predict how concentration changes over time.
In the above formula, \([A]_t\) and \([A]_0\) are the concentrations of the reactant at times \(t\) and 0 respectively. \(k\) stands for the rate constant, specific to this reaction's conditions. Grasping this equation lets you track how a second-order system is progressing over a period.
The form and variables within the rate equation differ depending on the reaction order, but they all aim to explain the transformation rate of reactants into products.

Rate Constant

The rate constant \(k\) is a crucial factor determining the speed of a reaction.
In the realm of chemical kinetics, it is a proportionality factor in the rate equation. It links the reaction rate to the concentrations of the reactants.
For a second-order reaction,

• \(k\) is expressed in units of \(\text{mol}^{-1}\text{dm}^3\text{s}^{-1}\), which helps maintain consistency in the rate's unit measure.

The value of \(k\) suggests how rapidly a reaction occurs; larger values indicate faster reactions under the same reactant concentrations.
Consistency in conditions like temperature is important because \(k\) can vary with different temperatures.
To determine \(k\) experimentally, as seen in our example, we use the concentration data at different time intervals combined with the rate equation. Calculating \(k\) facilitates numerous predictions about the reaction's progress and behavior.

Half-life

The concept of half-life in chemical kinetics refers to the time required for a reactant’s concentration to reduce to half its initial value.
For second-order reactions, half-life depends on both the initial concentration of the reactant and the rate constant \(k\).

• The formula to compute half-life in these reactions is: \( t_{1/2} = \frac{1}{k[A]_0} \)

This differs from first-order reactions, where half-life is independent of initial concentration.
Understanding half-life is essential, as it provides insight into how quickly a reaction reaches significant progress markers.
In our example, with an initial HI concentration of 1 mol/dm³ and a rate constant of \(1.25 \times 10^{-3} \text{mol}^{-1}\text{dm}^3\text{s}^{-1}\), the half-life is calculated as 800 seconds, showing us how rate constants and initial concentrations interplay to influence reaction timeframes.

Gas-phase Decomposition

Gas-phase decomposition is a common type of chemical reaction where a gaseous reactant breaks down into simpler molecules or elements.
In our exercise, the decomposition of hydrogen iodide (HI) exhibits second-order kinetics.
This indicates the rate at which HI decomposes is proportional to the square of its concentration.

• Understanding decomposition reactions is vital for grasping the broader topic of reaction mechanisms.
• Reaction conditions, particularly temperature, have substantial impacts on these types of reactions, influencing how quickly or slowly they proceed.

Gas-phase reactions are often studied under controlled temperatures, making calculations simpler as assumptions about volume and pressure remain stable.
Grasping the specifics of gas-phase decomposition allows learners to apply these concepts to more complex scenarios involving mixed-phase reactions.

Chemical Kinetics

Chemical kinetics involves studying the rates of reactions and the steps taken during conversion from reactants to products. In kinetic studies, we frequently assess factors involved in equations and constants to predict reaction outcomes.
Kinetics provides insight necessary for controlling reactions in industrial applications, leading to efficient processes in chemical manufacturing and energy sectors.

• It uses models like the second-order rate equation to simulate reaction dynamics.
• By manipulating conditions such as temperature and concentration, we can alter reaction rates according to desired outputs.

Studying kinetics helps chemists optimize these conditions to enhance reaction yields or reduce unwanted by-products.
Employing the principles of chemical kinetics enables a thorough understanding of both fundamental and advanced reactions in chemistry.