Question

The rate constant for the second-order reaction $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). (a) How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.62 M\) to \(0.28 M ?\) (b) Calculate the half-lives at these two concentrations.

Short answer

The time it takes for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.62 M\) to \(0.28 M\) is calculated to be 769 seconds. The half-life at the concentration of \(0.62 M\) is 3.034 seconds and the half-life at the concentration of \(0.28 M\) is 6.646 seconds.

Step-by-step solution

Write Down the Second-order Reaction Rate Law

A second-order reaction follows the rate law \(1/[A] = kt + 1/[A_0]\), where [A] is the concentration of the reactant at time \(t\), [A_0] is the initial concentration, and \(k\) is the rate constant. This equation can be rearranged to find the time duration: \(t = (1/k) * ([A] - [A_0])\) which will be used to solve for the time it takes the concentration to change in this exercise.

Substitute the Given Values into the Formula

Substitute the given values into the formula: \(t = (1/0.54 (M \cdot s) ) * (1/0.28 M - 1/0.62M )\). Simplify the equation to find the value of \(t\).

Calculate the Half-Lives at Each Concentration

For a second-order reaction, the half-life is calculated using the expression \( t_{1/2} = 1/(k [A_0]) \). Thus, the half-lives at the two concentrations are: \(t_{1/2} = 1/(0.54 M \cdot s * 0.62 M)\) and \(t_{1/2} = 1/(0.54 M \cdot s * 0.28 M)\). Solve these equations to find the half-lives at each concentration.

Key concepts

Reaction Rate Law for Second-Order Reactions

Understanding the reaction rate law for second-order reactions is pivotal for solving problems related to chemical kinetics. In these reactions, the rate of reaction is directly proportional to the square of the concentration of one reactant. This is represented by the integrated rate law:

• \( \frac{1}{[A]} = kt + \frac{1}{[A_0]} \)

Here, - \([A]\) is the concentration of the reactant at time \(t\), - \([A_0]\) is the initial concentration, and - \(k\) is the rate constant.
The equation helps in determining the concentration of reactants over time or the time required for a concentration change. This formula can be rearranged to find time (\(t = \left(\frac{1}{k}\right) \times \left(\frac{1}{[A]} - \frac{1}{[A_0]}\right)\)), which is essential for calculating how long it takes for a concentration to decrease in second-order reactions. By substituting known values into this equation, one can effortlessly solve for the unknown variable, typically time, related to concentration changes.

Half-life Calculation in Second-Order Reactions

For chemical reactions, the half-life is the time required for half of the reactant to be consumed. However, in second-order reactions, the half-life is dependent on the initial concentration of the reactant. This distinguishes it from first-order reactions, where the half-life is independent of the concentration. For a second-order reaction, the half-life (\(t_{1/2}\)) is calculated using:

• \( t_{1/2} = \frac{1}{k [A_0]} \)

Where \(k\) is the rate constant, and \([A_0]\) is the initial concentration. This equation shows that a higher initial concentration of reactant results in a shorter half-life.
Conversely, a lower initial concentration increases the half-life duration. In practical calculations, given the initial concentrations, you can directly place these values into the formula to compute the half-life for each starting concentration. This dependency on initial concentration means that as reactant concentrations decrease, subsequent half-lives will increase. This peculiarity of second-order reactions is important to grasp when studying their kinetics.

Understanding Rate Constant in Reaction Kinetics

The rate constant (\(k\)) is a fundamental component in the kinetics of chemical reactions. For a second-order reaction, the units of the rate constant are usually \(M^{-1}\cdot s^{-1}\), indicating its impact on concentration over time.
This constant is affected by factors such as temperature and the presence of catalysts.

• A higher rate constant implies a faster reaction at a given temperature.
• Conversely, a lower value suggests a slower reaction process.

The rate constant is vital when using the rate law equations since it directly influences calculations involving the concentration change over time or the half-life of a reactant.
For our exercise, the given rate constant was \(0.54 / M\cdot s\) at \(300^{\circ}\text{C}\). This must be input into rate law calculations to determine the time taken for concentration changes and the corresponding half-lives. Understanding this constant's role enhances the accuracy in predicting how a reaction progresses under specified conditions.