Question

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of \(\mathrm{NOBr}\) after \(22 \mathrm{~s}\) (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=\) \(0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\).

Short answer

The concentration of NOBr after 22 seconds is 0.041 M. The half-life when [NOBr]_0 = 0.072 M is 17.36 s, and it is 23.15 s when [NOBr]_0 = 0.054 M.

Step-by-step solution

Write the Rate Equation

The second-order rate equation for the reaction is \(\frac{d[NOBr]}{dt} = -k[NOBr]^2\), where [NOBr] is the concentration of NOBr, k is the rate constant, and t is time.

Integrate the Rate Equation

Integrating the rate equation with respect to time from 0 to t, we get \(\int_{[NOBr]_0}^{[NOBr]} \frac{d([NOBr])}{[NOBr]^2} = - \int_{0}^{t} k dt\). This results in \(-\frac{1}{[NOBr]} = kt - \frac{1}{[NOBr]_0}\), where [NOBr]_0 is the initial concentration of NOBr. This can be rearranged to give the expression for the concentration of NOBr at any time t: \([NOBr] = \frac{1}{\frac{1}{[NOBr]_0}+kt}\)

Calculate the concentration of NOBr after 22 s

The initial concentration [NOBr]_0 is 0.086 M, the rate constant k is 0.80 / M⋅s, and the time t is 22 s. Input these values into the equation [NOBr] = \(\frac{1}{\frac{1}{0.086}+0.80\times22}\). After calculation, we obtain [NOBr] = 0.041 M.

Calculate Half-Lives

The half-life of a second order reaction is given by the equation \(t_{1/2}=\frac{1}{k[A]_0}\) where [A]_0 is the initial concentration of the reactant. Use the given initial concentrations and the rate constant to calculate half-lives. For [NOBr]_0 = 0.072 M, \(t_{1/2}=\frac{1}{0.80\times0.072} = 17.36 s\). For [NOBr]_0 = 0.054 M, \(t_{1/2}=\frac{1}{0.80\times0.054} = 23.15 s\)

Key concepts

Chemical Kinetics

Chemical kinetics is a subject of chemistry that deals with the speeds or rates at which chemical reactions occur. Understanding kinetics is crucial for explaining how various reactions proceed and determining what affects the speed of chemical processes. The speed of a reaction often depends on conditions such as temperature, concentration of reactants, and the presence of catalysts.

In the given exercise, we look at the reaction involving nitrogen oxide bromide (NOBr) which breaks down into nitrogen oxide (NO) and bromine (Br2) in a second-order reaction. Here, the speed at which NOBr decomposes is directly proportional to the square of its concentration. This relationship is a key concept in kinetics, illustrating how the concentration of reactants can drastically influence the rate of a reaction.

Reaction Rate Equations

Reaction rate equations, or rate laws, express the relationship between the rate of a chemical reaction and the concentration of its reactants. For a second-order reaction, the rate is proportional to the square of the concentration of one reactant.

The general form of a second-order rate equation is \( \frac{d[A]}{dt} = -k[A]^2 \), where \( [A] \) is the concentration of the reactant, \( k \) is the second-order rate constant, and \( t \) is time. In the exercise, students are shown how to integrate this equation to find the remaining concentration of a reactant over time, \( [A] \), given its initial concentration and the rate constant. The integration process is central to finding solutions in chemical kinetics, demonstrating how differential equations model the dynamic changes in reactant concentrations.

Half-Life Calculation

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half its initial value. In chemical kinetics, half-life is an important concept as it gives a clear indication of how quickly a reactant is being consumed in a reaction. For a second-order reaction, the half-life is inversely proportional to the initial concentration of the reactant and the rate constant, according to the formula \( t_{1/2} = \frac{1}{k[A]_0} \).

The exercise shows how to calculate the half-life of the NOBr decomposition reaction given different initial concentrations. By applying the half-life formula, students can see the direct effect of changes in initial concentration on the duration of the half-life — a higher concentration results in a shorter half-life. This counter-intuitive concept highlights the distinctive characteristics of second-order reactions compared to first-order reactions where the half-life is constant.