Question

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?

Short answer

The reaction is second order in \(\mathrm{NO}_{2}\) as the units of the rate constant are \(\mathrm{M}^{-1}\mathrm{s}^{-1}\). To determine the time it takes for the concentration to decrease from 0.100 M to 0.025 M, use the integrated rate law for a second-order reaction: \[\frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} = kt\] Substitute the given values and solve for t: \[t = \frac{30 \mathrm{M}^{-1}}{0.63 \mathrm{M}^{-1}\mathrm{s}^{-1}} = 47.62 \mathrm{s}\] Therefore, it takes 47.62 seconds for the concentration to decrease to 0.025 M.

Step-by-step solution

Identify the order of the reaction using the units of k

The given units of the rate constant k are \(\mathrm{M}^{-1}\mathrm{s}^{-1}\). To determine the reaction order, we need to consider the form of the rate law: \[rate = k[\mathrm{NO}_{2}]^n\] Here, n represents the order of the reaction. Now we need to find out the unit of the rate which is generally expressed in \(\mathrm{M}\mathrm{s}^{-1}\). Equating the units can help us find the value of n: \[\mathrm{M}\mathrm{s}^{-1} = k \times \mathrm{M}^n\] \[\mathrm{M}\mathrm{s}^{-1} = \mathrm{M}^{-1}\mathrm{s}^{-1} \times \mathrm{M}^n\] Comparing the units, we get the value of n: \[n = 2\] Therefore, the reaction is second order in \(\mathrm{NO}_{2}\).

Use the integrated rate law for a second-order reaction

For a second-order reaction, the integrated rate law is given by: \[\frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} = kt\] We are given the initial concentration \([\mathrm{A}]_0\) and the final concentration \([\mathrm{A}]_t\), as well as the rate constant k. We need to find the time (t) it takes for the concentration to decrease from the initial to the final value.

Substitute the given values and solve for t

Now we can substitute the given values into the integrated rate law: \[\frac{1}{0.025 \mathrm{M}} - \frac{1}{0.100 \mathrm{M}} = (0.63 \mathrm{M}^{-1}\mathrm{s}^{-1})t\] Calculating the difference between the inverses of the concentrations, we get: \[ \frac{1}{0.025} - \frac{1}{0.1} = 40 - 10 = 30 \] Now we have: \[30 \mathrm{M}^{-1} = (0.63 \mathrm{M}^{-1}\mathrm{s}^{-1})t\] Finally, we solve for t: \[t = \frac{30 \mathrm{M}^{-1}}{0.63 \mathrm{M}^{-1}\mathrm{s}^{-1}}\] \[t = 47.62 \mathrm{s}\] So it would take 47.62 seconds for the concentration of \(\mathrm{NO}_{2}\) to decrease from 0.100 M to 0.025 M.

Key concepts

Rate Law

The concept of the rate law is crucial in understanding the kinetics of a chemical reaction. It is an equation that relates the rate of a reaction to the concentration of its reactants. This mathematical relationship is expressed as:

\[rate = k[Reactant]^n\]

Here, 'rate' is the speed at which the reactants turn into products, 'k' is the rate constant which varies with temperature, [Reactant] represents the concentration of the reactant, and 'n' is the order of the reaction with respect to that reactant. It is important to note that 'n' could be a whole number, a fraction, or zero and it must be determined experimentally.

Determining Reaction Order

Determining the reaction order requires careful examination of how the rate varies with changes in concentration. If doubling the concentration of a reactant leads to doubling the rate, the reaction is first order with respect to that reactant. However, if the rate becomes four times faster, it suggests a second-order dependency on that reactant.

In the exercise provided, the units of the rate constant 'k' are in \(\mathrm{M}^{-1}\mathrm{s}^{-1}\). This provides a direct hint toward the reaction order. In general, if the rate constant has units of \(\mathrm{M}^{-1}\mathrm{s}^{-1}\), it is characteristic of a second-order reaction with respect to the reactant.

Second-Order Reaction

A second-order reaction is one where the rate of reaction is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. This implies that the overall reaction order 'n' equals 2, which is the sum of the exponents in the rate law.

When dealing with a second-order reaction like \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO} + \mathrm{O}_{2}\) from our exercise, it is evident that the rate of the reaction will change significantly with small changes in concentration of \(\mathrm{NO}_{2}\). This sensitivity to concentration means that precise measurements are necessary to track the reaction progress.

Characteristic of Second-Order Reactions

The integrated rate law for second-order reactions, which links the concentrations of reactants to time, presents a unique curve when plotted. The plot of \(1/[\mathrm{A}]\) against time 't' is linear for a second-order reaction, which is not the case for zero or first-order reactions. The steepness of this line is directly related to the rate constant 'k'. A real-world implication of understanding second-order kinetics can be found in scenarios like the purification of gases where the removal rate of impurities depends on the square of their concentration.

Integrated Rate Law

The integrated rate law is a more advanced concept that allows us to relate the concentration of reactants to time more directly than the differential rate law. This is especially useful when you need to predict how long it will take for a reactant to reach a certain concentration.

For a second-order reaction, the integrated rate law is expressed as:

\[\frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} = kt\]

In this equation, \([\mathrm{A}]_0\) is the initial concentration of the reactant, \([\mathrm{A}]_t\) is the concentration at time 't', and 'k' is the second-order rate constant. By rearranging the equation and solving for 't', we can determine the time required for a change in concentration, as shown in the step-by-step solution.

Application in Predicting Reaction Time

To apply the integrated rate law practically, as in our exercise, one would rearrange the formula to solve for 't', the time variable, and then plug in the known values: initial and final concentrations, along with the rate constant 'k', to find out exactly how long the reaction needs to reach a certain concentration threshold. This reveals the kinetics of the reaction and how actionable insights can be taken for real-world chemical processes.