Question

The following data were collected for the rate of disappearance of \(\mathrm{NO}\) in the reaction $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow$ \(2 \mathrm{NO}_{2}(g)\) \begin{tabular}{llll} \hline & & Initial Rate \\ Experiment & {\([\mathrm{NO}](M)\)} & {\(\left[\mathrm{O}_{2}\right](M)\)} & $(M / s)$ \\ \hline 1 & \(0.0126\) & \(0.0125\) & \(1.41 \times 10^{-2}\) \\ 2 & \(0.0252\) & \(0.0125\) & \(5.64 \times 10^{-2}\) \\ 3 & \(0.0252\) & \(0.0250\) & \(1.13 \times 10^{-1}\) \\ \hline \end{tabular} (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of NO when \([\mathrm{NO}]=0.0750 \mathrm{M}\) and $\left[\mathrm{O}_{2}\right]=0.0100 \mathrm{M} ?(\mathrm{e})$ What is the rate of disappearance of \(\mathrm{O}_{2}\) at the concentrations given in part (d)?

Short answer

The rate law for the reaction is \(rate = k [\mathrm{NO}]^2 [\mathrm{O}_{2}]\). The units for the rate constant are \(M^{-1}s^{-1}\). The average value of the rate constant calculated from the three data sets is approximately \(2.83 \times 10^{4} \, M^{-1}s^{-1}\). The rate of disappearance of \(\mathrm{NO}\) is approximately \(0.127 \, M/s\), and the rate of disappearance of \(\mathrm{O}_{2}\) is approximately \(0.0635 \, M/s\).

Step-by-step solution

Analyze the given data

The given data includes the initial concentrations of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\), and the initial rates of the reaction for three different experiments. By comparing these experiments, we can find the rate law for the reaction.

Determine the order of the reaction with respect to each reactant

To determine the order of the reaction with respect to each reactant, we will compare the changes in the initial concentrations and their effects on the initial rates. We can start with comparing experiments 1 and 2, while the concentration of \(\mathrm{O}_{2}\) remains constant: $$\frac{rate_2}{rate_1} = \frac{5.64 \times 10^{-2}}{1.41 \times 10^{-2}} = 4$$ $$\frac{[\mathrm{NO}]_2}{[\mathrm{NO}]_1} = \frac{0.0252}{0.0126} = 2$$ As we can see, when the concentration of \(\mathrm{NO}\) is doubled, the rate quadruples. This gives us an order of reaction with respect to \(\mathrm{NO}\) of 2 (i.e., second-order in \(\mathrm{NO}\)). Now, compare experiments 2 and 3, where the concentration of \(\mathrm{NO}\) remains constant: $$\frac{rate_3}{rate_2} = \frac{1.13 \times 10^{-1}}{5.64 \times 10^{-2}} = 2$$ $$\frac{[\mathrm{O}_{2}]_3}{[\mathrm{O}_{2}]_2} = \frac{0.0250}{0.0125} = 2$$ Here, doubling the concentration of \(\mathrm{O}_{2}\) results in doubling the rate. This gives us an order of reaction with respect to \(\mathrm{O}_{2}\) of 1 (i.e., first-order in \(\mathrm{O}_{2}\)).

Write the rate law

Now that we know that the reaction is second-order in \(\mathrm{NO}\) and first-order in \(\mathrm{O}_{2}\), we can write the rate law: $$rate = k [\mathrm{NO}]^2 [\mathrm{O}_{2}]$$

Determine the rate constant using the given data

We can now use the given data from any experiment to determine the rate constant, k. For example, let's use experiment 1: $$1.41 \times 10^{-2} = k(0.0126)^2(0.0125)$$ $$k = \frac{1.41 \times 10^{-2}}{(0.0126)^2(0.0125)} \approx 2.84 \times 10^{4} \, M^{-1}s^{-1}$$

Calculate the average rate constant from all experiments

In order to find the average rate constant, we must first calculate them for all experiments and then take the average: For experiment 2: $$k_2 = \frac{5.64 \times 10^{-2}}{(0.0252)^2(0.0125)} \approx 2.82 \times 10^{4} \, M^{-1}s^{-1}$$ For experiment 3: $$k_3 = \frac{1.13 \times 10^{-1}}{(0.0252)^2(0.0250)} \approx 2.82 \times 10^{4} \, M^{-1}s^{-1}$$ The average rate constant is then: $$k_{avg} = \frac{k_1 + k_2 + k_3}{3} = \frac{2.84 \times 10^{4} + 2.82 \times 10^{4} + 2.82 \times 10^{4}}{3} \approx 2.83 \times 10^{4} \, M^{-1}s^{-1}$$

Calculate the rate of disappearance of NO and O2 at given concentrations

We can now use the rate law and the average rate constant to calculate the rate of disappearance of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) at the given concentrations: For \(\mathrm{NO}\): $$rate_{NO} = 2.83 \times 10^{4} (0.0750)^2(0.0100) \approx 0.127 \, M/s$$ For \(\mathrm{O}_{2}\): Since the stoichiometry shows that \(\mathrm{O}_{2}\) reacts in a 1:2 ratio with \(\mathrm{NO}\), the rate of disappearance of \(\mathrm{O}_{2}\) is half the rate of \(\mathrm{NO}\): $$rate_{O_2} = \frac{1}{2} \cdot 0.127 \approx 0.0635 \, M/s$$ #Summary# The solution includes the following findings: (a) The rate law for the reaction is: $$rate = k [\mathrm{NO}]^2 [\mathrm{O}_{2}]$$ (b) The units for the rate constant are: $$M^{-1}s^{-1}$$ (c) The average value of the rate constant calculated from the three data sets is approximately: $$2.83 \times 10^{4} \, M^{-1}s^{-1}$$ (d) The rate of disappearance of \(\mathrm{NO}\) is approximately: $$0.127 \, M/s$$ (e) The rate of disappearance of \(\mathrm{O}_{2}\) is approximately: $$0.0635 \, M/s$$