91Ó°ÊÓ

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.

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 \(\mathrm{O}_{2}\)).

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: $$rate = k [\mathrm{NO}]^2 [\mathrm{O}_{2}]$$

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

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: $$k_{avg} = \frac{k_1 + k_2 + k_3}{3} = \frac{2.84 \times 10^{4} + 2.82 \times 10^{4} + 2.82 \times 10^{4}}{3} \approx 2.83 \times 10^{4} \, M^{-1}s^{-1}$$

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: $$0.0635 \, M/s$$