The following data were collected for the rate of disappearance of \(\mathrm{NO}\) in the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) : $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{NO}](M)} & {\left[\mathrm{O}_{2}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.0126 & 0.0125 & 1.41 \times 10^{-2} \\ 2 & 0.0252 & 0.0125 & 5.64 \times 10^{-2} \\ 3 & 0.0252 & 0.0250 & 1.13 \times 10^{-1} \end{array} $$ (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 \(\mathrm{NO}\) 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 ( \(\mathrm{d}\) )?

To determine the order of the reaction with respect to each reactant, we need to compare the change in concentrations between pairs of experiments while keeping the other reactant concentration constant. Let's take a look at Experiments 1 and 2 since the concentration of \(\mathrm{O}_{2}\) is constant at 0.0125 M. If the order of the reaction with respect to \(\mathrm{NO}\) is 1 (first-order): $$\frac{\text{Rate}_2}{\text{Rate}_1}=\frac{[\mathrm{NO}]_2}{[\mathrm{NO}]_1}$$ Plugging in the values from the table: $$\frac{5.64 \times 10^{-2}}{1.41 \times 10^{-2}} = \frac{0.0252}{0.0126}$$ Solving for the ratio: $$3.998 = 2$$ Since the ratio obtained is 4, the order of the reaction with respect to \(\mathrm{NO}\) must be 2 (second-order). Next, let's compare Experiments 2 and 3, where the concentration of \(\mathrm{NO}\) is constant at 0.0252 M. If the order of the reaction with respect to \(\mathrm{O}_{2}\) is 1 (first-order): $$\frac{\text{Rate}_3}{\text{Rate}_2}=\frac{[\mathrm{O}_{2}]_3}{[\mathrm{O}_{2}]_2}$$ Plugging in the values from table: $$\frac{1.13 \times 10^{-1}}{5.64 \times 10^{-2}} = \frac{0.0250}{0.0125}$$ Solving for the ratio: $$2.00 = 2$$ Therefore, the order of the reaction with respect to \(\mathrm{O}_{2}\) is 1 (first-order).

Now that we know the order of the reaction with respect to each reactant, we can write the rate law as follows: $$\text{Rate} = k[\mathrm{NO}]^2[\mathrm{O}_{2}]$$

To determine the units of the rate constant k, we'll use the rate law equation above: $$\frac{M}{s} = k \cdot M^2 \cdot M$$ Solving for k, we get: $$k = \frac{1}{M^2 \cdot s}$$ The units of the rate constant k are \(\frac{1}{M^2 \cdot s}\).

To find the average value of the rate constant, we will use the rate law to calculate the rate constant for each experiment: For Experiment 1: $$k_1 = \frac{\text{Rate}_1}{[\mathrm{NO}]_1^2[\mathrm{O}_{2}]_1} = \frac{1.41 \times 10^{-2}}{(0.0126)^2(0.0125)}$$ For Experiment 2: $$k_2 = \frac{\text{Rate}_2}{[\mathrm{NO}]_2^2[\mathrm{O}_{2}]_2} = \frac{5.64 \times 10^{-2}}{(0.0252)^2(0.0125)}$$ For Experiment 3: $$k_3 = \frac{\text{Rate}_3}{[\mathrm{NO}]_3^2[\mathrm{O}_{2}]_3} = \frac{1.13 \times 10^{-1}}{(0.0252)^2(0.0250)}$$ Now calculate the average value of k: $$\bar{k} = \frac{k_1 + k_2 + k_3}{3}$$

Using the rate law and the average rate constant value, we can calculate the rate of disappearance of \(\mathrm{NO}\) when \([\mathrm{NO}] = 0.0750 M\) and \([\mathrm{O}_{2}] = 0.0100 M\): $$\text{Rate}_{\mathrm{NO}} = \bar{k}[\mathrm{NO}]^2[\mathrm{O}_{2}]$$ To find the rate of disappearance of \(\mathrm{O}_{2}\), we need to know the stoichiometry of the reactants. In this reaction, 2 moles of \(\mathrm{NO}\) react with 1 mole of \(\mathrm{O}_{2}\): $$2 \text{ moles NO} \longrightarrow 1 \text{ mole O}_{2}$$ Therefore, the rate of disappearance of \(\mathrm{O}_{2}\) is half the rate of disappearance of \(\mathrm{NO}\): $$\text{Rate}_{\mathrm{O}_{2}} = \frac{1}{2} \text{Rate}_{\mathrm{NO}}$$