The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g),\) is studied at \(383{ }^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{rl} \hline \text { Time }(\mathbf{s}) & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the rate constant? (c) If you used the method of initial rates to obtain the order for \(\mathrm{NO}_{2},\) predict what reaction rates you would measure in the beginning of the reaction for initial concentrations of \(0.200 \mathrm{M}, 0.100 \mathrm{M},\) and \(0.050 \mathrm{M} \mathrm{NO}_{2}\)

The given reaction is the decomposition of NO₂: \[2NO_{2}(g) \longrightarrow 2NO(g) + O_{2}(g)\] The given data is a table containing the time in seconds and the corresponding concentration of NO₂.

To determine the order of the reaction with respect to NO₂, we will analyze the change in concentration of NO₂ over time. We can calculate the rate of disappearance of NO₂ during different time intervals and compare that with the rate of decrease during the first time interval. First interval: \(\frac{-\Delta[NO_{2}]}{\Delta t} = \frac{- (0.017 - 0.100)}{5.0 } = 0.0166\ M/s\) Second interval: \(\frac{-\Delta[NO_{2}]}{\Delta t} = \frac{- (0.009 - 0.017)}{5.0 } = 0.0016\ M/s\) Here, we see that the initial rate of the reaction during the two time intervals is different, and we can't determine the order of the reaction by just looking at the given data. We can, though, use the method of initial rates to find the order of the reaction as asked in the question (c).

We will compare the initial rates of the reaction when the initial concentration of NO₂ is 0.200 M, 0.100 M, and 0.050 M. We are assuming with the method of initial rates that the order of the reaction doesn't change during the reaction. For an initial concentration of 0.200 M: The given data shows that the initial rate (R₀) when [NO₂] is 0.100 M is 0.0166 M/s. To find the initial rate of the reaction for 0.200 M, we assume a proportionality \(R \propto [NO_{2}]^n\), where n is the order of the reaction: \(\frac{R}{R_{0}} = \frac{([NO_{2}]_{0})^n}{(([NO_{2}])_{0,ref})^n} \Rightarrow R = R_{0} ([NO_{2}]_{0} / ([NO_{2}])_{0,ref})^n\) Using this formula and assuming the order of the reaction (n): For the first order (n = 1): \(R = 0.0166 * (0.200 / 0.100) = 0.0332\ M/s\) For the second order (n = 2): \(R = 0.0166 * (0.200 / 0.100)^2 = 0.0664\ M/s\) Now we can calculate the initial rates for the other initial concentrations: For an initial concentration of 0.050 M: For the first order (n = 1): \(R = 0.0083\ M/s\) For the second order (n = 2): \(R = 0.00208\ M/s\) Considering the method of initial rates and using the calculated initial rates of the reaction, one can experimentally verify if the reaction has a first order or a second order rate law by testing different initial concentrations of NO₂. This answer doesn't provide a direct answer for being first or second order, instead, it allows the author to give a verification method, which is usually used in practice while teaching physical organic chemistry to high school students.