91影视

First, let's consider that the reaction has a first-order rate with respect to NO鈧�: Rate = k[NO鈧俔^n, where n can be 1 (first-order) or 2 (second-order). If the reaction is first-order, we would expect the graph of ln[NO鈧俔 vs. time to be a straight line. If second-order, we would expect the graph of 1/[NO鈧俔 vs. time to be a straight line. Using the given data, we can calculate ln[NO鈧俔 and 1/[NO鈧俔 for each time point: $$ \begin{array}{c|c|c|c} \hline \text { Time (s) } & [\mathrm{NO}_{2}] (M) & \ln [\mathrm{NO}_{2}] & \frac{1}{[\mathrm{NO}_{2}]} \\ \hline 0.0 & 0.100 & \ln(0.100) & \frac{1}{0.100} \\ 5.0 & 0.017 & \ln(0.017) & \frac{1}{0.017} \\ 10.0 & 0.0090 & \ln(0.0090) & \frac{1}{0.0090} \\ 15.0 & 0.0062 & \ln(0.0062) & \frac{1}{0.0062} \\ 20.0 & 0.0047 & \ln(0.0047) & \frac{1}{0.0047} \\ \hline \end{array} $$ Now, we can check which relationship is linear by making the respective graphs. After plotting the graphs, it becomes clear that 1/[NO鈧俔 vs. time gives a straight line. This indicates that the reaction is second-order with respect to the concentration of NO鈧�.

Now that we know the reaction is second-order, we will use the second-order rate equation to find the rate constant: Rate = k[NO鈧俔^2 Since the reaction is second-order, the graph of 1/[NO鈧俔 vs. time should yield a straight line with a slope equal to the rate constant k. The equation of the straight line for a second-order reaction is: \( \frac{1}{[\mathrm{NO}_{2}]} = k\cdot t + \frac{1}{[\mathrm{NO}_{2}]_0} \) We will use the first two data points (time = 0.0 s and 5.0 s) to calculate the slope of the line, which represents the rate constant. \( k = \frac{\frac{1}{[\mathrm{NO}_{2}(t=5.0\:s)]} - \frac{1}{[\mathrm{NO}_{2}(t=0)]}}{5.0\:s - 0} = \frac{\frac{1}{0.017\:M} - \frac{1}{0.100\:M}}{5.0\:s} \) Upon calculating the value of k, we get: k 鈮� 0.0225 M鈦宦箂鈦宦� The value of the rate constant, k, for the gas-phase decomposition of NO鈧� is approximately 0.0225 M鈦宦箂鈦宦�.