12) The reaction 2NO_(2)arrow 2NO+O_(2) follows second-order kinetics.At 300^circ C,[NO_(2)] drops from 0.0100 M to 0.00650 M in 100.0 s. The rate constant for the reaction is __ M^-1s^-1 A) 0.096 B) 0.65 C) 0.81 D) 1.2 E) 0.54 Answer: E

To determine the rate constant for the reaction, we can use the integrated rate law for a second-order reaction:<br /><br />\[<br />\frac{1}{[A]} - \frac{1}{[A_0]} = kt<br />\]<br /><br />where:<br />- \([A]\) is the final concentration of the reactant,<br />- \([A_0]\) is the initial concentration of the reactant,<br />- \(k\) is the rate constant,<br />- \(t\) is the time.<br /><br />Given:<br />- \([A_0] = 0.0100 \, M\)<br />- \([A] = 0.00650 \, M\)<br />- \(t = 100.0 \, s\)<br /><br />Substitute these values into the integrated rate law equation:<br /><br />\[<br />\frac{1}{0.00650} - \frac{1}{0.0100} = k \cdot 100.0<br />\]<br /><br />Calculate the left-hand side:<br /><br />\[<br />\frac{1}{0.00650} = 153.85 \, M^{-1}<br />\]<br />\[<br />\frac{1}{0.0100} = 100.0 \, M^{-1}<br />\]<br />\[<br />153.85 - 100.0 = 53.85 \, M^{-1}<br />\]<br /><br />Now, solve for \(k\):<br /><br />\[<br />k = \frac{53.85}{100.0} = 0.5385 \, M^{-1}s^{-1}<br />\]<br /><br />Rounding to two decimal places, we get:<br /><br />\[<br />k \approx 0.54 \, M^{-1}s^{-1}<br />\]<br /><br />Thus, the correct answer is:<br /><br />E) 0.54