13.27 The decomposition reaction of NOBr is second order in NOBr , with a rate constant at \( 20^{\circ} \mathrm{C} \) of \( 25 \mathrm{M}^{-1} \mathrm{~min}^{-1} \). If the initial concentration of NOBr is 0.025 M , find (a) the time at which the concentration will be 0.010 M ; and (b) the concentration after 125 min of reaction.

To find the time at which the concentration of NOBr decreases from 0.025 M to 0.010 M, we can use the integrated rate law for a second-order reaction: \[ \frac{1}{[A]} = kt + \frac{1}{[A_0]} \] Where: - \( [A_0] = 0.025 \, \text{M} \) (initial concentration) - \( [A] = 0.010 \, \text{M} \) (final concentration) - \( k = 25 \, \text{M}^{-1} \, \text{min}^{-1} \) - \( t \) = time (in minutes) Substituting into the equation: \[ \frac{1}{0.010} = 25\, t + \frac{1}{0.025} \] Calculating the values: \[ 100 = 25\, t + 40 \] \[ 100 - 40 = 25\, t \] \[ 60 = 25\, t \] \[ t = \frac{60}{25} = 2.4 \, \text{min} \] So, the time at which the concentration will be 0.010 M is 2.4 minutes. For (b), to find the concentration after 125 minutes, we again use the integrated rate equation: \[ \frac{1}{[A]} = kt + \frac{1}{[A_0]} \] Substituting the values: \[ \frac{1}{[A]} = 25 \times 125 + \frac{1}{0.025} \] Calculating: \[ \frac{1}{[A]} = 3125 + 40 = 3165 \] Now, taking the reciprocal to find concentration: \[ [A] = \frac{1}{3165} \approx 0.000316 \, \text{M} \] Thus, the concentration of NOBr after 125 minutes of reaction is approximately \( 0.000316 \, \text{M} \).