Ace chemistry student Sidney Einstein (no relation) was carrying out a kinetics experiment in lab which produces iodine and uses starch as an indicator. In the first minute after mixing, Sidney spilled some of the solution. Rather than start over, he decided to pour out more solution until he had exactly half the original volume in the tube. Given that the particular reaction under investigation was second order in the reactant being investigated and was supposed to change color after 10 minutes, approximately how long (minutes) will it be before Sidney sees a color change? Explain briefly!

For a second-order reaction, the rate law can be written as: Rate = k[A]^{2}, where k is the rate constant, and [A] is the concentration of the reactant. The integrated rate law for a second-order reaction is: \(\frac{1}{[A]_{t}} = \frac{1}{[A]_{0}} + kt\), where [A]_{t} is the concentration at time t, [A]_{0} is the initial concentration, and t is the time. Since the reaction is second-order with respect to the reactant, the integrated rate law will help us relate the initial and final concentrations in relation to time and rate constant.

Let [A]_{10} be the concentration at the expected color change after 10 minutes. Using the integrated rate law, we can write: \(\frac{1}{[A]_{10}} = \frac{1}{[A]_{0}} + k(10)\), where [A]_{0} is the initial concentration before Sidney spilled the solution.

When Sidney pours out half of the solution, the initial concentration doubles because the volume is reduced to half. Let's write this as: \([A']_{0} = 2[A]_{0}\) Now Sidney has spilled some solution, so there will be a new time t' for the color change. We need to find t'. Using the integrated rate law with the new initial concentration [A']_{0}: \(\frac{1}{[A]_{10}} = \frac{1}{[A']_{0}} + kt'\), where t' is the new time after the spill. Note that [A]_{10} is still the concentration at the color change.

Substitute the expression for [A']_{0} from step 3: \(\frac{1}{[A]_{10}} = \frac{1}{2[A]_{0}} + kt'\) Substitute the expression for [A]_{10} from step 2: \(\frac{1}{\frac{1}{[A]_{0}} + k(10)} = \frac{1}{2[A]_{0}} + kt'\) Next, we will solve for t': Rearrange the equation: \(kt' = \frac{1}{2[A]_{0}} - \frac{1}{[A]_{0}} + k(10)\) The equation simplifies to: \(t' = \frac{1}{2k[A]_{0}} - \frac{1}{k[A]_{0}} + 10\) We see that the new time will depend on the rate constant k and the initial concentration [A]_{0}. Since the problem does not provide specific values for these parameters, we cannot calculate an exact value for t'. However, we can note that since Sidney poured out more solution, the color change should occur in less than 10 minutes.