Final answer:
The time at which 50% of a reactant remains in a second-order reaction cannot be directly obtained by doubling the half-life; the specific time must be calculated using the second-order rate equation.
Step-by-step explanation:
The student's question relates to determining the time required for a second-order reaction's reactant concentration to decrease to 50% of its initial amount, given that the reaction's half-life (t1/2) is 1 hour. A second-order reaction's half-life is inversely proportional to the initial concentration of the reactant and directly proportional to the rate constant. Unlike first-order reactions, for second-order reactions, the half-life will vary depending on the initial concentration.
For a second-order reaction, the half-life equation is t1/2 = 1 / (k[A]0), where k is the rate constant and [A]0 is the initial concentration of the reactant. If the half-life is given as 1 hour, we can infer that doubling the time to 2 hours does not equate to completing another half-life as it does in a first-order reaction.
To find the time when the concentration is reduced to 50% of the initial concentration, we need to use the integrated rate law for second-order reactions. This law is represented as 1/[A] = kt + 1/[A]0, where [A] is the concentration at time t. Solving for t when [A] is half of [A]0 leads us to establish that the time at which 50% of the reactant remains for a second-order reaction is not simply 2 × t1/2 (as might be the case in a first-order reaction) but will be calculated specifically based on rate constant (k) and the initial concentration ([A]0).