Answer:
Choice D. Approximately [tex]61.2[/tex] minutes.
Explanation:
The reaction rate of a second-order reaction is proportional to the square of one of its reactants.
Let [tex]k[/tex] denote the rate constant of this reaction. Assume that [tex]y_0[/tex] is the concentration of that reactant at the beginning of this reaction (when [tex]t = 0[/tex].) Because this reaction is of second order, it can be shown that:
[tex]\displaystyle y = \frac{1}{k\, t + (1/y_0)}[/tex].
The question states that the rate constant here is [tex]0.06\; \rm M\cdot min^{-1}[/tex]. Hence, [tex]k = 0.06[/tex]. For simplicity, assume that [tex]t = 0[/tex] when the concentration is [tex]0.13\; \rm M[/tex]. Therefore, [tex]y_0 = 0.13[/tex]. The equation for concentration [tex]y[/tex] at time [tex]t[/tex] would become:
[tex]\displaystyle y = \frac{1}{\underbrace{0.06}_{k}\, t + (1/\underbrace{0.13}_{y_0})}[/tex].
The goal is to find the time at which the concentration is [tex]0.088[/tex]. That's the same as solving this equation for [tex]t[/tex], given that [tex]y = 0.088[/tex]:
[tex]\displaystyle \frac{1}{0.06\; t + (1/0.13)} = 0.088[/tex].
[tex]t \approx 61.2[/tex].
In other words, it would take approximately [tex]61.2[/tex] minutes before the concentration of this second-order reactant becomes [tex]0.088\; \rm M[/tex].
