Question

‏A second - order reaction has a rate constant of 0.06 M min - 1 . It takes min for the reactant concentration to decrease from 0.13 M to 0.088 M. Select one : O a . 73.4 O b . 7.80 O c . 6.50 O d . 61.2

Answer 1

Answer:

Choice D. Approximately $61.2$ minutes.

Explanation:

The reaction rate of a second-order reaction is proportional to the square of one of its reactants.

• Let $y$ denote the concentration of that reactant (in $\rm M$.)
• Let $t$ denote time (in minutes.)

Let $k$ denote the rate constant of this reaction. Assume that $y_0$ is the concentration of that reactant at the beginning of this reaction (when $t = 0$.) Because this reaction is of second order, it can be shown that:

$\displaystyle y = \frac{1}{k\, t + (1/y_0)}$.

The question states that the rate constant here is $0.06\; \rm M\cdot min^{-1}$. Hence, $k = 0.06$. For simplicity, assume that $t = 0$ when the concentration is $0.13\; \rm M$. Therefore, $y_0 = 0.13$. The equation for concentration $y$ at time $t$ would become:

$\displaystyle y = \frac{1}{\underbrace{0.06}_{k}\, t + (1/\underbrace{0.13}_{y_0})}$.

The goal is to find the time at which the concentration is $0.088$. That's the same as solving this equation for $t$, given that $y = 0.088$:

$\displaystyle \frac{1}{0.06\; t + (1/0.13)} = 0.088$.

$t \approx 61.2$.

In other words, it would take approximately $61.2$ minutes before the concentration of this second-order reactant becomes $0.088\; \rm M$.