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

Choice D. Approximately
`61.2` minutes.

Step-by-step 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 = (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 (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`.