‏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

Choice D. Approximately minutes.

Explanation:

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

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

.

The question states that the rate constant here is . Hence, . For simplicity, assume that when the concentration is . Therefore, . The equation for concentration at time would become:

.

The goal is to find the time at which the concentration is . That's the same as solving this equation for , given that :

.

.

In other words, it would take approximately minutes before the concentration of this second-order reactant becomes .

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.