Question

The rate constant for the second order reaction: 2NO2------> 2NO + O2 is 0.54m^-1s^-1 at 300 degrees C. How long in seconds would it take for the concentration of NO2 to decrease from 0.62 M to 0.28 M ?

Answer 1

To calculate the time required for the concentration of NO2 to decrease from 0.62 M to 0.28 M, we use the integrated rate law for a second-order reaction. Plugging in the given values, we find that it would take 2 seconds for the concentration of NO2 to reach 0.28 M.

Step-by-step explanation:

To determine the time required for the concentration of NO2 to decrease from 0.62 M to 0.28 M, we can use the integrated rate law for a second-order reaction: rate = k[NO2]². Rearranging this equation, we have: dt = 1/(k[NO2]²) * d[NO2].

We know that the rate constant, k, is 0.54 M^-1s^-1. Plugging in the initial and final concentrations, we can calculate the time required: dt = 1/(0.54 M^-1s^-1 * (0.28 M)²) * (0.28 M - 0.62 M).

Simplifying the equation, we get dt = 1/(0.16632 M^-2s^-1) * (-0.34 M).

Using the given rate constant, we can calculate the desired time: dt = -2 seconds.