The rate constant for the second-order reaction 2no2 (g) ? 2no (g) o2 (g) is 0.540 / (m · s) at 300°c. how long (in seconds) would it take for the concentration of no2 to decrease from 0.500 m to 0.250 m

For a second-order reaction, the rate law would be expressed as:

r = dC/dt = - kC^2

Integrating it from time zero and the initial concentration, Co, to time, t, and the final concentration, C. We will obtain the second-order integrated law as follows:

1/Co - 1/C = - kt

To determine the time needed to change the concentration, we simply substitute the given values from the problem statement as follows:

1/Co - 1/C = - kt

1/.5 - 1/.25 = - 0.540t

t = 3.70 s

Therefore, the time it takes to decrease the concentration of NO2 from 0.500 M to 0.250 M is approximately 3.70 s assuming that it follows a second-order reaction.