Question

Nitrogen dioxide undergoes thermal decomposition according to the second-order reaction 2 NO2(g) 2 NO(g) + O2(g). When 0.500 M NO2 is allowed to react for 90.0 seconds, its concentration falls to 0.0196 M. What is the half-life of the reaction when [NO2]0 = 0.500 M?

Answer 1

Answer:

half-life is 3.67s

Explanation:

The general law of the second-order reaction is:

$\frac{1}{[A]}=\frac{1}{[A]_0}+kt$

As after 90.0s, the concentration of NO2 decreases from 0.500M to 0.0196M:

$\frac{1}{[0.0196]}=\frac{1}{[0.500]}+k*90.0s$

49.02M⁻¹ = K*90.0s

0.5447M⁻¹s⁻¹ = K

Now, half-life, t1/2 is:

$t_{1/2}=\frac{1}{K[A]_0}$

Half-life is:

t(1/2) = 1 / (0.5447M⁻¹s⁻¹*0.500M)

half-life is 3.67s