Nitrogen dioxide undergoes thermal decomposition according to the second-order reaction 2 NO₂(g)$\to$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?

Question

Quizplus · Accepted Answer

The rate law for the given reaction is:

rate = k[NOS]^2

We can calculate the rate constant (k) using the initial concentration of NOS (0.500 M) and the initial rate (when t=0):

rate = k[NOS]^2
0.118 = k(0.500)^2
k = 0.472 M^-1s^-1

Using the integrated rate law for a second-order reaction, we can calculate the half-life (t1/2) as:

t1/2 = 1/(k[NOS]o)

where [NOS]o is the initial concentration of NOS. Substituting the given values:

t1/2 = 1/(0.472 x 0.500)
t1/2 = 3.67 s

Therefore, the correct answer is C) 3.67 s.

Nitrogen Dioxide Undergoes Thermal Decomposition According to the Second-Order Reaction →\to→

Nitrogen Dioxide Undergoes Thermal Decomposition According to the Second-Order Reaction $\to$