Chemistry second order kinetics

Question

Consider the equation; NO2< O2 + NO

if the reaction proceeds at second order kinetics and has half-life of 100.0 seconds. If the reaction is carried out at some temperature and the initial temperature of NO2 is 1.225 M, what will the NO2 concentration be after 35.4s? At 175s? At 865 sec?

Answer 1

This is the decomposition of NO₂, which is a 2nd order reaction.

NO₂ ==> NO + O₂ .. unbalanced equation

2NO₂ ==> 2NO + O₂ .. balanced equation

Integrated rate equation for a 2nd order reaction is ...

1/[A] = 1/[A]_o + kt

Also, for a 2nd order reaction, t_1/2 = 1 / k[A]_o

We know [A]_o, which is 1.225 M (not the initial temperature as stated) and we know t_1/2 (100.0 s)

We can now find k (rate constant)

t_1/2 = 1 / k[A]_o

100.0 s = 1 / 1.225k

k = 8.16x10^-3 M^-1s^-1

Now use k in the integrated rate law and solve for [A] at the indicated times...

@35.4 s: 1/[A] = 1/1.225 + (8.16x10^-3)(35.4)

1/[A] = 0.8163 + 0.2890 = 1.105

[A] = 0.905 M (3 sig.figs.) = [NO₂] @ 35.4 s (be sure to check the math)

Repeat the above calculations for the other times by simply substituting175s and 865 s for 35.4 s.