Question

The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous question. One would expect the second half-life of this reaction to be ________ the first half-life of this reaction.

Answer 1

Step-by-step explanation:

For the given reaction
`2NO_(2) \rightarrow 2NO + O_(2)`

Now, expression for half-life of a second order reaction is as follows.

`t_(1/2) = (1)/([A_(0)]k)` ....... (1)

Second half life of this reaction will be
`t_(1/4)`. So, expression for this will be as follows.

`t_(1/4)` =
`(1)/(k) [(1)/([A]_(f)) - (1)/([A_(0)])]` ...(2)

where
`[A]_(f)` is the final concentration that is,
`([A]_(0))/(4)` here and
`[A]_(i)` is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

`t_(1/4)` =
`(1)/(k) [(4)/([A]_(0)) - (1)/([A_(0)])]`

=
`(3)/([A_(0)]k)` ...... (3)

Now, dividing equation (3) by equation (1) as follows.

`(t_(1/4))/(t_(1/2))` =
`(3)/([A_(0)]k) * [A_(0)]k`

= 3

or,
`t_(1/4)` = 3
`t_(1/2)`

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.