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

Explanation:

For the given reaction $2NO_{2} \rightarrow 2NO + O_{2}$

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

                  $t_{1/2} = \frac{1}{[A_{0}]k}$     ....... (1)

Second half life of this reaction will be $t_{1/4}$. So, expression for this will be as follows.

          $t_{1/4}$ = $\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]$  ...(2)

where $[A]_{f}$ is the final concentration that is, $\frac{[A]_{0}}{4}$ here and $[A]_{i}$ is the initial concentration.

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

        $t_{1/4}$ = $\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]$

                      = $\frac{3}{[A_{0}]k}$     ...... (3)

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

           $\frac{t_{1/4}}{t_{1/2}}$ = $\frac{3}{[A_{0}]k} \times [A_{0}]k$                  

                                    = 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.