Question

A reaction was observed for 15 days and the percentage of the reactant remaining after the days indicated was recorded in the following table:

Time(days)	0	2	4	6	8	10	12	14	15
% Reactant remaining	100	50	39	25	21	18	15	12.5	10

Which one of the following best describes the order and the half-life of the reaction?

Reaction order	Half-life (days)
A.First	2
B.First	6
C.Second	2
D.Zero	6
E.Third	2

Answer 1

Hint: In this question, the reaction is observed for 15 days and the percentage or reactant remaining is recorded. The half-life and the order of reaction is asked. Half-life is defined as the time at which the percentage of reactant is reduced to $50\% $ . Here by observing the given table, we can see that on the 2nd day the reactant is reduced to half. Hence, the half-life day is 2 days. To determine the order of reaction we will use the order equation.

Complete answer:
By observing the table, it is clear that the percentage of reactant is reduced by $50\% $ on the 2nd day. Hence the half-life days is 2days.
Now we will evaluate the order of reaction.
For that, let us assume the equation for ${n^{th}}$ order reaction,
$k = \dfrac{1}{{t(n - 1)}}\left[ {\dfrac{1}{{{{(a - x)}^{(n - 1)}}}} + \dfrac{1}{{{{(a)}^{(n - 1)}}}}} \right]$
 Here,
$x$ is concentration of reactant consumed
$a$ is the initial concentration of reactant
For half-life, $t = 2$ days, \[x = \dfrac{a}{2}\]
Hence,
$k = \dfrac{1}{{2(n - 1)}}\left[ {\dfrac{1}{{{{(a - \dfrac{a}{2})}^{(n - 1)}}}} + \dfrac{1}{{{{(a)}^{(n - 1)}}}}} \right]$
$k = \dfrac{1}{{2(n - 1)}}\left[ {\dfrac{{{2^{(n - 1)}} - 1}}{{{a^{(n - 1)}}}}} \right]$ - (i)
For t-6 days, the reactant has reduced to $25\% $ of the reactant.
Hence, $t = 6$ days and $x = \dfrac{{3a}}{4}$
\[k = \dfrac{1}{{2(n - 1)}}\left[ {\dfrac{1}{{{{(a - \dfrac{{3a}}{4})}^{(n - 1)}}}} + \dfrac{1}{{{{(a)}^{(n - 1)}}}}} \right]\]
$k = \dfrac{1}{{2(n - 1)}}\left[ {\dfrac{{{4^{(n - 1)}} - 1}}{{{a^{(n - 1)}}}}} \right]$ … (ii)
We will divide (ii) by (i),
$1 = 3({2^{n - 1}} + 1)\left[ {\dfrac{{{2^{(n - 1)}} - 1}}{{{2^{(n - 1)}} - 1}}} \right]$
$1 = \dfrac{3}{{({2^{(n - 1)}} + 1)}}$
By cross multiplication,
${2^{(n - 1)}} + 1 = 3$
${2^{(n - 1)}} = 3 - 1$
On simplification,
${2^{(n - 1)}} = {2^1}$
Here LHS=RHS,
By comparing both the sides, we get
$n - 1 = 1$
$n = 1 + 1$
On simplifying we get,
$ \Rightarrow n = 2$
Therefore, the order of reaction is 2. Hence the correct answer is option (C) Second order and 2 days.

Note:
We have to know that the order of reaction is defined as the power dependence of the rate in accordance to the concentration of each reactant. By using the nth order equation and the percentage amount of reactant left with the exceeding number of days, we found the order of reaction. While determining the half-life of the reaction was quite easy. As the half-life days of a reaction is basically the number of days after which the reactant becomes half of the actual mass.