Question

A and B decompose via first-order kinetics with half-lives \[{\mathbf{54}}.{\mathbf{0}}\]min and \[{\mathbf{18}}.{\mathbf{0}}\]min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become \[{\mathbf{16}}\] times that of B is __________ min. (Round off to the nearest integer).

Answer 1

Hint: Half life is defined as the time at which half of the initial concentration of reactant has decomposed or half of the amount of initial reactant is left. Half live is inversely proportional to \[{a_0}\left( {n - 1} \right)\]where n is the order of the reaction and ao is the initial amount of concentration.

Formula Used:
For first order reaction, \[n = 1\]
\[kt = \ln \frac{{{a_o}}}{a}\]
 \[k{t_{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = \ln \frac{{{a_o}}}{{{\raise0.5ex\hbox{$\scriptstyle {{a_o}}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}\]
\[k{t_{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = \ln 2\]

Complete Step by Step Answer:
 Rate constants for A, ${k_1} = \frac{{\ln 2}}{{{t_1}}}$where t1 is the half life of A.
${k_1} = \frac{{\ln 2}}{{54}} = 0.0128$
Similarly, \[{k_2} = \frac{{\ln 2}}{{{t_2}}}\] where t2 is the half life of B.
${k_2} = \frac{{\ln 2}}{{18}} = 0.0385$
Equimolar mixture means both A and B are present in same concentration. ‘Equi’ means equal and ‘molar’ means molarity (that is a concentration term). So,\[{A_0} = {B_0}\]. Let the time be ‘t’ when \[\left[ A \right]{\text{ }} = 16[B]\]
\[{k_1}t = \ln \frac{{{A_o}}}{A}\] ----------(1)
 \[{k_2}t = \ln \frac{{{B_o}}}{B}\] ----------(2)
Put \[\left[ A \right]{\text{ }} = 16[B]\]and \[{A_0} = {B_0}\]in eq (1)
\[{k_1}t = \ln \frac{{{B_o}}}{{16B}}\] ----------(3)
\[{k_1}t = \ln \frac{{{B_o}}}{B} - \ln 16\]
Using eq(2)
\[{k_1}t = {k_2}t - \ln 16\]
\[ - {k_1}t + {k_2}t = \ln 16\]
\[t({k_2} - {k_1}) = \ln 16\]
$t(0.0385 - 0.0128) = \ln 16$
$t(0.0257) = \ln 16$
$t = 107.88 = 108$
Hence, the time when concentration of A will be 16 times that of B is 108 min (rounded to the nearest integer).

Note: Half-life of a first order reaction is independent of the initial concentration of reactant. It only depends on the rate constant. Hence two substances can have the same half-life if the two have the same rate constant if they may have different initial concentration. The unit of rate constant for first order reaction is min-1 and the order of the reaction can be identified using the unit of rate constant if the order is not mentioned.