A flask contains a mixture of compounds A and
B. Both compounds decompose by first-order
kinetics. The half-lives for A and B are 300 s
and 180 s, respectively. If the concentrations
of A and B are equal initially, the time required
for the concentration of A to be four times that
of B(in s) :
(Use ln 2 = 0.693)
Solution
A<sub>t</sub> = A<sub>0</sub>.e<sup>-k<sub>1</sub>t</sup>
<br><br>B<sub>t</sub> = B<sub>0</sub>.e<sup>-k<sub>2</sub>t</sup>
<br><br>k<sub>1</sub> = ${{\ln 2} \over {300}}$
<br><br>k<sub>2</sub> = ${{\ln 2} \over {180}}$
<br><br>Given, A<sub>0</sub> = B<sub>0</sub>
<br><br>and A<sub>t</sub>
and B<sub>t</sub>
are related as [A] = 4[B]
<br><br>$\therefore$ A<sub>0</sub>.e<sup>-k<sub>1</sub>t</sup> = 4B<sub>0</sub>.e<sup>-k<sub>2</sub>t</sup>
<br><br>$\Rightarrow$ ${e^{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}}$ = 4
<br><br>$\Rightarrow$ ${\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}$ = ln 4 = 2ln 2
<br><br>$\Rightarrow$ ${\left( {{1 \over {180}} - {{\mathop{\rm l}\nolimits} \over {300}}} \right)t}$ = 2
<br><br>$\Rightarrow$ t = ${{2 \times 180 \times 300} \over {120}}$ = 900 sec
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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