A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer).
Answer (integer)
108
Solution
Initially : $\left[ {{A_0}} \right] = \left[ {{B_0}} \right] = a$<br><br>After time 't' min : $\left[ A \right] = 16\left[ B \right]$<br><br>$\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}$<br><br>$\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}$<br><br>$\Rightarrow a\,.\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}$<br><br>$\Rightarrow {e^{ - \left( {{k_A} - {k_B}} \right)t}} = 16$<br><br>$\Rightarrow \left( {{k_B} - {k_A}} \right)t = \ln 16$<br><br>$\Rightarrow \ln 2\left( {{1 \over {18}} - {1 \over {54}}} \right)t = 4\ln 2$<br><br>$\Rightarrow t = {{54 \times 18 \times 4} \over {36}} = 108$ min
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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