JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 19)
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
108
Explanation
Initially : $$\left[ {{A_0}} \right] = \left[ {{B_0}} \right] = a$$
After time 't' min : $$\left[ A \right] = 16\left[ B \right]$$
$$\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}$$
$$\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}$$
$$ \Rightarrow a\,.\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}$$
$$ \Rightarrow {e^{ - \left( {{k_A} - {k_B}} \right)t}} = 16$$
$$ \Rightarrow \left( {{k_B} - {k_A}} \right)t = \ln 16$$
$$ \Rightarrow \ln 2\left( {{1 \over {18}} - {1 \over {54}}} \right)t = 4\ln 2$$
$$ \Rightarrow t = {{54 \times 18 \times 4} \over {36}} = 108$$ min
After time 't' min : $$\left[ A \right] = 16\left[ B \right]$$
$$\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}$$
$$\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}$$
$$ \Rightarrow a\,.\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}$$
$$ \Rightarrow {e^{ - \left( {{k_A} - {k_B}} \right)t}} = 16$$
$$ \Rightarrow \left( {{k_B} - {k_A}} \right)t = \ln 16$$
$$ \Rightarrow \ln 2\left( {{1 \over {18}} - {1 \over {54}}} \right)t = 4\ln 2$$
$$ \Rightarrow t = {{54 \times 18 \times 4} \over {36}} = 108$$ min
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