JEE MAIN - Chemistry (2020 - 5th September Morning Slot - No. 3)
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)
(Use ln 2 = 0.693)
180
120
300
900
Explanation
At = A0.e-k1t
Bt = B0.e-k2t
k1 = $${{\ln 2} \over {300}}$$
k2 = $${{\ln 2} \over {180}}$$
Given, A0 = B0
and At and Bt are related as [A] = 4[B]
$$ \therefore $$ A0.e-k1t = 4B0.e-k2t
$$ \Rightarrow $$ $${e^{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}}$$ = 4
$$ \Rightarrow $$ $${\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}$$ = ln 4 = 2ln 2
$$ \Rightarrow $$ $${\left( {{1 \over {180}} - {{\mathop{\rm l}\nolimits} \over {300}}} \right)t}$$ = 2
$$ \Rightarrow $$ t = $${{2 \times 180 \times 300} \over {120}}$$ = 900 sec
Bt = B0.e-k2t
k1 = $${{\ln 2} \over {300}}$$
k2 = $${{\ln 2} \over {180}}$$
Given, A0 = B0
and At and Bt are related as [A] = 4[B]
$$ \therefore $$ A0.e-k1t = 4B0.e-k2t
$$ \Rightarrow $$ $${e^{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}}$$ = 4
$$ \Rightarrow $$ $${\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}$$ = ln 4 = 2ln 2
$$ \Rightarrow $$ $${\left( {{1 \over {180}} - {{\mathop{\rm l}\nolimits} \over {300}}} \right)t}$$ = 2
$$ \Rightarrow $$ t = $${{2 \times 180 \times 300} \over {120}}$$ = 900 sec
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