JEE MAIN - Chemistry (2021 - 18th March Evening Shift - No. 20)
A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is _________ min. (Round off to the Nearest Integer). [Use : ln 2 = 0.69; ln 10 = 2.3]
Answer
10
Explanation
Given t1/2 = 1 min
$$ \therefore $$ $$K = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over 1}$$
From formula we know,
$$Kt = \ln \left( {{{100} \over {100 - x}}} \right)$$
$$ \Rightarrow {{\ln 2} \over 1} = {1 \over t}\ln \left( {{{100} \over {0.1}}} \right)$$
$$ \Rightarrow 0.3 = {1 \over t}\ln \left( {{{10}^3}} \right)$$
$$ \Rightarrow 0.3 = {1 \over t} \times 3$$
$$ \Rightarrow t = 10$$ min
$$ \therefore $$ $$K = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over 1}$$
From formula we know,
$$Kt = \ln \left( {{{100} \over {100 - x}}} \right)$$
$$ \Rightarrow {{\ln 2} \over 1} = {1 \over t}\ln \left( {{{100} \over {0.1}}} \right)$$
$$ \Rightarrow 0.3 = {1 \over t}\ln \left( {{{10}^3}} \right)$$
$$ \Rightarrow 0.3 = {1 \over t} \times 3$$
$$ \Rightarrow t = 10$$ min
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