JEE MAIN - Chemistry (2021 - 27th July Evening Shift - No. 21)
For the first order reaction A $$\to$$ 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is __________ min. (Round off to the nearest integer).
[Use : ln 2 = 0.69, ln 10 = 2.3]
Properties of logarithms : ln xy = y ln x;
$$\ln \left( {{x \over y}} \right) = \ln x - \ln y$$
(Round off to the nearest integer)
[Use : ln 2 = 0.69, ln 10 = 2.3]
Properties of logarithms : ln xy = y ln x;
$$\ln \left( {{x \over y}} \right) = \ln x - \ln y$$
(Round off to the nearest integer)
Answer
600to700
Explanation
_27th_July_Evening_Shift_en_21_1.png)
Now, $$t = {{{t_{1/2}}} \over {\ln 2}} \times {{[{A_0}]} \over {[{A_t}]}}$$
$$100 = {{{t_{1/2}}} \over {\ln 2}} \times \ln {1 \over {0.9}} \Rightarrow {t_{1/2}} = 690$$ min (taking ln 3 = 1.11)
Ans. 600 to 700
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