For the first order reaction A → 2B, 1 mole of reactant A gives 0 ....

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{x}^{y} = ylnx $\leftln \left(\dfrac{x}{y}\right) = \ln x - \ln y]$

(Round off to the nearest integer)

Correct answer

Step-by-step solution

$A ⟶ 2Bt = 01t = 100\text{min}1 - x2x$

$2x = 0 . 2 ⇒ x = 0 . 1 [ {A}_{t} ] = 1 - x = 0 . 9$

Now, $t = \dfrac{{t}_{1/2}}{\ln 2} \times \dfrac{\left⌊{A}_{0}\right]}{\left[{A}_{t}\right]}$

$100 = \dfrac{{t}_{1/2}}{\ln 2} \times ln\dfrac{1}{0 . 9} ⇒ {t}_{1/2} = 655min$

(taking $\ln 3 = 1 . 11 )$

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