Find the two third life $(t_{2 / 3})$ of a first order reaction in which $K = 5.48 \times 10^{- 14} s e c^{- 1}$ .

t_{2 / 3} = 5 \times 10^{13}

sec

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t_{2 / 3} = 2 \times 10^{13}

sec

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t_{2 / 3} = 1.25 \times 10^{13}

sec

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none of these

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Solution

The correct option is B $t_{2 / 3} = 2 \times 10^{13}$ sec
1st Order Reaction,

$t = \frac{2.303}{k} l o g \frac{[100]}{[100 - x]}$

Substituting the values, we get
$t_{2 / 3} = \frac{2.303}{5.48 \times 10^{- 14}} \times l o g [3]$

$t_{2 / 3} = 2 \times 10^{13} s e c$

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