For a first order reaction A→B the rate constant, k=5.5×10−14s−1. The time required for 67% completion of reaction is x×10−1times the half life of reaction. The value of x_________is (Nearest integer)(Given : log 3 = 0.4771)

t67%=1kln⁡11−0.67=t1/2ln⁡2×ln⁡11−23t67%=t1/2log⁡2×log⁡3=t1/2×0.47710.301⇒t67%=1.585×t1/2X×10−1=1.585⇒X=15.85 Ans.16

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For a first order reaction $A \to B$ the rate constant, $k = 5.5 \times 10^{- 14} s^{- 1}$ . The time required for 67% completion of reaction is $x \times 10^{- 1}$ times the half life of reaction. The value of x_________is (Nearest integer)
(Given : log 3 = 0.4771)

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answer is 16.

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Detailed Solution

$\begin{array}{l} t_{67 %} = \frac{1}{k} \ln (\frac{1}{1 - 0.67}) = \frac{t_{1 / 2}}{\ln 2} \times \ln (\frac{1}{1 - \frac{2}{3}}) \\ t_{67 %} = \frac{t_{1 / 2}}{\log 2} \times \log 3 = \frac{t_{1 / 2} \times 0.4771}{0.301} \\ \Rightarrow t_{67 %} = 1.585 \times t_{1 / 2} \\ X \times 10^{- 1} = 1.585 \\ \Rightarrow X = 15.85 \\ Ans.16 \end{array}$