JEE Main 2022 · 28 Jun · Shift 1 · Chemistry Q88

← JEE Main 2022 · 28 Jun · Shift 1

ChemistryInteger+4 / -1Question 88 of 90

For a first order reaction A $\rightarrow$ B, the rate constant, k = $5.5\times10^{-14}$ s $^{-1}$ . The time required for 67% completion of reaction is $x\times10^{-1}$ times the half life of reaction. The value of $x$ is ______ (Nearest integer)

(Given : log 3 = 0.4771)

Your answer

Answer and worked solution

Official answer: 16

First order. $t_{67\%}=\dfrac{2.303}k\log\dfrac{100}{33}\approx\dfrac{2.303\times0.4771}k$ and $t_{1/2}=\dfrac{0.693}k$ .

Ratio. $\dfrac{1.0988}{0.693}\approx1.59=15.9\times10^{-1}$ , so $x\approx16$ .

Answer: 16

Topics:Integrated rate equations Half-life

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JEE Main 2022 · 28 Jun · Shift 1, Chemistry question 88