For a first order reaction, A arrow P, t_ 1/2 (half-life) is 10 days. The time required for (1)/(4) ^ th conversion of A (in days) is: ( 2=0.693, 3=1.1) .

Correct answer: C. 4.1 The half life t_ 1/2 =10 days The decay constant k= 0.693 t_ 1/2 = 0.693 10 days = 0.0693 days ^ -1 The time required for one fourth conversion aligned &t= 2.303 k _ 10 a a-x &= 2.303 0.0693 day ^ -1 _ 10 (1)/(1-(1 / 4))=4.1 days aligned

JEE Main 201815 AprilEvening ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Evening Shift
Mode: Online
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2018 · 15 April · Evening Shift · Online · Chemistry · Chemical Kinetics

For a first order reaction, A arrow P, t_ 1/2 (half-life) is 10...

For a first order reaction,

\mathrm{A} \rightarrow \mathrm{P},

\mathrm{t}_{1/2}

(half-life) is 10 days. The time required for

\frac{1}{4}

\text{th }

}

conversion of \mathrm{A} (in days) is:

(\ln 2=0.693, \ln 3=1.1)

.

Options

A $3.2$
B $2.5$
C $4.1$
D5

Correct answer

C. $4.1$

Step-by-step solution

The half life

\mathrm{t}_{1/2}=10

days The decay constant

\mathrm{k}=\frac{0.693}{\mathrm{t}_{1/2}}=\frac{0.693}{10 \text{ days }}=

0.0693

\text{ days }^{-1}

The time required for one fourth conversion

\begin{aligned}&\mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\\ &=\frac{2.303}{0.0693 \text{ day }^{-1}} \log _{10} \frac{1}{1-(1 / 4)}=4.1 \text{ days } \end{aligned}

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