The reaction 2 X arrow B is a zeroth order reaction. If the initial concentration of X is 0.2 M , the half-life is 6 ~h . When the initial concentration of X is 0.5 M , the time required to reach its final concentration

Correct answer: C. 18.0 ~h For the reaction 2 X arrow B , follow zeroth order Rate equation is Kt=[A]_ 0 -[A] For the half-life; t=t_ 1/2 and [A]=0.1 Kt_ 12 =0.2-0.1 (0.2-0.1)/(6)=(0.1)/(6) Mhr^ -1 Time required to reach from 0.5 M to 0.2 M Kt=[A]_ 0 -[A] (0.1)/(6) × t=(0.5-0.2) ; t=18 hour

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2019 · 11 January · Evening Shift · Chemistry · Chemical Kinetics

The reaction 2 X arrow B is a zeroth order reaction. If the initial...

The reaction

2 \mathrm{X} \rightarrow \mathrm{B}

is a zeroth order reaction. If the initial concentration of

\mathrm{X}

0.2 \mathrm{M}

, the half-life is

6 \mathrm{~h}

. When the initial concentration of

X

0.5 \mathrm{M}

, the time required to reach its final concentration of

0.2 \mathrm{M}

will be

Options

A $9.0 \mathrm{~h}$
B $12.0 \mathrm{~h}$
C $18.0 \mathrm{~h}$
D $7.2 \mathrm{~h}$

Correct answer

C. $18.0 \mathrm{~h}$

Step-by-step solution

For the reaction

2 \mathrm{X} \rightarrow \mathrm{B}

, follow zeroth order Rate equation is

\mathrm{Kt}=[\mathrm{A}]_{0}-[\mathrm{A}]

For the half-life;

t=t_{1/2}

and

[A]=0.1

\mathrm{Kt}_{12}=0.2-0.1

\frac{0.2-0.1}{6}=\frac{0.1}{6} \mathrm{Mhr}^{-1}

\therefore

Time required to reach from

0.5 \mathrm{M}

0.2 \mathrm{M}

\mathrm{Kt}=[\mathrm{A}]_{0}-[\mathrm{A}]

\frac{0.1}{6} \times \mathrm{t}=(0.5-0.2) ; \mathrm{t}=18

hour

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