QJEEM18Cha463: JEE Main 2018 Chemistry - Class 12 JEEM18Cha463 Solution: For a first order reaction, A → P, t 1/2 (half-life) is 10 days …

Question

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For a first order reaction, A

\to

P, t_1/2 (half-life) is 10 days The time required for

\frac{1}{4}

^th conversion of A (in days) is : (ln 2 = 0.693, ln 3 = 1.1)

Accepted Answer

For first order reaction,

The half life, t_{$\frac{1}{2}$} =

\frac{0.693}{k}

Here given t_{$\frac{1}{2}$} = 10 days

∴

k =

\frac{0.693}{10}

= 0.0693 days^{$-$ 1}

Now, the time required for

\frac{1}{4}

th conversion of A is ,

t =

\frac{2.303}{k}

log₁₀

(\frac{a}{a - x})

=

\frac{2.303}{0.0693}

log

(\frac{1}{1 - \frac{1}{4}})

=

\frac{2.303}{0.0693}

log

(\frac{4}{3})

= 4.1 days.

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