Question

For a first order reaction, $A⟶P{t}_{1/2}$ ( half- life ) is $10$ days. The time required for $\frac{1}{4}th$ conversion of $A$ ( in days) is :

($ln2=0.693$, $ln3=1.1$)

• A. $5$
• B. $3.2$
• C. $4.1$
• D. $2.5$

Answer 1

The correct option is C $4.1$

Given:

Half-life of a $I^{st}$ order reaction, $t_{1/2}=10$ days

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10}=0.0693/day$

Time required for $\frac{1}{4}th$ conversion of $A$ is :

$t=\frac{2.303}{k}lo{g}_{10}\frac{InitialConcentration}{InitialConcentration-FinalConcentration}$

$\Rightarrow t=\frac{2.303}{0.0693}lo{g}_{10}\left(\frac{1}{1-\frac{1}{4}}\right)$

$\Rightarrow t=\frac{2.303}{0.0693}lo{g}_{10}\left(\frac{1}{\frac{3}{4}}\right)$

$\Rightarrow t=\frac{2.303}{0.0693}lo{g}_{10}\left(\frac{4}{3}\right)$

Now, $2.303logx=lnx$

$\Rightarrow t=\frac{1}{0.0693}\times ln\frac{4}{3}$

$\Rightarrow t=\frac{1}{0.0693}\times (ln4-ln3)$

$\Rightarrow t=\frac{1}{0.0693}\times (2ln2-ln3)$

$\Rightarrow t=\frac{1}{0.0693}\times (2\times 0.693-1.1)$

$\Rightarrow t=\frac{1}{0.0693}\times 0.286$

$\Rightarrow t=4.1$ days

Therefore, time needed for $\frac{1}{4}th$ conversion of $A$ is $4.1$ days.

Hence, option (C) is correct.