Question

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

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

Answer 1

The correct option is C $4.1$

The half life $t_{1/2}=10$ days

The decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10days}=0.0693da{y}^{-1}$

The time required for one fourth conversion

$t=\frac{2.303}{k}lo{g}_{10}\left(\frac{a}{a-x}\right)$

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

$t=4.11$ days