Question

In a first order reaction, $75\%$ of the reactants disappeared in $1.386hr$. The value of the rate constant is :

• A. $0.693h{r}^{-1}$
• B. $1h{r}^{-1}$
• C. $1.386h{r}^{-1}$
• D. $2.772h{r}^{-1}$

Answer 1

The correct option is B

$1h{r}^{-1}$

Time taken for completion of $75\%$ of the reaction is double than that of the time taken for half of the reaction to complete.
So, $t_{75\%}=2\times t_{\frac{1}{2}}$
But ,
$t_{75\%}=1.386hr$
$\therefore t_{\frac{1}{2}}=\frac{1.386}{2}=0.693hr$
Now,
$k=\frac{\ln 2}{t_{\frac{1}{2}}}=\frac{\ln 2}{0.693}\approx 1h{r}^{-1}$
Hence, (b) is the correct option.