Question

Half-life is independent of the concentration of $A$. After $10$ min volume of $N_2$ gas is $10$ L and after the complete reaction is $50$ L.

Hence, the rate constant is :

• A. $\frac{2.303}{10}log5mi{n}^{-1}$
• B. $\frac{2.303}{10}log1.25mi{n}^{-1}$
• C. $\frac{2.303}{10}log2mi{n}^{-1}$
• D. $\frac{2.303}{10}log4mi{n}^{-1}$

Answer 1

The correct option is B $\frac{2.303}{10}log1.25mi{n}^{-1}$

It follows first-order kinetics since $T_{50}$ is independent of concentration.

$\underset{Att=0,aAtt=10min.(a-x)Atcompletion0}{A}\to \underset{0xa}{PhCl}+\underset{0xa}{N_2}$

Hence, x = 10 L, a = 50 L
$\therefore k=\frac{2.303}{t}log\frac{\left[A_o\right]}{\left[A\right]}$

$\therefore k=\frac{2.303}{10}log\frac{50}{40}=\frac{2.303}{10}log1.25mi{n}^{-1}$