Question

Half-life is independent of the concentration of $A$. After $10$ minutes, the volume of $N_2$ gas is $10L$ and after completion of reaction, it is $50L$.

Hence, rate constant is:

• A. $\frac{2.303}{10}\log 5$ min${}^{-1}$
• B. $\frac{2.303}{10}\log 1.25$ min${}^{-1}$
• C. $\frac{2.303}{10}\log 2$ min${}^{-1}$
• D. $\frac{2.303}{10}\log 4$ min${}^{-1}$

Answer 1

The correct option is B $\frac{2.303}{10}\log 1.25$ min${}^{-1}$

$t_{50\%}$ is independent of the concentration of $A$. Hence first-order reaction.

$A$ $\to N_2\left(g\right)$

At $t=0$, $a=0$

At time $t,(a-x);x=$ $10L$
(after 10 minutes)

At complete reaction, $a$ $=$ $50L$

$\therefore (a-x)=40L$

$\therefore k=\frac{2.303}{10}log\frac{50}{40}$

$k=\frac{2.303}{10}log\left(1.25\right)mi{n}^{-1}$

Option B is correct.