Question

The half-life of this reaction is independent of the concentration of reactant. After 10 minutes volume of $N_2$ gas is 10 liter and after complete reaction 100 liters. The rate constant of the reaction in ${min}^{-1}$ unit is:

• A. $\frac{2.303}{10}$
• B. $\frac{2.303}{10}\log 5.0$
• C. $\frac{2.303}{10}\log 2.0$
• D. $\frac{2.303}{10}\log 4.0$

Answer 1

The correct option is A $\frac{2.303}{10}$

Given, the half-life of the reaction is independent of the concentration of the reactant. So, it is a 1st order reaction.

Given,

Time (t) = 10 minutes.
Initial reactant (Vo) = 100 litre.
Final reactant (V) = 10 litre.

So, putting the values in the formula :

$k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$

$k=\frac{2.303}{10}\log \left(\frac{100}{10}\right)$

$k=\frac{2.303}{10}\times 1$