Question

The half-life of this reaction is independent of the concentration of reactant.
After 10 minutes the volume of ${N_2}$ gas is 10 litre and after complete reaction 100 litres. The rate constant of the reaction in ${\min ^{ - 1}}$ is
A. $ \dfrac{{2.303}}{{10}} $
B. $\dfrac{{2.303}}{{10}}\log \,5.0 $
C.$\dfrac{{2.303}}{{10}}\log 2.0 $
D.$\dfrac{{2.303}}{{10}}\log 4.0$

Answer 1

Hint:Rate constant is a coefficient of proportionality relating to a chemical reaction at a given temperature of the concentration of reactant (in a unimolecular reaction) or to the product of the concentrations of reactants.
In this problem formula of rate constant will be used.Formula of rate constant:
$K = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)$.
Here,
K is rate constant.
a is the initial amount of reactant.
(a-x) is the amount left after t time.
x is the amount of reactant reacted. Now proceed accordingly.

Complete step by step answer:
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 amount of reactant $\left( {{V_0}} \right) = 10\;litres$
Final amount of reactant (V) = 10 litres
So, putting the values in the formula
\[K = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)\]
\[K = \dfrac{{2.303}}{{10}}\log \left( {\dfrac{{100}}{{10}}} \right)\]
\[K = \dfrac{{2.303}}{{10}}\log 10\] [log 10 =1]
\[K = \dfrac{{2.303}}{{10}} \times 1\]
\[K = \dfrac{{2.303}}{{10}}\]

Therefore, option (A) is the correct answer.

Note:
The half life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration (that is the time taken for the reactant concentration to reach half of its initial value). It is denoted by the symbol ${{\rm{t}}_{1/2}}$ and is usually expressed in seconds. Half-life is constant over the lifetime of an exponentially decaying quantity, and it is a characteristic unit for the exponential decay equation.