Question

Half life period of a first order reaction is 10 min. starting with initial concentration 12 M, the rate after 20 min:
A. 0.0693 $M {\min ^{ - 1}}$
B. 0.693 $ \times $ 2 $M{\min ^{ - 1}}$
C. 0.693 $ \times $ 3 $M{\min ^{ - 1}}$
D. 0.693 $ \times $ 4 $M{\min ^{ - 1}}$

Answer 1

Hint: Rate law is depiction of molar concentration of reactants in the reaction raised to the exponents which may or may not be equal to its stoichiometric exponents hence rate law can not be determined theoretically.

Complete step by step solution:
Half life time of a reaction is defined as the time required for the concentration of reactant to be half decomposed and is denoted by ${t_{1/2}}$ .
Since we are given that the reaction is a first order reaction therefore half life period is given by:
${t_{1/2}} = \dfrac{{0.693}}{k}$
Where k is the rate constant and the constant of proportionality. Further for the first order equation we know that
$
\dfrac{{dx}}{{dt}} = {k_1}[A] = {k_1}(a - x) \\
\Rightarrow {k_1} = \dfrac{1}{t}\ln \left( {\dfrac{a}{{a - x}}} \right) \\
$
Therefore as we are given with the initial concentration we can calculate its final concentration after some given minutes by,
${A_t} = {A_ \circ }.{e^{ - {k_1}t}}$
Let us therefore first find the value of k for this reaction by the half life time period equation,
$\Rightarrow k = \dfrac{{0.693}}{{10}} = 0.0693\,{\min ^{ - 1}}$
Now final concentration can be found out and then we can find the rate using the rate law. Hence we get,
 $
\Rightarrow {A_t} = {A_ \circ }.{e^{ - {k_1}t}} = 12 \times {e^{ - \dfrac{{\ln 2}}{{10}} \times 20}} = 3\,M \\
\Rightarrow Rate = k[{A_t}] = 0.0693 \times 3\,M{\min ^{ - 1}}
$

Hence we can conclude that the correct option is C. 0.693$ \times $ 3M ${\min ^{ - 1}}$

Note:
Kinetic studies not only help us to determine the speed or rate of the reaction but also describes the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction.