Question

Half life of a first order reaction
40s calculate time,
required for 60% of reaction to complete​

Answer 1

Answer:

Notice the the half-life is independent of initial concentration. This is not the case with other reaction orders. The half-life of a first-order reaction was found to be 10 min at a certain temperature.The absorption half life can be calculated from KA using the natural log of 2 (which is approximately 0.7) i.e. absorption half-life=0.7/KA. The equation above predicts the time course of drug concentration in the blood from a first-order input process.

Answer 2

Given :

Half life of a first order reaction = 40s

To Find :

Time required for 60% of reaction to complete.

Solution :

❖ First of all we have to find decay constant of the chemical reaction.

Relation between half life of reaction and decay constant is given by

$\dag\:\underline{\boxed{\bf{\orange{t_{\frac{1}{2}}=\dfrac{\ln 2}{\lambda}=\dfrac{0.693}{\lambda}}}}}$

Here λ denotes decay constant

By substituting the given values;

➙ 40 = 0.693 / λ

➙ λ = 0.693 / 40

➙ λ = 0.017 $\sf{s^{-1}}$

♦ Rate constant of 1st order reaction :

$\bf:\implies\:\lambda=\dfrac{2.303}{t}\:\log\:\dfrac{a_o}{a_t}$

• a｡denotes initial concentration of reactant
• $\sf{a_t}$ denotes final concentration of reactant

Final concentration will be 0.4 a｡

$\sf:\implies\:0.017=\dfrac{2.303}{t}\:\log\:\dfrac{a_o}{0.4a_o}$

$\sf:\implies\:t=\dfrac{2.303}{0.017}\:\left[\log(10)-\log(4)\right]$

$\sf:\implies\:t=135.47\times\left[1-0.602\right]$

$\sf:\implies\:t=135.47\times0.398$

$:\implies\:\underline{\boxed{\bf{\gray{t=53.9\:s}}}}$