# What is the relation between t_(1//2) and t_(3//4) in first order reaction?

Jul 14, 2018

${t}_{\textcolor{n a v y}{\frac{1}{2}}} = \frac{\ln \left(2\right)}{2 \ln \left(2\right) - \ln \left(3\right)} \cdot {t}_{\textcolor{n a v y}{\frac{3}{4}}} \approx 2.41 \cdot {t}_{\textcolor{n a v y}{\frac{3}{4}}}$

#### Explanation:

The expression

${\text{N" (t) = "N}}_{0} \cdot {e}^{- \frac{t}{\lambda}}$

gives the quantity of the reactant remaining at time $t$ into a first-order reaction. (The activity regarding the number of reactant particles converted per unit time is directly related to its concentration in the mixture, or quantity if the volume stays constant, for a particular reaction under a specific temperature. This relationship accounts for the exponential relationship between the number of reactant particles remaining and time since the first derivative of an exponential function is mostly identical to the original function.)

Rearranging and taking the natural logarithm of both sides relates the time after the reaction has begun to the ratio between the number quantity of the reactant remaining and that at the beginning of the process.

$\left({\text{N"(t)) / ("N}}_{0}\right) = {e}^{- \frac{t}{\lambda}}$

$\ln \left(\left({\text{N"(t)) / ("N}}_{0}\right)\right) = \frac{t}{- \lambda}$

$t = - \lambda \cdot \ln \left(\left({\text{N"(t))/ ("N}}_{0}\right)\right)$

In this expression ${\text{N}}_{0}$ resembles the number of reactant particles remaining at the beginning of the process (i.e. $t = 0$)

This number should have halved one half-life ${t}_{\frac{1}{2}}$ into the reaction. That is:

$\left({\text{N"(t_(1/2))) / ("N}}_{0}\right) = \frac{1}{2}$

${t}_{\frac{1}{2}} = - \lambda \cdot \ln \left(\left({\text{N"(t_(1/2)))/ ("N}}_{0}\right)\right) = - \lambda \cdot \ln \left(\frac{1}{2}\right) = \lambda \cdot \ln \left(2\right)$

Similarly, ("N"(t_(3/4))) = 3/4 * "N"_0 and

${t}_{\frac{3}{4}} = - \lambda \cdot \ln \left(\left({\text{N"(t_(3/4)))/ ("N}}_{0}\right)\right)$
$= - \lambda \cdot \ln \left(\frac{3}{4}\right) = \lambda \cdot \left(2 \ln \left(2\right) - \ln \left(3\right)\right)$

$\lambda$ the decay constant is independent of the progress of the reaction process and shall be uniform throughout the reaction as long as the temperature of the reaction system stays unchanged. It thus cancels out in ratio calculations.

(t_(1/2)) / (t_(3/4)) = (color(red)(cancel(color(black)(lambda))) * ln(2)) / (color(red)(cancel(color(black)(lambda))) * (2ln(2) - ln(3))) = (ln(2)) / (2 ln(2) - ln(3)) ~~ 2.409

Hence the conclusion:

${t}_{\textcolor{n a v y}{\frac{1}{2}}} \approx 2.41 \cdot {t}_{\textcolor{n a v y}{\frac{3}{4}}}$