Question

If the half-life period of a first-order reaction is x and 3/4th life period of the same reaction is y, how are x and y related to each other?

• A. x=y
• B. 2x=y
• C. x=2y
• D. None of these

Answer 1

Answer: (B)

2x=y

Half life of a reaction is defined at the time during which the concentration of the reactant becomes half of initial value. If $a_o$ is the concentration of the reactant initially then at half life i.e. $t_{1/2}$ the concentration will be $a_{o/2}$.

Now, $3/4^{th}$ life period of a reaction is the time during which the concentration of the reactant reacted becomes $3/4^{th}$of the initial concentration.

Now, according to question,

$t_{1/2}=x$ and $t_{3/4}=y$

For first order reaction:-

$K=\frac{1}{t}ln\left(\frac{a}{a-x}\right)$

where, $K$=rate constant

$t$= time

$a$= initial concentration of reactant

$x$= amount of reactant reacted at time $t$

Case I:- $t=t_{1/2}$

$K=\frac{1}{t_{1/2}}ln\left(\frac{a}{a/2}\right)$

$\Rightarrow$ $K=\frac{1}{t_{1/2}}ln$ $2$ $-\left(i\right)$

Case II: $t=t_{3/4}$

$K=\frac{1}{t_{3/4}}ln$ $\left(\frac{a}{\frac{1}{4}a}\right)$

$\Rightarrow$$K=\frac{1}{t_{3/4}}ln$ $\left(\frac{4}{1}\right)$ $-\left(ii\right)$

Equating $\left(i\right)$ and $\left(ii\right)$ and putting values of $t_{1/2}$ and $t_{3/4}$:-

$\frac{1}{x}ln$ $2$= $\frac{1}{y}ln$ $\left(\frac{4}{1}\right)$

$\Rightarrow$ $\frac{1}{x}ln$ $2$= $\frac{1}{y}\times 2ln$ $2$ $[\because ln(a^b)=b$ $ln$ $a]$

$\Rightarrow$ $y=2x$