Q54 / 180 · COMEDK 2025 (25 May Shift) · Chemistry
The half- life of a first order reaction is ' $p$ ' and its $3 / 4^{\text {th }}$ life period is ' $q$ '. $p$ and $q$ are related to each other by: [Given: $\log 3=0.4771, \log 4=0.6040$ ]
- A$p=2 q$
- Correct answerB$2 p=q$
- C$q=\dfrac{p}{4}$
- D$\dfrac{p}{q}=3$
Solution
For a first order reaction, the rate constant $k$ is given by $k = \dfrac{\ln 2}{p}$, where $p$ is the half-life.
The $3/4^{\text{th}}$ life period $q$ is the time taken for the concentration to reduce to $1/4$ of its initial value. The integrated rate equation is $k = \dfrac{1}{q} \ln \left( \dfrac{[A]_0}{[A]_t} \right)$.
Substituting $[A]_t = \dfrac{1}{4}[A]_0$, we get $k = \dfrac{1}{q} \ln(4) = \dfrac{2 \ln 2}{q}$.
Equating the two expressions for $k$: $\dfrac{\ln 2}{p} = \dfrac{2 \ln 2}{q}$.
Simplifying this, we get $\dfrac{1}{p} = \dfrac{2}{q}$, which implies $q = 2p$ or $2p = q$.
Answer: $2 p=q$
The $3/4^{\text{th}}$ life period $q$ is the time taken for the concentration to reduce to $1/4$ of its initial value. The integrated rate equation is $k = \dfrac{1}{q} \ln \left( \dfrac{[A]_0}{[A]_t} \right)$.
Substituting $[A]_t = \dfrac{1}{4}[A]_0$, we get $k = \dfrac{1}{q} \ln(4) = \dfrac{2 \ln 2}{q}$.
Equating the two expressions for $k$: $\dfrac{\ln 2}{p} = \dfrac{2 \ln 2}{q}$.
Simplifying this, we get $\dfrac{1}{p} = \dfrac{2}{q}$, which implies $q = 2p$ or $2p = q$.
Answer: $2 p=q$
Hence, the correct answer is option: 'B'.