Question

For a chemical reaction, half-life period ($t_{1/2}$) is 10 minutes. How much reactant will be left after 20 minutes if one starts with 100 moles of reactant and the order of the reaction be (i) zero, (ii) one and (iii) two?

• A. 0, 25, 33.33
• B. 25, 0, 33.33
• C. 33.33, 25, 0
• D. 25, 33.33, 0

Hint:

Half-life memory trick: \begin{itemize} \item First order: constant halving \item Zero order: linear decay \item Second order: concentration decreases slowly \end{itemize} More order $\Rightarrow$ slower depletion at long times.

Answer 1

The Correct Option is D

Concept: Half-life dependence on order: \begin{itemize} \item Zero order: $t_{1/2} = \frac{[A]_0}{2k}$ \item First order: constant half-life \item Second order: $t_{1/2} = \frac{1}{k[A]_0}$ \end{itemize} Given: \[ t_{1/2} = 10 \text{ min}, \quad t = 20 \text{ min}, \quad [A]_0 = 100 \] (i) Zero order reaction For zero order: \[ [A] = [A]_0 - kt \] From half-life: \[ 10 = \frac{100}{2k} \Rightarrow k = 5 \] After 20 min: \[ [A] = 100 - 5(20) = 0 \] (ii) First order reaction Two half-lives elapsed (20 min = 2 $\times$ 10 min). \[ [A] = 100 \times \left(\frac{1}{2}\right)^2 = 25 \] (iii) Second order reaction Integrated law: \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \] From half-life: \[ 10 = \frac{1}{k \cdot 100} \Rightarrow k = \frac{1}{1000} \] After 20 min: \[ \frac{1}{[A]} = \frac{1}{100} + \frac{20}{1000} \] \[ = 0.01 + 0.02 = 0.03 \] \[ [A] = 33.33 \] Final order: \[ 0, 25, 33.33 \] Thus correct sequence corresponds to option (D).