Q6 / 155 · WBJEE 2025 · Chemistry

For a chemical reaction, half-life period $\left(\mathrm{t}_{\frac{1}{2}}\right)$ 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?

  • Correct answer
    A
    $0,25,33.33$
  • B
    $25,0,33.33$
  • C
    $33.33,25,0$
  • D
    $25,33.33,0$

Solution

For zero order, $k=\frac{x}{t}$ $\begin{aligned} & \mathrm{t}_{\frac{1}{2}}=\frac{\mathrm{a}}{2 \mathrm{~K}} \\ & 10=\frac{100}{2 \times \mathrm{K}} \Rightarrow \mathrm{~K}=5 \\ & 5=\frac{\mathrm{x}}{20} \Rightarrow \mathrm{x}=100 \therefore(\mathrm{a}-\mathrm{x})=0 \end{aligned}$ For 1st order, No. of half lives $=\frac{20}{10}=2$ $\therefore(a-x)=\frac{100}{2^2}=25$ For 2nd order, $\mathrm{k}=\frac{\mathrm{x}}{\mathrm{at}(\mathrm{a}-\mathrm{x})}$ $\begin{aligned} & \therefore \mathrm{t}_{\frac{1}{2}}=\frac{1}{\mathrm{ak}} \\ & 10=\frac{1}{100 \times \mathrm{k}} \Rightarrow \mathrm{k}=10^{-3} \\ & \therefore 10^{-3}=\frac{\mathrm{x}}{100 \times 20 \times(100-\mathrm{x})} \\ & \Rightarrow \mathrm{x}=66.66 \\ & \therefore(\mathrm{a}-\mathrm{x})=33.33 \end{aligned}$

Hence, the correct answer is option: 'A'.

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