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MCQEasyJEE 2025Integrated Rate Laws

JEE Chemistry 2025 Question with Solution

Drug X becomes ineffective after 50%50\% decomposition. The original concentration of drug in a bottle was 16 mg/mL16 \, \text{mg/mL} which becomes 4 mg/mL4 \, \text{mg/mL} in 1212 months. The expiry time of the drug in months is _____ . Assume that the decomposition of the drug follows first order kinetics.

  • A

    1212

  • B

    22

  • C

    33

  • D

    66

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Initial concentration of drug is 16 mg/mL16 \, \text{mg/mL} and after 1212 months it becomes 4 mg/mL4 \, \text{mg/mL}. The decomposition follows first-order kinetics. The drug expires after 50%50\% decomposition.

Find: The expiry time in months.

For a first-order reaction,

ln⁡([A]0[A]t)=kt\ln \left( \frac{[A]_0}{[A]_t} \right) = kt

Using [A]0=16 mg/mL[A]_0 = 16 \, \text{mg/mL}, [A]t=4 mg/mL[A]_t = 4 \, \text{mg/mL}, and t=12t = 12 months,

ln⁡(164)=12k\ln \left( \frac{16}{4} \right) = 12k ln⁡(4)=12k\ln(4) = 12k 1.386=12k1.386 = 12k k=0.1155 month−1k = 0.1155 \, \text{month}^{-1}

Using half-life condition

At expiry, 50%50\% of the drug has decomposed, so the concentration becomes half of the initial value:

[A]t=[A]02=8 mg/mL[A]_t = \frac{[A]_0}{2} = 8 \, \text{mg/mL}

Now apply the first-order equation again:

ln⁡([A]0[A]t)=ktexpiry\ln \left( \frac{[A]_0}{[A]_t} \right) = kt_{\text{expiry}} ln⁡([A]0[A]0/2)=ktexpiry\ln \left( \frac{[A]_0}{[A]_0/2} \right) = k t_{\text{expiry}} ln⁡(2)=0.1155⋅texpiry\ln(2) = 0.1155 \cdot t_{\text{expiry}} 0.693=0.1155⋅texpiry0.693 = 0.1155 \cdot t_{\text{expiry}} texpiry=6 monthst_{\text{expiry}} = 6 \, \text{months}

Therefore, the expiry time of the drug is 6 months6 \, \text{months} and the correct option is D.

Common mistakes

  • Using the concentration 4 mg/mL4 \, \text{mg/mL} as the expiry concentration is incorrect because the drug becomes ineffective after 50%50\% decomposition, not after 75%75\% decomposition. At expiry, the concentration should be 8 mg/mL8 \, \text{mg/mL}.

  • Applying a zero-order or second-order formula is incorrect because the question explicitly states first-order kinetics. Use the first-order relation ln⁡([A]0[A]t)=kt\ln \left( \frac{[A]_0}{[A]_t} \right) = kt or its logarithmic equivalent.

  • Confusing the given 1212 months with the expiry time is wrong. The 1212 months correspond to the drop from 1616 to 4 mg/mL4 \, \text{mg/mL}, whereas expiry occurs earlier at 8 mg/mL8 \, \text{mg/mL}.

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