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NVAEasyJEE 2023Integrated Rate Laws

JEE Chemistry 2023 Question with Solution

The rate constant for a first order reaction is 20 min−120 \, \text{min}^{-1}. The time required for the initial concentration of the reactant to reduce to its 132\frac{1}{32} level is _____ ×10−2 min\times 10^{-2} \, \text{min}. (Nearest integer) (Given: ln⁡10=2.303\ln 10 = 2.303, log⁡2=0.3010\log 2 = 0.3010)

Answer

Correct answer:17

Step-by-step solution

Standard Method

Given: The reaction is first order with rate constant k=20 min−1k = 20 \, \text{min}^{-1}.

The concentration falls to 132\frac{1}{32} of its initial value, so

CC0=132\frac{C}{C_0} = \frac{1}{32}

Find: The value of time in the form _____ ×10−2 min\times 10^{-2} \, \text{min}.

For a first-order reaction, the integrated rate law is

ln⁡(C0C)=kt\ln \left( \frac{C_0}{C} \right) = kt

Substituting CC0=132\frac{C}{C_0} = \frac{1}{32} gives

ln⁡(C0C)=ln⁡32=5ln⁡2=5×0.693=3.465\ln \left( \frac{C_0}{C} \right) = \ln 32 = 5\ln 2 = 5 \times 0.693 = 3.465

Now,

t=3.46520=0.17325 mint = \frac{3.465}{20} = 0.17325 \, \text{min}

Therefore,

0.17325 min=17.325×10−2 min0.17325 \, \text{min} = 17.325 \times 10^{-2} \, \text{min}

So, the nearest integer is 17.

Using half-life idea

Given: A first-order reaction with k=20 min−1k = 20 \, \text{min}^{-1}.

Find: Time to reach 132\frac{1}{32} of the initial concentration.

Since

132=125\frac{1}{32} = \frac{1}{2^5}

the concentration becomes one-half of itself 5 times. Hence,

t=5t1/2t = 5t_{1/2}

For a first-order reaction,

t1/2=0.693k=0.69320t_{1/2} = \frac{0.693}{k} = \frac{0.693}{20}

Therefore,

t=5×0.69320=0.17325 min=17.325×10−2 mint = 5 \times \frac{0.693}{20} = 0.17325 \, \text{min} = 17.325 \times 10^{-2} \, \text{min}

Thus, the required nearest integer is 17.

Common mistakes

  • Using the first-order formula incorrectly as a linear decrease. The concentration does not drop uniformly with time; use the integrated law ln⁡(C0C)=kt\ln\left(\frac{C_0}{C}\right)=kt instead.

  • Writing ln⁡32\ln 32 incorrectly. Since 32=2532 = 2^5, we must use ln⁡32=5ln⁡2\ln 32 = 5\ln 2, not ln⁡2/5\ln 2 / 5 or 32ln⁡232\ln 2.

  • Stopping at 0.17325 min0.17325 \, \text{min} and entering 0.17325 as the answer. The question asks for the coefficient of ×10−2 min\times 10^{-2} \, \text{min}, so convert it to 17.325×10−2 min17.325 \times 10^{-2} \, \text{min} and then take the nearest integer.

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