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Chemical Kinetics and Nuclear Chemistry question

2023 · 31 Jan · Shift 2 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2023 · 31 Jan · Shift 2 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The rate constant for a first order reaction is 20 min−120 \mathrm{~min}^{-1}20 min−1. The time required for the initial concentration of the reactant to reduce to its 132\frac{1}{32}321​ level is ‾\underline{\hspace{2cm}}​×10−2 min\times 10^{-2} \mathrm{~min}×10−2 min. (Nearest integer) (Given : ln⁡10=2.303\ln 10=2.303ln10=2.303 and log⁡2=0.3010 )\log 2=0.3010 \text { )}log2=0.3010 )
Numerical answer
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Correct answer: 17

  1. For a first-order reaction, [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}[A]t​=[A]0​e−kt so [A]t[A]0=e−kt\frac{[A]_t}{[A]_0} = e^{-kt}[A]0​[A]t​​=e−kt

  2. It is given that the concentration becomes its 132\frac{1}{32}321​ of the initial value: [A]t[A]0=132\frac{[A]_t}{[A]_0} = \frac{1}{32}[A]0​[A]t​​=321​ Hence, e−kt=132e^{-kt} = \frac{1}{32}e−kt=321​

  3. Taking natural logarithm: kt=ln⁡32kt = \ln 32kt=ln32

  4. Since 32=2532 = 2^532=25 therefore, ln⁡32=5ln⁡2\ln 32 = 5\ln 2ln32=5ln2

  5. Using the given values: log⁡2=0.3010\log 2 = 0.3010log2=0.3010 and ln⁡10=2.303\ln 10 = 2.303ln10=2.303

    Convert log⁡2\log 2log2 to ln⁡2\ln 2ln2: ln⁡2=(log⁡2)(ln⁡10)=0.3010×2.303\ln 2 = (\log 2)(\ln 10) = 0.3010 \times 2.303ln2=(log2)(ln10)=0.3010×2.303 ln⁡2≈0.6932\ln 2 \approx 0.6932ln2≈0.6932

    So, ln⁡32=5×0.6932=3.466\ln 32 = 5 \times 0.6932 = 3.466ln32=5×0.6932=3.466

  6. Given rate constant, k=20 min−1k = 20\ \text{min}^{-1}k=20 min−1

    Therefore, t=ln⁡32k=3.46620=0.1733 mint = \frac{\ln 32}{k} = \frac{3.466}{20} = 0.1733\ \text{min}t=kln32​=203.466​=0.1733 min

  7. The question asks in the form: t=‾×10−2 mint = \underline{\hspace{1cm}} \times 10^{-2}\ \text{min}t=​×10−2 min

    Now, 0.1733 min=17.33×10−2 min0.1733\ \text{min} = 17.33 \times 10^{-2}\ \text{min}0.1733 min=17.33×10−2 min

  8. Nearest integer = 171717

Therefore, the required answer is 171717.

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