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Chemical Kinetics and Nuclear Chemistry question

2005 · Shift 0 · Q40
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Chemical Kinetics and Nuclear Chemistry question

2005 · Shift 0 · Q40

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
t1/4 can be taken as the time taken for the concentration of a reactant to drop to 343 \over 443​ of its initial value. If the rate constant for a first order reaction is K, the t1/4 can be written as
  1. A
    0.10 / K
  2. B
    0.29 / K
  3. C
    0.69 / K
  4. D
    0.75 / K
View written solutionFree

Correct answer: B

  1. Use the integrated rate law for a first-order reaction

For a first-order reaction,

[A]t=[A]0e−Kt[A]_t = [A]_0 e^{-Kt}[A]t​=[A]0​e−Kt

or equivalently,

Kt=ln⁡([A]0[A]t)Kt = \ln\left(\frac{[A]_0}{[A]_t}\right)Kt=ln([A]t​[A]0​​)
  1. Apply the condition for t1/4t_{1/4}t1/4​

Here, t1/4t_{1/4}t1/4​ is defined as the time when the concentration becomes 34\tfrac{3}{4}43​ of the initial concentration:

[A]t=34[A]0[A]_t = \frac{3}{4}[A]_0[A]t​=43​[A]0​

Substitute into the first-order equation:

Kt1/4=ln⁡([A]0(3/4)[A]0)Kt_{1/4} = \ln\left(\frac{[A]_0}{(3/4)[A]_0}\right)Kt1/4​=ln((3/4)[A]0​[A]0​​) Kt1/4=ln⁡(43)Kt_{1/4} = \ln\left(\frac{4}{3}\right)Kt1/4​=ln(34​)

So,

t1/4=1Kln⁡(43)t_{1/4} = \frac{1}{K}\ln\left(\frac{4}{3}\right)t1/4​=K1​ln(34​)
  1. Evaluate the logarithm
ln⁡(43)≈0.2877≈0.29\ln\left(\frac{4}{3}\right) \approx 0.2877 \approx 0.29ln(34​)≈0.2877≈0.29

Hence,

t1/4≈0.29Kt_{1/4} \approx \frac{0.29}{K}t1/4​≈K0.29​
  1. Match with the options

Thus the correct option is:

B: 0.29K\boxed{\text{B: } \frac{0.29}{K}}B: K0.29​​
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