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Chemical Kinetics and Nuclear Chemistry question

2012 · Shift 1 · Q1
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Chemical Kinetics and Nuclear Chemistry question

2012 · Shift 1 · Q1

JEE AdvancedChemistryChemical Kinetics and Nuclear ChemistryNumerical+3 / −1
An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the value of [t1/8t1/10]×10\left[ {{{{t_{1/8}}} \over {{t_{1/10}}}}} \right] \times 10[t1/10​t1/8​​]×10? (log⁡102=0.3{\log _{10}}2 = 0.3log10​2=0.3)
Numerical answer
View written solutionFree

Correct answer: 9

  1. Use the first-order integrated rate law

For a first-order reaction,

kt=2.303log⁡([A]0[A]t)kt = 2.303\log\left(\frac{[A]_0}{[A]_t}\right)kt=2.303log([A]t​[A]0​​)

where [A]0[A]_0[A]0​ is initial concentration and [A]t[A]_t[A]t​ is concentration at time ttt.

  1. Time for concentration to become 1/81/81/8 of initial

Here,

[A]t=[A]08[A]_t = \frac{[A]_0}{8}[A]t​=8[A]0​​

So,

kt1/8=2.303log⁡8kt_{1/8} = 2.303\log 8kt1/8​=2.303log8

Since

8=238 = 2^38=23

and given log⁡2=0.3\log 2 = 0.3log2=0.3,

log⁡8=3log⁡2=3(0.3)=0.9\log 8 = 3\log 2 = 3(0.3)=0.9log8=3log2=3(0.3)=0.9

Thus,

kt1/8=2.303(0.9)kt_{1/8} = 2.303(0.9)kt1/8​=2.303(0.9)
  1. Time for concentration to become 1/101/101/10 of initial

Here,

[A]t=[A]010[A]_t = \frac{[A]_0}{10}[A]t​=10[A]0​​

So,

kt1/10=2.303log⁡10kt_{1/10} = 2.303\log 10kt1/10​=2.303log10

Since

log⁡10=1\log 10 = 1log10=1

we get,

kt1/10=2.303kt_{1/10} = 2.303kt1/10​=2.303
  1. Find the ratio
t1/8t1/10=2.303(0.9)/k2.303/k=0.9\frac{t_{1/8}}{t_{1/10}} = \frac{2.303(0.9)/k}{2.303/k} = 0.9t1/10​t1/8​​=2.303/k2.303(0.9)/k​=0.9
  1. Required value
(t1/8t1/10)×10=0.9×10=9\left(\frac{t_{1/8}}{t_{1/10}}\right)\times 10 = 0.9 \times 10 = 9(t1/10​t1/8​​)×10=0.9×10=9

Final Answer

9\boxed{9}9​

The derived answer matches the stored correct answer.

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