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Chemical Kinetics and Nuclear Chemistry question

2023 · 30 Jan · Shift 2 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2023 · 30 Jan · Shift 2 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
An organic compound undergoes first-order decomposition. If the time taken for the 60%60 \%60% decomposition is 540 s540 \mathrm{~s}540 s, then the time required for 90%90 \%90% decomposition will be ‾\underline{\hspace{2cm}}​ s. (Nearest integer). Given: ln⁡10=2.3;log⁡2=0.3\ln 10=2.3 ; \log 2=0.3ln10=2.3;log2=0.3
Numerical answer
View written solutionFree

Correct answer: 1350

  1. Use the first-order rate law

For a first-order reaction,

kt=2.303log⁡(aa−x)kt = 2.303 \log\left(\frac{a}{a-x}\right)kt=2.303log(a−xa​)

where:

  • aaa = initial amount
  • xxx = amount decomposed
  • a−xa-xa−x = amount remaining
  1. For 60% decomposition

If 60%60\%60% is decomposed, then 40%40\%40% remains.

So,

aa−x=10040=52\frac{a}{a-x} = \frac{100}{40} = \frac{5}{2}a−xa​=40100​=25​

Given t=540 st=540\,\text{s}t=540s,

k(540)=2.303log⁡(52)k(540) = 2.303\log\left(\frac{5}{2}\right)k(540)=2.303log(25​)

Now,

log⁡(52)=log⁡5−log⁡2\log\left(\frac{5}{2}\right)=\log 5 - \log 2log(25​)=log5−log2

Using log⁡2=0.3\log 2 = 0.3log2=0.3,

log⁡5=log⁡(102)=1−0.3=0.7\log 5 = \log\left(\frac{10}{2}\right)=1-0.3=0.7log5=log(210​)=1−0.3=0.7

Hence,

log⁡(52)=0.7−0.3=0.4\log\left(\frac{5}{2}\right)=0.7-0.3=0.4log(25​)=0.7−0.3=0.4

So,

k(540)=2.303×0.4k(540)=2.303\times 0.4k(540)=2.303×0.4 k=2.303×0.4540k=\frac{2.303\times 0.4}{540}k=5402.303×0.4​
  1. For 90% decomposition

If 90%90\%90% is decomposed, then 10%10\%10% remains.

Thus,

aa−x=10010=10\frac{a}{a-x} = \frac{100}{10}=10a−xa​=10100​=10

For time t=t90t=t_{90}t=t90​,

kt90=2.303log⁡10kt_{90}=2.303\log 10kt90​=2.303log10

Given log⁡10=1\log 10 = 1log10=1 (or equivalently ln⁡10=2.3\ln 10 = 2.3ln10=2.3),

kt90=2.303kt_{90}=2.303kt90​=2.303

Substitute kkk:

t90=2.303kt_{90}=\frac{2.303}{k}t90​=k2.303​ t90=2.3032.303×0.4540t_{90}=\frac{2.303}{\frac{2.303\times 0.4}{540}}t90​=5402.303×0.4​2.303​ t90=5400.4=1350 st_{90}=\frac{540}{0.4}=1350\,\text{s}t90​=0.4540​=1350s
  1. Final Answer
1350 s\boxed{1350\ \text{s}}1350 s​

The derived answer matches the stored correct answer.

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