An organic compound undergoes first order decomposition. The time taken for its decomposition to 1 8 and 1 10 of its initial concentration are t 1 8 and t 1 10 respectively. What is the value of t 1 8 t 1 10 × 10 ? (take lo g 10 2 = 0.3 )

As we know k t = ln ⁡ A 0 A t t 1 8 t 1 10 = ln ⁡ 8 ln ⁡ 1 0 = 0.9 t 1 8 t 1 10 × 10 = 9

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ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

An organic compound undergoes first order decomposition. The time taken for its decomposition to $\frac{1}{8}$ and $\frac{1}{10}$ of its initial concentration are $t_{\frac{1}{8}}$ and $t_{\frac{1}{10}}$ respectively. What is the value of $\frac{[t_{\frac{1}{8}}]}{t_{\frac{1}{10}}} \times 10$ ?

(take $lo g_{10} 2 = 0.3$ )

Answer:

9.00

Solution:

As we know

$k t = \ln \frac{A_{0}}{A_{t}}$

$\frac{t_{\frac{1}{8}}}{t_{\frac{1}{10}}} = \frac{\ln 8}{\ln 1 0} = 0.9$

$\frac{t_{\frac{1}{8}}}{t_{\frac{1}{10}}} \times 10 = 9$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical KineticsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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