For a certain first order reaction 32 % of the reactant is left after 570 s . The rate constant of this reaction is × 10 - 3 s - 1 . (Round off to the Nearest Integer). [Given: log 10 2 = 0 . 301 , ln 10 = 2 . 303 ]

For 1 st order reaction, K = 2 . 303 t · log A 0 A t = 2 . 303 570 sec · log 100 32 = 1 . 999 × 10 - 3 sec - 1 ≈ 2 × 10 - 3 sec - 1

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ChemistryChemical KineticsFirst Order ReactionsMedium2 minPYQ_2021

ChemistryMediumnumerical

For a certain first order reaction $32 %$ of the reactant is left after $570 s .$ The rate constant of this reaction is $\times 10^{- 3} s^{- 1} .$ (Round off to the Nearest Integer).

[Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303$ ]

Answer:

2.00

Solution:

For $1^{st}$ order reaction,

$K = \frac{2.303}{t} \cdot \log \frac{[A_{0}]}{[A_{t}]} = \frac{2.303}{570 \sec} \cdot \log (\frac{100}{32})$

$= 1.999 \times 10^{- 3} \sec^{- 1} \approx 2 \times 10^{- 3} \sec^{- 1}$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:First Order Reactions

⏱ 2mℹ️ Source: PYQ_2021

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