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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Consider a certain reaction A -> Products with k = 2.0 × 10^−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.

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प्रश्न

Consider a certain reaction \[\ce{A -> Products}\] with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.

संख्यात्मक
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उत्तर १

Given: k = 2.0 × 10−2 s1

T1 = 0

[A]1 = 1.0 M

T2 = 100 s

[A]2 = ?

k = `1/(T_2 - T_1) ln  ([A]_1)/([A]_2)`

2.0 × 10−2 = `1/(100 - 0) ln  1.0/([A]_2)`

(2.0 × 10−2) × 100 = `ln  1.0/([A]_2)`

2 = `ln  1/([A]_2)`

7.389 = `1/([A]_2)`

∴ [A]2 = `1/7.389`

= 0.135 mol L1

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उत्तर २

The units of k (s1) indicate that the reaction is of first order. For a first-order reaction,

k = `2.303/t log_10  [A]_0/([A])`

∴ 2.0 × 10−2 = `2.303/100 log_10  1.0/([A])`

2.0 × 10−2 = `-2.303/100 log_10 [A]`

⇒ log10 [A] = `-(2.0 xx 10^-2 xx 100)/2.303`

= `-2/2.303`

⇒ log10 [A] = −0.8684

⇒ [A] = antilog10 (−0.8684)

= 0.1354 mol L1

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पाठ 3: Chemical Kinetics - Exercises [पृष्ठ ८७]

APPEARS IN

एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 3 Chemical Kinetics
Exercises | Q 3.24 | पृष्ठ ८७
नूतन Chemistry [English] Class 12 ISC
पाठ 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.24 | पृष्ठ २८१

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