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

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s−1. Calculate k at 318 K and Ea. - Chemistry

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प्रश्न

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s−1. Calculate k at 318 K and Ea.

संख्यात्मक
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उत्तर

t = `2.303/k_1 log  [R]_0/(90/100 [R]_0)`

t = `2.303/k_2 log  [R]_0/(75/100 [R]_0)`

t = `2.303/k_1 log  10/9`,

t = `2.303/k_2 log  4/3`

`2.303/k_1 log  10/9 = 2.303/k_2 log  4/3`

`=> k_2/k_1 = (log(4/3))/(log(10/9))`

= `log 1.333/log 1.111`

= `0.1249/0.0457`

= 2.733

`log  k_2/k_1 = E_a/(2.303 R)((T_2 - T_1)/(T_1 T_2))`

⇒ log 2.733 = `E_a/(2.303 xx 8.314) ((308 - 298)/(298 xx 308))`

E= `(2.303 xx 8.314 xx 308 xx 298)/10 xx 0.4367`

= `(19.147 xx 308 xx 298)/10 xx 0.4367`

= 76.75 kJ mol−1

ln k = ln A `- E_a/(RT)`

log k = `log A - E_a/(2.303 RT)`

= `log (4 xx 10^10) - (76.75 xx 1000)/(2.303 xx 8.314 xx 318)`

= `10.6021 - 76750/6088.791`

= 10.6021 − 12.6051

= −2.003

k = Antilog (−2.003) = 9.93 × 10−3

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Notes

The answer in the textbook is incorrect.

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पाठ 3: Chemical Kinetics - Exercises [पृष्ठ ८८]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 3 Chemical Kinetics
Exercises | Q 3.29 | पृष्ठ ८८
नूतन Chemistry Part 1 and 2 [English] Class 12 ISC
पाठ 4 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.29 | पृष्ठ २८३

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