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Karnataka Board PUCPUC Science 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 × 10^10 s−1. Calculate k at 318 K and Ea.

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Question

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.

Numerical
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Solution

Let [R]0 = 100

t = `2.303/k_1 log  100/90 = 2.303/k_2 log  100/75`

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

⇒ `1/k_1 log 1.111 = 1/k_2 log 1.333`

⇒ `k_2/k_1 = (log 1.333)/(log 1.111)`

⇒ `k_2/k_1 = 0.1249/0.0457`

⇒ `k_2/k_1` = 2.733

⇒ `log  k_2/k_1` = log 2.733 = 0.4367

`log  k_2/k_1 = E_a/(2.303 R) ((T_2 - T_1)/(T_1T_2))`

⇒ 0.4367 = `E_a/(2.303 xx 8.314) ((308 - 298)/(298 xx 308))`

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

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

= 76743.13 J mol−1

⇒ Ea = `76743.13/1000`

= 76.74 kJ mol−1

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

⇒ log k = `log (4 xx 10^10) - (76.74 xx 1000)/(2.303 xx 8.314 xx 318)`

⇒ log k = `10.6021 - 76740/6088.79`

⇒ log k = 10.6021 − 12.6034

⇒ log k = −2.0013

⇒ k = Antilog(−2.0013)

= Antilog(3.9987)

⇒ k = 9.97 × 10−3 s−1

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Notes

The answer in the textbook is incorrect.

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Chapter 3: Chemical Kinetics - 'NCERT TEXT-BOOK' Exercises [Page 283]

APPEARS IN

Nootan Chemistry [English] Class 12 ISC
Chapter 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.29 | Page 283
NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 3 Chemical Kinetics
Exercises | Q 3.29 | Page 88

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