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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.
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उत्तर
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
Notes
The answer in the textbook is incorrect.
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