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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
t = `2.303/k_1 log [R]_0/(90/100 [R]_0)`
= `2.303/k_2 log [R]_0/(75/100 [R]_0)`
= `2.303/k_1 log 10/9`
= `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))`
Ea = `(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
Notes
The answer in the textbook is incorrect.
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