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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.
Solution
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))`
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 × 1010) `- (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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