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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

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 × 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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अध्याय 4: Chemical Kinetics - Exercises [पृष्ठ १२०]

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एनसीईआरटी Chemistry [English] Class 12
अध्याय 4 Chemical Kinetics
Exercises | Q 29 | पृष्ठ १२०

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