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Question
The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?
Solution
Given: k1 = 4.5 × 103 s−1
T1 = 273 + 10 = 283 K
k2 = 1.5 × 104 s−1
Ea = 60 kJ mol−1 = 6.0 × 104 J mol−1
From Arrhenius equation, we obtain
log `"k"_2/"k"_1 = "E"_"a"/(2.303"R")(("T"_2-"T"_1)/("T"_1"T"_2))`
log `(1.5xx10^4)/(4.5xx10^3) = (60000)/(2.303xx8.314) (("T"_2-283) /(283"T"_2))`
or, log 3.333 = 3133.63`("T"_2 - 283)/(283"T"_2)`
or, `0.5228/3133.63 = ("T"_2 - 283)/(283 "T"_2)`
or, 0.0472T2 = T2 − 283
or, T2 = `283/0.9528`
= 297 K
= 297 − 273
= 24°C
Hence, k would be 1.5 × 104 s−1 at 24°C.
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