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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?
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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
T2 = ?
From Arrhenius equation, we obtain,
`log k_2/k_1 = E_a/(2.303 R)((T_2 - T_1)/(T_1T_2))`
⇒ log `(1.5 xx 10^4)/(4.5 xx 10^3) = (60000)/(2.303 xx 8.314) ((T_2 - 283) /(283 xx T_2))`
⇒ log 3.333 = `3133.63 (T_2 - 283)/(283 xx T_2)`
⇒ 0.5228 = `3133.63 (T_2 - 283)/(283 xx T_2)`
⇒ `0.5228/3133.63 = (T_2 - 283)/(283 xx T_2)`
⇒ 1.6683 × 10−4 = `(T_2 - 283)/(283 xx T_2)`
⇒ 1.6683 × 10−4 × 283 × T2 = T2 − 283
⇒ 0.0472 T2 = T2 − 283
283 = T2 − 0.0472 T2
283 = (1 − 0.0472) T2
283 = 0.9528 T2
⇒ T2 = `283/0.9528`
= 297 K
= 297 − 273
= 24°C
Hence, k would be 1.5 × 104 s−1 at 24°C.
Notes
The answer in the textbook is incorrect.
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