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Questions
A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [R = 8.314 J K−1 mol−1]
[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
A first order reaction is 50% completed in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate the activation energy of the reaction. (R = 8.314 J K−1 mol−1)
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Solution
Given: T1/2 = 30 min, T1 = 300 K
T1/2 = 10 min, T2 = 320 K
Formula: `log k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`
k = `0.693/(t_(1//2))`
k1 at 300 K = `0.693/30` min−1
k2 at 320 K = `0.693/10` min−1
`log (k_2/k_1) = E_a/(2.303 R) [1/T_1 - 1/T_2]`
`log ((0.693/10)/(0.693/30)) = E_a/(2.303 R) [1/300 - 1/320]`
log (3) = `E_a/(2.303 R) [20/(300 xx 320)]`
0.4771 = `E_a/(2.303 xx 8.314) [20/(96000)]`
0.4771 = `E_a/(2.303 xx 8.314) [20/(96000)]`
Ea = `(0.4771 xx 2.303 xx 8.314 xx 96000)/20`
= `876969.73/20`
= 43848.48 J/mol
Ea = 43.8 kJ mol−1
Hence, the activation energy of the reaction is 43.8 kJ−1.
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