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प्रश्न
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
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उत्तर
Given: T1 = 298 K
T2 = 308 K
`K_2/k_1` = 2
R = 8.314 JK−1 mol−1
Ea = ?
According to the Arrhenius equation,
`log k_2/k_1 = (E_a)/(2.303 R) [1/T_1 - 1/T_2]`
∴ log 2 = `E_a/(2.303 xx 8.314) [1/298 - 1/308]`
0.3010 = `E_a/(2.303 xx 8.314)^-1 xx 10/(298 xx 308)`
∴ Ea = `(0.3010 xx 2.303 xx 8.314 xx 298 xx 308)/10`
= `528977.78/10`
= 52897.7 J mol−1
= 52.897 kJ mol−1
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