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प्रश्न
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
The rate of reaction becomes four times when the temperature changes from 293 K to 313 K.
Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature.
(R = 8.314 J K–1 mol–1)
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.
[Given: log 4 = 0.602, log 2 = 0.301, R = 8.314 J K–1mol]
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उत्तर १
Given: T1 = 293 K,
T2 = 313 K
`log k_2/k_1 = E_a/(2.303 R) [(T_2 - T_1)/(T_1 T_2)]`
Ea = `2.303 R (T_1 T_2)/(T_2 - T_1) log k_2/k_1`
`(T_1 T_2)/(T_2 - T_1) = (293 xx 313)/(313 - 293)`
`91709/20`
= 4585.45 K
`log k_2/k_1 = log 4/1` = log 4 = 0.6021
Ea = 2.303 × 8.314 JK−1 mol−1 × 4585.45 K × 0.6021
= 52863 J mol−1
= 52.8 kJ mol−1
उत्तर २
Given: T1 = 293 K,
T2 = 313 K,
k2 = 4 × k1,
R = 8.314 × 10−3 kJ K−1 mol−1
Formula: According to the Arrhenius equation,
`log_10 k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`
or, `log_10 (4 xx k_1)/k_1 = E_a/(2.303 xx 8.314 xx 10^-3) xx [1/293 - 1/313]`
or, log10 (4) = `E_a/0.0191 xx (313 - 293)/(293 xx 313)`
or, 0.6021 = `E_a/0.0191 xx 20/(293 xx 313)`
or, Ea = `(0.6021 xx 0.0191 xx 293 xx 313)/20`
= `1054.66/20`
= 52.73 kJ mol−1
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