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प्रश्न
The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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उत्तर
Given:
k1 = 4 × 10−2
k2 = 8 × 10−2
T1 = 300 K
T2 = 310 K
Solution:
`log(k_2/k_1)=E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`log((8xx10^(-2)|)/(4xx10^(-2)))=E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`0.301= E_a/(2.303xx 8.314JK^(-1)mol^(-1))[(310-300)/(310xx300)]`
`E_a=(0.301 × 2.303 × 8.314 × 93000)/10`
Ea = 53598.5 J
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