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प्रश्न
The rate constant of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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उत्तर
According to the Arrhenius equation,
k=Ae(−Ea/RT)
From this, we get
`"log"k_2/k_1=E_a/(2.303R)((T_2-T_1)/(T_1T_2))`
We are given that
Initial temperature, T1=300 K
Final temperature, T2=310 K
Rate constant at initial temperature, k1=2×10−2
Rate constant at final temperature, k2=4×10−2
Gas constant, R=8.314 J K−1 mol−1
Substituting the values, we get
`"log"((4xx10^(-2))/(2xx10^(-2)))=E_a/(2.303xx8.314)((310-300)/(300xx310))`
`therefore " activation energy of the reaction, "E_a=(log2xx2.303xx8.314xx300xx310)/10`
`=535985.94" J mol"^(-1)`
`=535.98" kJ mol"^(-1)`
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