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प्रश्न
The decomposition of a hydrocarbon has value of rate constant as 2.5×104s-1 At 27° what temperature would rate constant be 7.5×104 × 3 s-1if energy of activation is 19.147 × 103 J mol-1 ?
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उत्तर
log `K_2/K_1= Ea /(2.303 R) [(T_2 - T_1)/(T_1T_2)]`
`T_1 -> 300k-> 2.5 xx 10^4 s^-2 → K_2`
`T_1 -> 300k-> 2.5 xx 10^4 s^-2 → K_2`
log `(7.5xx10^4(T-300))/(19.15 (300T))`
⇒ `0.477/1000 xx 300T = (T-300)1000`
⇒ 143T = 1000T - 300000
⇒ 300000 = 1000T - 143T
= 857 T
`T = 300000/857 = 350 K`
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