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mojhsa [17]
3 years ago
7

For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the

rate constant at 666 K is 8.28×10-3 /s. The activation energy for the gas phase decomposition of 1-bromopropane is
Chemistry
1 answer:
Radda [10]3 years ago
6 0

Answer is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁ = 0,000643 1/s.<span>
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

<span> 1/T</span>₁ = 1/622 K = 0,0016 1/K.<span>
1/T</span>₂ = 1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

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ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

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