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發問:
The activation energy of an uncatalyzed reaction is 92kJ/mol . The addition of a catalyst lowers the activation energy to 65kJ/mol . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 16answer: kc/k = 7.6×104Assuming that the... 顯示更多 The activation energy of an uncatalyzed reaction is 92kJ/mol . The addition of a catalyst lowers the activation energy to 65kJ/mol . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 16 answer: kc/k = 7.6×104 Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 132 answer: kc/k = 3000 請問怎樣得出這答案,,,求詳細過程,THANKS
最佳解答:
Rate constant k = Ae^(Ea/RT) where A is the collision factor, Ea the activation energy, R is gas constant (8.314) and T in kelvin. At T=(273+16)=289 (kc/k) = Ae^(-65000/RT) / Ae^(-92000/RT) ln(kc/k) = (-65000/RT) - (-92000/RT) = 27000/(8.314*289) = 11.2371428357 (kc/k) = 75897.79 => 7.6 x 10^4 At T=(273+132)=405 (kc/k) = Ae^(-65000/RT) / Ae^(-92000/RT) ln(kc/k) = (-65000/RT) - (-92000/RT) =27000/(8.314*405) = 8.0186031593 (kc/k) = 3036.93 => 3 x 10^3
其他解答:0D7DAC4E4B9703B0
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化學(reaction+catalyst)發問:
The activation energy of an uncatalyzed reaction is 92kJ/mol . The addition of a catalyst lowers the activation energy to 65kJ/mol . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 16answer: kc/k = 7.6×104Assuming that the... 顯示更多 The activation energy of an uncatalyzed reaction is 92kJ/mol . The addition of a catalyst lowers the activation energy to 65kJ/mol . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 16 answer: kc/k = 7.6×104 Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 132 answer: kc/k = 3000 請問怎樣得出這答案,,,求詳細過程,THANKS
最佳解答:
Rate constant k = Ae^(Ea/RT) where A is the collision factor, Ea the activation energy, R is gas constant (8.314) and T in kelvin. At T=(273+16)=289 (kc/k) = Ae^(-65000/RT) / Ae^(-92000/RT) ln(kc/k) = (-65000/RT) - (-92000/RT) = 27000/(8.314*289) = 11.2371428357 (kc/k) = 75897.79 => 7.6 x 10^4 At T=(273+132)=405 (kc/k) = Ae^(-65000/RT) / Ae^(-92000/RT) ln(kc/k) = (-65000/RT) - (-92000/RT) =27000/(8.314*405) = 8.0186031593 (kc/k) = 3036.93 => 3 x 10^3
其他解答:0D7DAC4E4B9703B0
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