達人に解決をお願いします

2015-06-16View Original

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1. ある化学反応の開始温度は25℃で、温度が12℃上昇すると反応速度は2倍になる。この反応の活性化エネルギー（J/mol）は、次のうちどれか？(A) 44364 (B) 49986 (C) 52928 (D) 63769

Reply #22015-06-16

アレニウスの式 lnk=lnA-Ea/RTによって計算する

Reply #32015-06-16

Aでしょう。lnk1=lnA-Ea/RT1……（1）；lnk2=lnA-Ea/RT2……（2）；（2）-（1）でEaを求める

Reply #42015-07-28

ln(k2/K1)=-Ea(1/T2-1/T1)/R、すなわち：ln2=-Ea(1/310-1/298)/8.314、Ea=44364

Reply #52015-07-28

上記の計算は正しいでしょうか、専門家の方々にご教示をお願いします

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達人に解決をお願いします

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