某药物在90℃，分解10%需要10天，设其活化能为20kcal/mol，求25℃时的储存期与半衰期。(lg
3.2851=1928，1g3.6872=4886)
正确答案:药物在90℃，分解10%需要10天，有
t=0.1054/k
k
=0.1054/1.0=0.01054
logk/k
=e(t
-t
)/(2.303r×t×t)
logk=logk-e(t-t)/2.303×1.987×298×338
logk=log0.01054-20×10
(90-25)/2.303×8.319×298×363
logk=-1.977-2.626
logk=-4.603
t=0.1054/k
t=0.1054/k=4220天
t
=0.693/k=27720天
=0.1054/1.0=0.01054
logk/k=e(t-t)/(2.303r×t×t)
logk=logk-e(t-t)/2.303×1.987×298×338
logk=log0.01054-20×10(90-25)/2.303×8.319×298×363
logk=-1.977-2.626
logk=-4.603
t=0.1054/k
t=0.1054/k=4220天
t=0.693/k=27720天答案解析:有
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