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简答题某药物在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-13.2851=1928,1g-13.6872=4886)
  • 药物在90℃,分解10%需要10天,有 t=0.1054/ K90=0.1054/1.0=0.01054 logK90/K25=E(T90-T25)/(2.303R×T90×T25) logK25=logK90E(T90-T25)/2.303×1.987×298×338 logK25=log0.01054-20×103(90-25)/2.303×8.319×298×363 logK25=-1.977-2.626 logK25=-4.603 t=0.1054/K t25=0.1054/K=4220天 t1/2=0.693/K=27720天

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