光化学文献中的问题求助?如图,这是一个静态猝灭过程,当 六甲基苯 (HMB)过量的时候,基态复合物(Ex)与[CoIII(TPP)]+*的浓度比例为Ex / [[CoIII(TPP)]+*]=Kex/(1+Kex[HMB]),这一步是怎么得来的?HMB过量的话应该是一级反应吧,如果此式成立,那么Kex就应该是反应平衡常数,而不是速率或者形成常数才对。他之后又说Ex / [[CoIII(TPP)]+*]对应于(I0 − I)/(I0 −I∞),这一步又是怎么来的?I0是没有加荧光 猝灭剂 的荧光强度,I是加了荧光猝灭剂时的的荧光强度,那么I∞是什么意思?原文如下: Indeed, the fluorescence lifetime of [CoIII(TPP)]+ remains the same with increasing concentration of HMB up to 50 mM (Figure S9). In the presence of an excess amount of HMB, the ratio of concentration of the exciplex (formed via eq 7) to [CoIII(TPP)]+* is given by eq 8: Ex / [[CoIII(TPP)]+*]=Kex/(1+Kex[HMB]) Since [Ex]/[[CoIII(TPP)]+*] corresponds to (I0 − I)/(I0 −I∞), eq 8 can be rewritten as eq 9, (I0 − I)/(I0 −I∞)=Kex/(1+Kex[HMB]) eq 9 which predicts a linear correlation between (I0 − I)−1 and[HMB]−1. The formation constant of the exciplex of HMB (Kex) was determined to be 65 ± 5 M−1 from the slope and the intercept of a plot of (I0 − I)−1 versus [HMB]−1 原文图片电脑好像截不了图,所以只能上传文献了。具体文献我已经上传了。 图片.png查看更多1个回答 . 16人已关注