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TD cinétique chimique QCM 3


Go to solution Solved by Le_Polâuné,

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Posted

Salut 👋

Tu peux te servir de cette relation la pour répondre à cet item, [A]t =[A]0 e ^-a.k.t

Si 30% est décomposé au bout de de 9minutes, alors le rapport [A]t /[A]0 vaut 0,7 (70%), soit 0,7 = e ^-a.k.9

Donc ln(0,7)/9 = -a.k

Or a=1, donc k= 0,0396, soit à peu près 0,04 u.v.

 

J’espère que c’est plus claire maintenant 😊

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