CHMA11H3 Lecture 5: CHMA11H3 Lecture 5 Chapter 13
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CHMA11H3 Full Course Notes
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The equation above can relate the two rate constants at two different temperatures. Solution: assign the given values: (cid:1837)(cid:2869)=(cid:884). (cid:887)(cid:889) (cid:1838)(cid:1865)(cid:1867)(cid:1864) (cid:1871);(cid:2869)=(cid:889)(cid:882)(cid:883) (cid:1837);(cid:1837)(cid:2870)=(cid:887)(cid:888)(cid:889) (cid:1838)(cid:1865)(cid:1867)(cid:1864) (cid:1871);(cid:2870)=(cid:890)(cid:891)(cid:887) (cid:1837) Then, substitute the values to the equation given above. ln((cid:1837)(cid:2870)(cid:1837)(cid:2869))=(cid:1831)(cid:3028)((cid:883)(cid:2869) (cid:883)(cid:2870)) ln(cid:4684)(cid:887)(cid:888)(cid:889) (cid:1838)(cid:1865)(cid:1867)(cid:1864) (cid:1871) (cid:884). (cid:887)(cid:889) (cid:1838)(cid:1865)(cid:1867)(cid:1864) (cid:1871)(cid:4685)=(cid:1831)(cid:3028)( (cid:883)(cid:889)(cid:882)(cid:883) (cid:1837) (cid:883)(cid:890)(cid:891)(cid:887) (cid:1837)) Increasing the temperature raises the kinetic energy of the reactant molecules: there is a minimum amount of kinetic energy needed for the collision to be converted into enough potential energy to form the activated complex. Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy: bond needs energy to break or to form bonds. Maximum rate: the rate of collisions with enough energy. Activation energy: the amount of energy needed to convert reactants into the activated complex. It is a chemical species with partially broken and partially formed bonds. This is always high in energy because of its partial bonds.