CHEM 104 1st Edition Lecture 17When a mechanism’s first step is the slow/rate-determining step, this is automatically the rate – when this is not the case, solving for the rate is more complicated (*we’ll only ever solve for rates when the mechanism involves a fast equilibrium step followed by the slow step) - Example of mechanism with fast equilibrium step followed by slow step: 2NO + Br2 2NOBr Experimental rate law: rate = k[NO]2[Br2]Proposed mechanism: Step 1) NO + Br2(k1)(k-1) NOBr2Step 2) NOBr2 + NO (k2) 2NOBrRate = k2[NOBr2][NO] Can’t use this rate because there’s an intermediate involved in it Rate of NOBr2 formation = k1[NO][Br2] Rate of NOBr2 consumption = k2[NOBr2][NO] + k-1[NOBr2] First portion of consumption rate is small compared to the second, so it can be ignored Rate of NOBr2 formation = rate of NOBr2 consumptionK1[NO][Br2] = k-1[NOBr2] [NOBr2] = k1/k-1[NO][Br2]Rate = k2[NOBr2][NO] = k2(k1/k-1[NO][Br2])[NO] = k[NO]2[Br2] Catalysis - Catalyst = participates in reaction and is regenerated at the end; changes mechanism; lowers activation energy (causing reaction to occur faster) - Example: rate of cis butane to trans butane without a catalyst = k[cis], rate with I2 catalyst = k’[cis][I2]1/2- Catalysis can be homogenous or heterogeneous (i.e. reaction between gases can occur on a solid) - Enzymes = highly efficient catalysts (speed up reactions 107-1014 times); very selective for one reaction or one type of reaction o Example: aldehyde dehydrogenase catalyzes oxidation of aldehydeso Substrate = reactant, molecule whose reaction is catalyzed by the enzymeo Active site = part of the enzyme that interacts with the
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