CSU CHEM 113 - More Practice with Reaction Mechanisms (3 pages)

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More Practice with Reaction Mechanisms



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More Practice with Reaction Mechanisms

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More Practice with Reaction Mechanisms


Lecture number:
8
Pages:
3
Type:
Lecture Note
School:
Colorado State University- Fort Collins
Course:
Chem 113 - General Chemistry II

Unformatted text preview:

CHEM 113 9 12 Lecture 8 Outline of Last Lecture I Using Example 1 to practice problems of reaction mechanisms Outline of Current Lecture II More practice of reaction mechanisms Current Lecture In class more practice with reaction mechanisms were discussed All practice problems and solutions can be found on the course website Here is an outline of the types of problems we have been doing Here are the elementary steps of a reaction 1 A B X fast 2 X C Y slow 3 Y D fast a What is the overall balanced equation By canceling out the intermediates and adding up the reactants and products the following equation was determined A B C D b Identify the intermediates These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute X Y are variables that act as both reactants and products and are therefore the intermediates Some problems may ask about transition states which are determined by n 1 where n is the number of intermediates present in the reaction c What is the molecularity of each step The molecularity is how many molecules are interacting at each step of the reaction Step 1 bimolecular because there are two variables A B Step 2 also bimolecular Step 3 Unimolecular d Is this consistent with the overall observed rate law Here we are comparing the observed rate law and the rate law of the rate determining step to make sure they are consistent So our observed rate law is Rate k A B C This can be deduced using the overall balanced equation Now we take out rate law for step 2 which is the rate determining step because it is the slowest Rate k2 X C However we need to rearrange this because there are intermediates included To do this we use the rate law for step 1 Since the first step is described with we can set the two sides equal to each other k1 A B k 1 X Then we manipulate the equation to equal X or the intermediate concentration X k1 A B k 1 We then substitute this into the rate law



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