CH353 - Physical Chemistry I
Spring 2015, Unique 51170

Lecture Summary, 23 April  2015

Solving rate laws:  Any chemical reaction can be broken down into a series of steps which proceed in order from reactant to product.  These steps are called elementary reactions in equilibrium, and the rate law from each step can be written down from the stoichiometrically balanced elementary reaction.  These elementary reactions demonstrate that during the reaction, short-lived, high energy products are formed and then consumed essentially immediately.  These are called intermediates, and are not shown in either the reactants or products.  By writing down a correct series of elementary reactions, and then figuring out what happens to the intermediates, we can solve a rate law analytically.  By comparing this rate law to the experimentally observed rate law, we can determine if our series of elementary steps and our assumption about the intermediates are correct.  If our solved rate law does not match the experimentally determined rate law, then we have either written down an incorrect series of elementary reactions, or made an incorrect assumption about the nature of the intermediates, or both.

Principle of Detailed Balance:  Since our elementary equations are in equilibrium and all species are represented by the stoichiometric amounts:

   k(1) / k(-1) = [P] / [R] = K(c)

or the ratio of the forward and backward rate constants is equal to the equilibrium constant of that elementary reaction.  This is called the principle of detailed balance. 

Steady-State Approximation:  The steady-state approximation says that the intermediate generated by the rate-determining step of the reaction will be consumed essentially as soon as it is formed:

  d[I] / dt = 0

This can happen in two ways.  Either the backwards reaction that created [I] in the first place is very fast or the forward reaction that consumes [I] and takes it further to products is very fast.  That in turn means that one rate constant in the reaction will be significantly greater than all the others, and it dominates any equation we set up for the rate of the reaction.