Hydronium and Hydroxide Ions Part 1

   When a chemical reaction reaches equilibrium it is not that the reactions terminate, it is that the rates at which the forward and reverse reactions continue are equal. This means that relative amounts of reactants and products stay constant at the equilibrium concentrations - both reactions continue but the concentrations are stable.
   If you add one or more of the reactants or products to a mixture in equilibrium, thus changing the relative concentrations, actions will occur until a new equilibrium is reestablished (i.e. the rates will gradually go back to equal after being changed to unequal by the addition.)
   The rate of a reaction is correlated directly with the number of individual reactions that take place. In order for an individual reaction to take place, the required molecules, in the right ratios, must be in close enough proximity and have high enough energy for the reaction (i.e. the redistribution of electrons, changing bonds holding electrons to their nuclei) to occur.
   The  temperature is a measure of the average kinetic energy (KE) of the molecules in our mixture. The actual kinetic energy of the individual molecules is probabilistically distributed about this average - some having greater (KE) and some less (KE) than the average. This distribution of KE for the molecules in our reactions is determined by the temperature of the experiment so the probability of our
molecules having high enough energy to react is a constant determined by the temperature.
   This means that the rate of a reaction, at a given temperature, is proportional to the probability of getting the right molecules in close enough proximity.
   Suppose our reactants are A and B and we need m-molecules of A and n-molecules of B, in close enough proximity and with high enough energy for a reaction to occur. Suppose we have a small volume we call dV that satisfies the requirement for "close enough proximity". The probability that a reaction will occur in dV is the probability of having m-molecules of A and n-molecules of B in dV times the probability that a reaction will occur given they are there (i.e. the probability they have high enough energy). The probability of having the required molecules in dV depends, by definition, on the concentration of A and B molecules in our experiment. Assume a uniform distribution of A (and B) in our mixture. The concentration of A, written [A], is given in moles per liter.  The number of A molecules, N(A), in dV is:
    N(A) = ([A] mol/Liter)  (6.022x10^23 molecules/mole) (k Liter/dV)
where k is a constant (1/k=how many dV's in a Liter). Suppose k was small enough to cancel Avogadro's number. Then we would have N(A) = [A] molecules/dV. That is, we would expect this concentration, a number between zero and one, to be the number of A molecules in dV - verses the required m such molecules. This means we would not have a concentration dense enough for the reaction to occur.
     If P(A) is the probability of getting one A-molecule in a given dV, then the probability of getting two would be P(A)*P(A) and the probability of getting m-molecules of A in dV would be P(A)^m. Similarly, the probability of getting n-molecules of B in dV is P(B)^n and the probability of getting them all in dV is the product.
    To Be Continued…..