Shifting the position of an equilbrium

[Moogie]

Hi

At equilibrium there is a constant ratio of products to reactants and the rates of the forward and back reactions are the same.

When you distrub the equilibrium by say adding more reactants the equilibrium position shifts to the right so as to consume more reactants and produce more products to restore the equilibrium ratio.

Question 1
During this time of readjustment the rates of the forward and back reactions are the different, but once the new equilibrium position has been attained the rates of reaction of the forward and reverse reactions will once again be equal but will they be the same rates that they were before? In theory they could be slower or faster?

Question 2
Does the 'position of the equilibrium' refer to the specific ratio of products/reactants that equals the equilibrium constant? So if you are shifting the equilibrium to the right you are increasing the product concentration and decreasing the reactant concentration in this ration? If you are shifting the equilibrium to the left you are doing the reverse

thanks

 

[jbsteele]

Answer 1: Yes, the rates of the forward and reverse reactions will be equal once the equilibrium is re-established, but they may be different than the original rates. The rate of reaction is determined by the concentration of reactants and products, so if the concentrations are different, the rate of reaction will also be different. Answer 2: Yes, the position of the equilibrium refers to the ratio of products/reactants that equals the equilibrium constant. If you shift the equilibrium to the right, you are increasing the product concentration and decreasing the reactant concentration in this ratio. If you shift the equilibrium to the left, you are doing the reverse.