Chemical Equilibrium and Chemical KineticsLearning Goal:To understand the relationship between the equilibrium constant and rate constants.For a general chemical equationA+B?C+D the equilibrium constant can be expressed as a ratio of the concentrations:K c =[C][D] [A][B]If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows:forward rate reverse rate = = k f [A][B] k r [C][D]where k f and k r are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal:k f [A][B]=k r [C][D]Thus, the rate constants are related to the equilibrium constant in the following manner:Kc =k f k l =[C][D] [A][B]Part AFor a certain reaction, K c = 5.21×1010 and k f = 0.237M ?2 ?s ?1 .

Part A

Calculate the value of the reverse rate constant, k r , given that the reverse reaction is of the same molecularity as the forward reaction? .Express your answer with the appropriate units.

Part B

For a different reaction, K c = 1.47×104, k f =9.28×10 5 s ?1 , and k r = 63.3s ?1 . Adding a catalyst increases the forward rate constant to 2.38×108s ?1 . What is the new value of the reverse reaction constant, k r , after adding catalyst?

Express your answer with the appropriate units.

Part C

Yet another reaction has an equilibrium constant Kc =4.32×10 5 at 25 ? C . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 ? C , what will happen to the equilibrium constant,

The equilibrium constant willYet another reaction has an equilibrium constant at 25 . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 , what will happen to the equilibrium constant,