All forward chemical reactions are accompanied by their reverse reactions (8,14,15). at the start of a reaction, when little or no product is present, the rate of the reverse reaction is negligible [8.14,15). however, as the concentration of products increases, the rate at which they decompose into reactants becomes greater [8,14,15). at equilibrium, the reverse rate matches the forward rate and the reactants and products are present in abundances given by the equilibrium constant for the reaction (8,14,15). we can analyze this behavior by thinking of a very simple reaction of the form (8,14,15]: forward: a- b r ate of the forward reaction=k, [a] (57) reverse: b-a rate of the reverse reaction = kx[b] (58) where ky is the rate constant for the forward reaction, and ke-rate constant for the reverse reaction. recall that the equilibrium constant k for the overall reaction is related to the rate constants k, and k, as k=k (59) the concentration of a is reduced by the forward reaction (at a rate k, [a]) but it is increased by the reverse reaction (at a rate k [b] [8.14,15). the net rate of change is therefore [8,14,15): = k [a] – kz[b] (60) where x represents either the decrease of the concentration of a or increase of the concentration of b. if we denote the initial concentration of a as e, and the initial concentration of b as b. equation (60) can be written as: =k: (@- x) - kz(b + x) (61) introducing the auxiliary parameters landk, ke- (64) k=k, +k equation (61) can be simplified as k(l -1) using your mathematica notebook for example 6.2 as a template, (a) verify that k(a - x) - kz(b + x) = k(l - x). rearrange equation (64) in such a way as to collect all terms related to r on the right hand side (rhs) of the equation, and all terms related to time and k on the left hand side (ihs), (b) print out the left hand side of the rearranged equation, (c) print out the right hand side of the rearranged equation,