Suppose the reaction A₂ + B₂ ⇆ 2 AB proceeds via a one-step mechanism involving a collision between one A₂ molecule and one B₂ molecule. Suppose also that this reaction is reversible, and that the forward reaction is inherently much faster than the reverse reaction.

(a) Does the equilibrium lie to the left or to the right? Explain your choice in terms of the reactant and product concentrations necessary to establish equal forward and reverse rates.

(b) Does an analysis in terms of the relationship K_eq = k_f/k_r yield the same answer as in (a)? Explain.