Consider the reactionN2O4(g)→2NO2(g) where N2O4 is colorless and NO2 is brown. At a temperature of 5∘C, the reaction is colorless at equilibrium. This indicates that the equilibrium constant for this reaction at this temperature is Consider the reactionwhere is colorless and is brown. At a temperature of 5, the reaction is colorless at equilibrium. This indicates that the equilibrium constant for this reaction at this temperature is_______.a. negative.b. large.c. small (less than 1).d. zero, because equilibrium can only be obtained at room temperature.

When a reversible reaction is at equilibrium, the reactants are forming the products at the same speed that the products are forming the reactants, so there is no visual difference. The equilibrium can be characterized by a constant (Kc). For a generic reaction:

aA + bB ⇄ cC + dD

Kc = [tex]\frac{[C]^c*[D]^d}{[A]^a*[B]^b}[/tex]

For the given reaction: N₂O₄(g) ⇄ 2NO₂(g)

Kc = [NO₂]²/[N₂O₄]

If only NO₂ has color, and the solution is colorless, there are a few NO₂ in it, and [N₂O₄]>[NO₂], so Kc will be small, and less the 1, because for Kc = 1, [N₂O₄] = [NO₂]².