The equilibrium constant for the dissociation of molecular iodine, I₂(g) ⇌ 2 I(g), at 800 K is K_c = 3.1 × 10^–5. (a) Which species predominates at equilibrium I₂ or I?

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1. Understand the equilibrium expression for the reaction: The equilibrium constant, Kc, is defined for the reaction I2(g) ⇌ 2 I(g) as Kc = [I]^2 / [I2]. This expression relates the concentrations of the products and reactants at equilibrium.
2. Interpret the value of Kc: A small value of Kc (3.1 × 10–5) indicates that the equilibrium position favors the reactants. In this case, it suggests that the concentration of I2 is much greater than that of I at equilibrium.
3. Consider the stoichiometry of the reaction: The reaction shows that one molecule of I2 dissociates into two atoms of I. Therefore, if the concentration of I2 is high, the concentration of I will be relatively low.
4. Analyze the equilibrium position: Since Kc is much less than 1, the reaction does not proceed significantly towards the formation of products (I atoms). This implies that I2 is the predominant species at equilibrium.
5. Conclude based on the equilibrium constant: Given the small Kc value, the equilibrium mixture contains mostly I2 molecules, with very few I atoms present. Thus, I2 predominates at equilibrium.