If the prepared solution was made with 4.00 × 10⁻⁴ M lithium fluoride and 4.00 × 10⁻⁴ M of zinc chloride, evaluate the reaction quotient, Q, for zinc fluoride at these values.

Answer 1

Interesting! How does one have a REACTION quotient (Q) for a PRODUCT? ZnF₂ is a product of the reaction, and even if you look at the reverse reaction, where ZnF2 is a reactant, how do you calculate Q for a single species. The value of Q (reaction quotient) is the ratio of the products to reactant, each raised to a power of the stoichiometric coefficient.

First, we must write the correctly balanced equation for the reaction taking place:

2LiF(aq) + ZnCl₂(aq) → ZnF₂(s) + 2LiCl(aq) .. balanced equation

Q (reaction quotient) = [LiCl]²[ZnF₂] / [LiF]²[ZnCl₂] and since ZnF₂ is a solid, it should be omitted

Furthermore, LiF will be limiting in this reaction since equal concentrations of LiF and ZnCl₂ are present.

The actual value of Q would be (4x10^-4)² / 2x10^-4

Q = 8.0x10^-4

So, maybe I misunderstood the question, but I don't see how one can calculate a reaction quotient for a single species taking place in the reaction.

Answer 2

The reaction between lithium fluoride (LiF) and zinc chloride (ZnCl₂) can be represented by the following equation:

LiF + ZnCl₂ → ZnF₂ + 2 LiCl

The reaction quotient, Q, is the product of the concentrations of the products raised to their stoichiometric coefficients, divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. In other words:

Q = [ZnF₂]^1 [LiCl]^2 / [LiF]^1 [ZnCl₂]^1

At the given concentrations of 4.00 × 10⁻⁴ M LiF and 4.00 × 10⁻⁴ M ZnCl₂, the reaction quotient is:

Q = ([ZnF₂]^1) (2 x [LiCl]^2) / ([LiF]^1) ([ZnCl₂]^1) = ([ZnF₂]^1) (2 x (4.00 x 10⁻⁴ M)^2) / ((4.00 x 10⁻⁴ M)^1) ((4.00 x 10⁻⁴ M)^1) = ([ZnF₂]^1) (3.2 x 10⁻⁷) / (1.6 x 10⁻⁷) = 2 [ZnF₂]

Therefore, the reaction quotient for zinc fluoride at these concentrations is 2 times the concentration of zinc fluoride, [ZnF₂].