Question

Consider the reaction below that has a K_eq of 7.94⋅10^-2. 3.4 mol of A and 6.7 mol of B are added to a 5 L container. Calculate the equilibrium concentration of C. Enter your answer in scientific notation using 2 significant figures.

2 A + 1 B ↔ 1 C

Answer 1

To solve this problem, we'll use the stoichiometry and equilibrium expression of the given reaction.

The stoichiometry of the reaction tells us that for every 2 moles of A and 1 mole of B, 1 mole of C is produced.

Given: Initial moles of A (A0) = 3.4 mol Initial moles of B (B0) = 6.7 mol Volume of the container (V) = 5 L Equilibrium constant (Keq) = 7.94 × 10^(-2)

Let's assume the equilibrium concentration of C is x (in moles/L).

According to the stoichiometry, the change in concentration of A will be -2x (since 2 moles of A react to produce 1 mole of C), and the change in concentration of B will be -x (since 1 mole of B reacts to produce 1 mole of C).

Using the equilibrium expression: Keq = [C] / ([A]^2 * [B])

Substituting the given values: 7.94 × 10^(-2) = x / ((3.4 - 2x)^2 * (6.7 - x))

Simplifying the equation: 7.94 × 10^(-2) = x / (11.56 - 13.6x + 4x^2)

Rearranging the equation: 0 = 4x^2 - 13.6x + 11.56 - (7.94 × 10^(-2))

Solving this quadratic equation, we find: x ≈ 0.0055 moles/L

Therefore, the equilibrium concentration of C is approximately 5.5 × 10^(-3) moles/L.