Consider the reaction. 2HF(g) +

H2(g) + F2(g)

At equilibrium at 600 K, the concentrations are as follows.

[HF] = 5.82 x 10-2 M

[H2] = 8.4 x 10-3M

[F2] = 8.4 x 10-3M

What is the value of Kea for the reaction expressed in scientific notation?

O 2.1 x 10-2

102.1 x 102

1.2 x 103

0 1.2 x 10-3

0.021

Explanation:

Step 1: Write the balanced equation

2 HF(g) ⇄ H₂(g) + F₂(g)

Step 2: Calculate the value of the concentration equilibrium constant

The concentration equilibrium constant (K) is equal to 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.

K for this reaction is:

K = [H₂] × [F₂] / [HF]²

K = (8.4 × 10⁻³) × (8.4 × 10⁻³) / (5.82 × 10⁻²)² = 0.021

the answer is c.the valence of the atoms combine .