Question

The ΔG°f of atomic oxygen is 230.1 kJ/mol. Find ΔG° for the following dissociation reactionO2 (g) <--> 2O (g)then calculate its equilibrium constant at 298 K.

Answer 1

Kc = 2.145 × 10⁻⁸¹

Step-by-step explanation:

Let's consider the following reaction:

O₂(g) ⇄ 2O(g)

The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:

ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)

where,

ni are the moles of products and reactants

ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants

In this case,

ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)

ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol

ΔG° = 460.2 kJ/mol

With this information, we can calculate the equilibrium constant (Kc) using the following expression:

`Kc=e^(-\Delta G \°/R.T ) = e^{-460.2 kJ/mol/(8.314 * 10^(-3)kJ/mol.K)  * 298K }=2.145 * 10^(-81)`