What is the equilibrium constant of this reaction as written, in forward direction at 298k you can find out equilibrium constant and pressure also,

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## What is Gibbs free energy?

Mainly, Gibbs free energy is used to measure the maximum work done in thermodynamics when temperature and pressure are kept constant. it is denoted by symbol “G”. it is also called Gibbs energy. it is represent in j/kj. the Gibbs energy of reaction is represent by [∆G°]

### What is the equilibrium constant of this reaction as written, in forward direction at 298k

m2o3 – 2m(s) + 3/2 O2 (g)

Where,

M2O3 = -6.70

M(s) = O

o2(g) = O

this information regarding equilibrium constant of this reaction. and this equation for, Gf(kj/mole).

According to question, you can find the different value,

1. What is the Gibbs energy of this reaction?
2. What is the equilibrium constant of this reaction?
3. What is the equilibrium pressure of this reaction?

### What is the Gibbs energy of this reaction?

First, We find Gibbs energy of reaction?;

The basic formula of Gibbs energy of reaction is represent by,

∆G° = Gf products – Gf reactant

∆G° = [nM(s)×∆G°M(s) + no2(g) × ∆G°o2(g))] – [

Whaere,

∆G° – Gibbs energy of reaction

n – number of moles

Now, with this equation, you can find the Gibbs energy of the reaction. First you put all the value in this equation, we get

∆G° = [3mole × (0kj/mole) + 2moles × (0kj/mole)] – [

∆G° = 9.50 kj/mole

Therefore, the Gibbs energy of this reaction is, 9.50kj/mole.

### What is the equilibrium constant of this reaction?

equilibrium constantthe equilibrium constant of a chemical reaction provide the relationship between product and reactant when chemical reaction reach equilibrium. equilibrium constant is denoted by K.

So you can write the equation;

K = equilibrium constant

it is represent the relationship between the product and reactant if chemical reaction reached to the equilibrium.

the relationship between the equilibrium constant and Gibbs free energy is,

∆G° = -RT × In K

∆G° – Gibbs free energy = 9.50kj/ mole = 9500j/mole

R – gas constant = 8.314 j/k, mole

T – Temperature = 298k

K – equilibrium constant = ?

9500 = -8. 314 j/k mole × 298k × In K

K = 0.0216

therefore, the equilibrium constant of this reaction is, 0.0216.

### What is the equilibrium pressure of this reaction?

Now, we can calculate the value of equilibrium pressure of o2(g).

you know that,

K = (Po2)2

0.0216 = (Po2)2

Po2 = 0.0147 atm

Therefore, the equilibrium pressure of this reaction is, 0.0147 atm.