**Consider the decomposition the a steel oxide to its elements, wherein M to represent a generic metal.You are watching: What is the equilibrium pressure of o2(g) over m(s) at 298 k?**

**M203(s)—> 2M(s) + 3/2 O2(g)**

**info provided for Gf(kJ/mol):****M203= -6.70****M(s)=0****O2(g)= 0**

**what is the standard adjust in Gibbs power for rxn as composed in front direction? (kJ/mol)****What is the equilibrium continuous (K) that this rxn, as written in front direction at 298K?**

**What is the equilibrium push of O2(g) over M(s) in ~ 298K? (atm)**

**Concepts and reason**

**Fundamentals**The typical Gibb’s complimentary energy readjust can be calculated by:

**Answer:**Given reaction :

From the given temperature, the equilibrium consistent K is calculate by substituting the given values in Gibb’s cost-free energy formula through respect to equilibrium constant

The standard readjust in Gibb’s complimentary energy for the reaction in forward direction is

.The equilibrium continuous K of this reaction is 0.0669.The equilibrium pressure of over M(s) at 298K is 0.164 atm.See more: How To Change Light Bulbs In High Chandelier &Ndash; Upgraded Home

From the offered reaction, only oxygen is gas. Therefore, the equilibrium press of oxygen is calculation in terms of partial pressure using equilibrium consistent value.