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= -6.70M(s)=0O2(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 reasonThe difference between complimentary energy of assets and cost-free energy of reactants is stood for as standard free energy adjust of the reaction.The ratio of concentration of a product come the proportion of concentration of a reactant is represented as equilibrium constant. The equilibrium constant is expressed as.The equilibrium constant, in regards to partial pressure, is stood for as equilibrium pressure.
FundamentalsThe 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
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.