Consider the decomposition the a steel oxide come its elements, whereby M represents a generic metal.
You are watching: What is the equilibrium constant of this reaction, as written, in the forward direction at 298 k?
M203(s)—> 2M(s) + 3/2 O2(g)
info given for Gf(kJ/mol):M203= -6.70M(s)=0O2(g)= 0
what is the standard change in Gibbs energy for rxn as written in front direction? (kJ/mol)What is the equilibrium consistent (K) of this rxn, as created in front direction in ~ 298K?
What is the equilibrium push of O2(g) end M(s) at 298K? (atm)
Concepts and also reasonThe distinction between free energy of commodities and totally free energy of reaction is stood for as standard cost-free energy adjust of the reaction.The ratio of concentration of a product come the ratio of concentration that a reactant is stood for as equilibrium constant. The equilibrium constant is expressed as.The equilibrium constant, in regards to partial pressure, is stood for as equilibrium pressure.
FundamentalsThe conventional Gibb’s cost-free energy readjust can it is in calculated by:
Answer:Given reaction :
From the provided temperature, the equilibrium continuous K is calculated by substituting the offered values in Gibb’s cost-free energy formula with respect to equilibrium constant
The standard readjust in Gibb’s complimentary energy for the reaction in front direction is
See more: How Long To Beat Ni No Kuni 2, How Long Is Ni No Kuni Ii: Revenant Kingdom
From the offered reaction, only oxygen is gas. Therefore, the equilibrium press of oxygen is calculation in regards to partial pressure using equilibrium continuous value.