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 reason

The 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 :

The Gibb’s totally free energy readjust of products and also reactants to be given. Gibb’s free change because that reaction in front direction have the right to be calculated by substituting the offered values in Gibb’s totally free energy formula.


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

.The equilibrium consistent K the this reaction is 0.0669.The equilibrium pressure of
over M(s) in ~ 298K is 0.164 atm.

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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.