Consider the decomposition the a steel oxide to its elements, wherein M to represent a generic metal.

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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 reason

The 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:

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Answer:Given reaction :

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

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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

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

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.The equilibrium continuous K of this reaction is 0.0669.The equilibrium pressure of
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over M(s) at 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 terms of partial pressure using equilibrium consistent value.