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Partial pressure of reacting gas

Using equations (15) we can see that the equilibrium constant is related to the partial pressures of reacting gases:

$${K_P} = {{{p_C}} \over {{p_A}{p_B}}}$$

for the reaction \({\rm{A}} + {\rm{B}} \to {\rm{C}}\). (Remember that these pressures must be related to a standard state.)

For a metal oxidation reaction ,

2M (s) + O2 (g) = 2MO (s) ,

the equilibrium constant has the form

$${K_P} = {1 \over {p_{{O_2}}^{}}}$$

Click here for a more detailed discussion of the equilibrium constant.

We can therefore find the equilibrium partial pressure of oxygen at a particular temperature from the value of ΔG°:

$${\left. {{p_{{O_2}}}} \right|_{eq.,T}} = \exp {{\Delta G^\circ } \over {RT}}$$

The equilibrium partial pressure of oxygen is the pressure at which the driving force for the reaction is zero. From equation 14 we see that if the partial pressure of oxygen is greater than this value, the free energy change for the reaction is negative and there is a driving force for the reaction to take place. Metal will be oxidised, and the partial pressure of oxygen will drop until it reaches equilibrium. This is effect described by Le Chatelier’s principle.

If the partial pressure of oxygen is below the equilibrium value, oxidation is avoided. (In fact, the metal oxide will disassociate to form metal plus oxygen gas-this is because there is a driving force for the reaction to proceed backward. For this reason the equilibrium partial pressure is often known as the dissociation pressure.)


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