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What happens to the equilibrium when a inactive gas included? It is said that Partial...

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shihan | Student, Undergraduate | Salutatorian

Posted June 20, 2013 at 5:03 AM via web

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What happens to the equilibrium when a inactive gas included? It is said that Partial pressure of each gas doesn't change because inactive gas doesn't change the equlibrium point.but as far as i think when a inactive gas included then the total moles will be changed.there by fraction of moles will be changed so the partial pressure of each gas should change. 

eg:- 2X(g) <---> Y(g) + Z(g) the question is when 10 moles of inactive gas S is included what is the partial pressure of X(g),Y(g),Z(g),S(g)?in the answer it is said that partial pressure of X,Y,Z, wont change.and the partial pressure of S is 1*10^5 Pa (earlier it is said that total moles is 10 and total pressure is 10^5Pa )

i need to slove this.please help

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llltkl | College Teacher | Valedictorian

Posted June 20, 2013 at 7:29 AM (Answer #1)

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The inert gases do not participate in the reaction. Still addition of an inert gas has some effect on the position of a gaseous equilibrium. This is ascribable on the basis of Le Chatelier’s principle as follows:

a) When the inert gas is added at constant volume: the total gas pressure will increase but the partial pressures of individual components will remain the same. Addition of inert gas will not therefore affect the position of equilibrium.  

b) When the inert gas is added at constant pressure: the volume will increase, resulting in a fall of overall pressure and the equilibrium will shift towards the side where greater number of moles of gaseous species is formed.

This implies that when ∆n(g)  = +ve, the equilibrium will shift to the product side and when ∆n(g)  = -ve, the equilibrium will shift to the reactant side.

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