Hi all, i've encountered this question below during my coursework...
In an experiment, x
moles of A placed in a flask underwent a reaction to produce B.
A(g) <-> 2B(g) [reversible reaction]
At equilibrium, y
moles of A had reacted and the total pressure in the flask was P
moles of inert
C is added to the closed flask, how will the value of equilibrium constant and the position of equilibrium be affected?
I've worked my answer out for this and my answer differs from my tutor's regarding the position of equilibrium. I strongly think that the position of equilibrium will shift left as by Le Chatelier's Principle, the closed system will relieve the stress (due to the addition of C molecules, therefore the backward reaction is favoured.
However, my tutor claimed otherwise and said because C is inert, it doesn't affect the system's equilibrium position.
Please advise and convince me that I'm wrong.