Here is a question from my practice problems:
I know that the answer is A, but I don't know why. Is it because A has the most moles of gas on the left? thanks
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Shifts in equilibrium position of chemical reactions can be addressed by Le Chatelier's Principle. In short, the response of a system will be such that it counteracts the change that you are applying.
For a closed system, an increase in pressure at equal temperature will result in a decrease in volume according to the ideal gas law: $$\frac{P}{RT}=\frac{n}{V} $$
This change will cause an upward shift in partial pressures which the system will counteract. In this case the equilibrium will shift to the right for reaction A, because this will lower the partial pressures. Indeed this means that for an increase in pressure the reaction will shift to the side with the lower number of moles.
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I think the most intuitive way to think about this is rates of reaction. A reaction involving 2 molecules in the gas phase only happens when they collide. At higher pressures there will be more collisions between molecules.
This means that the reaction with the most molecules will be sped up more, so if a reaction has 2 molecules reacting to give 1 molecule of product, the forward reaction will be sped up more than the reverse reaction by increasing the pressure.
To do the kinetic calculations properly will require all the mechanistic steps involved. The general principle is that that at equilibrium, both reactions are happening at the same rate, but when you increase the pressure you will speed up the reaction which has more moles of gas, resulting in that reaction happening faster than the other reaction, shifting the equilibrium until the rates of the two reactions are balanced again.
