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I understand how LCP works in favouring the side with more moles of gas, if the imposed change is a decrease in pressure. Or vice versa, if the imposed change is an increase in pressure the side with less moles of gas is where equilibrium shifts. I'm also aware of how a dilution (addition of H2O) means the equilibrium shifts to favour the side with more aqueous species. However I struggle with this equilibrium shifting concept in relation to their corresponding reaction rate vs time graph.

I was taught that in general, the side with more species is affected the most by the imposed change. However this is my confusion. For eg. In the following equilibrium,

2SO2(g) + O2 (g) >>><<< 2SO3 (g)

Ignore the H2 Removed if the imposed change is an decrease in pressure (increase in volume), it will shift towards the reactants, favouring the reverse reaction, to counteract the change. In the reaction rate/time graph (treat H2 removed = doubling of volume ), the forward reaction rate line drops significantly more than the reverse reaction rate line...why is this so? Why is it that side with MORE moles of gas, will be affected more by the increase in volume, when there is more reactant particles colliding compared to the products?? Shouldn't the reverse rate drop more as there are less mole of gas and thus a lower partial pressure??

Will appreciate every answer, thankyou.

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  • $\begingroup$ What is the relation of H2 to reaction of SO2 oxidation ? $\endgroup$
    – Poutnik
    Sep 15 at 8:11
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See, I will explain my concept to you, it might be helpful In a reversible reaction in equilibrium, N2 + 3 H2 —— 2 NH3 For example, if you apple pressure in this system, then reactant had more than products, so reactant molecules will aggregate and increase in amount at a particular place (saying just to explain), then according to Le-Chateliar Principle, if reactant had more moles than product, then reaction will shift towards product… This is my explanation Hope it’s helpful 😀

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