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### Decreasing volume

Posted: Sat Jan 25, 2020 4:12 pm
why is it that decreasing volume (and conversely increasing pressure) and having more moles of gas on the reactant side, favors product formation? viceversa?

### Re: Decreasing volume

Posted: Sat Jan 25, 2020 4:23 pm
The reaction would shift in a way that wants to minimize the resulting increase in pressure, so that reaction would shift to the side with fewer gas molecules because it would decrease the pressure of the system.

### Re: Decreasing volume

Posted: Sat Jan 25, 2020 4:48 pm
Due to Le Chatelier's Principle, the reaction would want to minimize pressure and would thus shift to the side with fewer moles.

### Re: Decreasing volume

Posted: Sat Jan 25, 2020 4:54 pm
It's helpful to think of decreasing/increasing V or increasing/decreasing P in terms of Q and K. When you change the volume (basically changing the pressure), you're changing the Q value and whether the new Q is larger or lesser than K, you know which way the reaction will shift.

### Re: Decreasing volume

Posted: Sat Jan 25, 2020 5:22 pm
I like to think of it as the system relieving itself of the increase in pressure in order to balance itself out or reach equilibrium again by shifting to the side with less moles of gas.

### Re: Decreasing volume

Posted: Sat Jan 25, 2020 6:20 pm
I saw a Khan Academy video that explained this pretty well. The main idea of it was that if having 4 moles on the reactant side can turn into 2 moles on the product side, then you have less on the products. If you increase pressure by decreasing volume, you are making the space that everything is in smaller. Therefore, to compensate for this, the reaction wants to put less things into that smaller space to make the pressure even back at to less. Because of this, the reaction will make more products because it is taking the 4 moles and turning it into two moles.