Each of the following reactions is allowed to reach equilibrium in a sealed container. For which of the reactions could you shift the equilibrium to the right by decreasing the pressure?

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Each of the following reactions is allowed to reach equilibrium in a sealed container. For which of the reactions could you shift the equilibrium to the right by decreasing the pressure?

Chemistry
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These are the general cases you can have, when you increase pressure the equilibrium will shift according to minimize the amount of gases molecules present. The stoichiometric coefficients are what's important. \(\huge A_{(g)}\rightleftharpoons \color{red}2 B_{(g)}\) shift left \(\huge \color{red}2 A_{(g)}\rightleftharpoons B_{(g)}\) shift right \(\huge A_{(g)}\rightleftharpoons B_{(g)}\)no change
@aaronq ok so this is so confusing for me. I'm horrible at chemistry. I'm sorry 🙈

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@aaronq explained it perfectly, but I think you might get confused by those equations by seeing it like that for the first time @gemini_babe18
Try to follow along with this analogy Imagine an elevator ; where you have 15 people crammed into it. Imagine that this is your chemical reaction: let's say in the elevator in the figure below: There are 15 people: Let's say (inside our elevator is the products) and outside the elevator with the say 2 people are the reactants. the number of people = number of moles. So inside the elevator we have 15 moles/ products ; outside we have 2 moles reactants. If this were a chemical reaction it would be a case where we have 15 moles of products and 2 moles of reactants. now let's say you increase the pressure; by cramming more people into the elevator? what do you think would happen? what would you do in a crammed elevator? would you want to get off as soon as possible? well these people, you Including myself would feel very uncomfortable in the elevator and we would want to get off as soon as possible to where there aren't as many people. |dw:1440968962196:dw| |dw:1440969444115:dw| So the same thing in a chemical reaction; if you increase the pressure it's going to favor the side with the fewer number of moles (If I remember this correctly). for the equation let's say from aaronq's example we have |dw:1440969646164:dw| Before we determine what side will be favored we have to see which side has more moles. in this case it's the products, so when we increase the pressure, the side with the less moles is going to be favored and that's A. reactants. Let's say if we had the following |dw:1440969816144:dw|
Ohh ok I see! Thank you guys both! :) @aaronq @photon336

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