<p>LF61 and hihihi1368, I was merely commenting that I don't think that increasing pressure alone is enough to cause the reaction to move towards the side with less moles of gas. However, the question was indeed worded with the term, "compressing", which suggests that volume was decreased, therefore, resulting in an increase in concentration for both products and reactants.</p>
<p>I think mass of water is definitely needed. Its molar mass for goodness sake. (grams/mole)</p>
<p>I don't think both the products and the reactants can increase in concentration, the substance that is less in moles relative to the other side will increase in concentration.</p>
<p>If the volume is decreased, assuming both sides have the same number of moles of gas, the concentration will be increasing for both reactants + products. If the number of moles differ, than the change in concentration will vary depending on many factors.</p>
<p>The only factor we are looking at here is the decrease in volume
Everything else should be a constant.</p>
<p>One factor would just be the mole ratio of the gases. Another factor would be if the shift in equilibrium causing a shift towards either the reactants or products be large enough to neutralize the increase of the concentration due to the decrease in volume.</p>
<p>You mentioned the mole ratio as well, so what are the many factors you were referring to?</p>
<p>You said "I don't think both the products and the reactants can increase in concentration, the substance that is less in moles relative to the other side will increase in concentration."</p>
<p>If it was something like xA(g) + yB(g) ->/<- zC(g)
And the volume was reduced by x percent. This means that in order to have a decrease in concentration for let's say, the product, then the shift in towards either side would need to cause the number of moles of the product would have to be small enough such that this new number will be 1/5 of the orignal number of moles or less in order to have a smaller concentration than before.</p>
<p>It is common that both the reactants and products will experience an increase or a decrease in concentration. Otherwise, how would the equilibrium constant stay constant?</p>
<p>Well, you've lost me there with the mole stuff.
You don't chose what you want to decrease the concentration of, the concentration of products and reactants increases/decreases according to the factor given.
The conc. of products will <em>not</em> decrease with the given factor.
Think about it. what is reacting together - A(g) and B(g) - you make the volume smaller - they are bumping into eachother more frequently and colliding more and more - producing more C(g) - concentration of C increases.
Volume decrease, pressure increase - basically the same factor.
Don't really understand how the mole stuff works out but to get back to the original topic - pressure does affect concentrations in equilibrium reactions involving gases.
Also, editing posts gets confusing.</p>
<p>my impression of the sat II: meh</p>
<p>Explanation worth2try:</p>
<p>1A(g) + 1B(g) ->/<- 1C(g)</p>
<p>There are 2 moles of gas on the left side, and 1 mole of gas on the right side. When gases are compressed, they become more pressurized, so they want to relieve that pressure. In order to do this, the 2 moles of reactants can be significantly decreased to 1 mole, thereby relieving the pressure.</p>
<p>And yes, pressure is the same factor as volume in a closed container. Increasing the pressure "compresses" the gas. You cannot weigh the effect of the change in volume against the effect of the change in pressure because pressure is what causes a change in volume. An increase in pressure has the exact same effect as an equal decrease in volume, and since the two occur simultaneously, they create the same effect.</p>
<p>Well, since you both agree, it seems that you two are correct. Hehe, chemistry is confusing as heck...I need to study more :P</p>
<p>An increase of PARTIAL pressure of the REACTANTS is the same factor as the volume. But increasing total pressure through some inert gas does nothing.</p>
<p>
</p>
<p>All you need from water is its vapor pressure at the temperature and volume it takes up so you can find volume taken by gas, etc. Do a seach for determining molar mass vapor. You don't need mass of water.</p>