Confusing Chem SAT II quest.

Archive SAT Subject Test Preparation
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<p>This was on the sparknotes practice test. I got it wrong by overthinking it, but I think I might actually be right…</p>

<p>X2(g) + 2H2(g)2H2X(g) + energy</p>

<li><pre><code>Addition of argon to the above equilibrium will
</code></pre>

<p>(A) Increase the equilibrium concentration of <a href=“B”>H2</a> Increase the equilibrium concentration of <a href=“C”>H2X</a> Decrease the equilibrium concentration of <a href=“D”>H2X</a> Have no effect on the equilibrium concentrations
(E) Decrease the amount of heat energy given off by the system </p></li>
</ol>

<p>The correct answer was D because “Adding an inert gas will not effect equilibrium concentrations since it is not involved in the reaction.” But if you add an inert gas it will increase the pressure of the system right? According to Dalton’s law the total pressure is the sum of all of the partial pressures. So if you add an Argon, the pressure increases, and according to Le Chatelier’s principle the equilibrium will shift to the direction which minimizes the stress, which in this case is to the right because it has less moles, making the answer B.</p>

<p>thoughts? Or am I totally off. A response would be nice so I don’t bomb the SAT II saturday with illogical thoughts.</p>

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<p>Argon wouldnt participate in the reaction… therefore it doesnt act as a stressor. If it were an compound that had X2 or any other substances in the reaction, the common ion effect will happen</p>

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<p>Doesn’t adding Argon (g) to the system increase the pressure of the system? According to Dalton’s law the pressure of the system = sum of partial pressures, so if more molecules of Argon gas were added, pressure would increase. </p>

<p>And if so, if the question had simply said if the pressure was increased on the system what would occur, the correct answer would be B.</p>

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<p>that would only be true if you add more moles of X2 or H2 to the system. Whichever side the equilibrium will shift will have a less total number of moles. If you add argon, it would not effect the system at all because if it did, it wouldnt produce the same result, and also argon is an inert gas… meaning it doesnt really react with anything. so it’ll just be… around doing nothing</p>

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<p>well think about it.</p>

<p>remember the dalton’s partial gas thing? that each gas exert pressure as if the other ones are not there? same thing applies. adding a inert gas will exert a pressure of its own. others will also exert pressure of its own. so it’s just not Keq is not affected.</p>

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<ol>
<li><p>The OP is correct. The answer should be B.</p></li>
<li><p>The reasoning is due to LeChateler’s principle. If you don’t remember this concept, then try google or wikipedia:
<a href="http://en.wikipedia.org/wiki/Le_Chatelier’s_principle#Total_pressure"&gt;http://en.wikipedia.org/wiki/Le_Chatelier’s_principle#Total_pressure&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;
</ol>

<p>It even has a section saying that inert gases can still cause shifts.</p>

<ol>
<li>@MSU, no offense, but I think you have some studying to do for Saturday test. I’m willing to help you though if you post questions (preferably very hard ones to exercise my chem knowledge).</li>
</ol>

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<p>no it’s okay I’m not offended. I actually studied… A LOT. But it’s weird, I keep getting the really obvious ones wrong and the hard ones right.</p>

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<p>okay so when will adding inert gases affect it because it doesn’t always shift it</p>

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<p>well, with everything i’ve seen, the assumption is that the addition of inert gases does NOT cause a change in equilibrium concentrations. so i agree with spark notes in that it should be D. because even if you look at the wiki link, it says that inert gases MAY change the concentrations, but all the practice ap questions i’ve ever done regarding always say no change. </p>

<p>also, i just checked back on the link. under the “effect of adding an inert gas” section, it has a subsection “volume held constant,” which is our assumption, where it states that the partial pressures should not change. so sunpenguin’s explanation is perfect.</p>

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<p>and to answer msu’s question, it is affected when volume changes. because the partial pressures are directly affected.</p>

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<p>Sometimes Wikipedia is full of wrong stuff… Wikipedia can be changed by ANYONE… don’t always trust wiki. But I think you can make a case for B or D.</p>

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<p>from my AP chem textbook… hope this clears things up:</p>

<p>It is possible to change the total pressure of the system without changing its volume. For example, pressure increases if additional amounts of any of the reacting components are added to the system. We have already seen how to deal with a change in concentration of a reactant or product. The total pressure within the reaction vessel might also be increased by adding a gas that is not involved in the equilibrium. For example, argon might be added to the ammonia equilibrium system. The argon would not alter the partial pressures of any of the reacting components and therefore would not cause a shift in equilibrium.</p>

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<p>So it all depends on the particles in the equilibrium equation only.</p>

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<p>Sparknotes is right. There have been similar questions on old AP Chem MC, and in every case, adding a gas that is not involved in the equation has no effect.</p>

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<p>if you think it will increase the pressure, don’t be tricked, it won’t for the sat 2 chem test
any noble gas won’t have any affect on equilibrium</p>

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<p>No, the test answer is right. It is true that by adding argon the pressure increases. However, the pressure would increase for the whole system, not just part of it. Therefore, le chatlier’s principle will not work because you are not affecting one portion of the reaction, instead it is affecting the whole system.</p>

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<p>@afruff23 LOL</p>