Equilibrium Reactions: Which is Least Affected by Volume Change?

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SUMMARY

The discussion centers on identifying which equilibrium reaction is least affected by a change in volume, with the correct answer being option C: H2 (g) + S (l) = H2S (g). The reasoning is based on Le Chatelier's principle, which states that changes in volume primarily affect gaseous reactions. In this case, the presence of a solid (S) and a liquid (H2O) means that the reaction is less sensitive to volume changes compared to those involving only gases.

PREREQUISITES
  • Understanding of Le Chatelier's principle
  • Knowledge of chemical equilibrium
  • Familiarity with states of matter (solid, liquid, gas)
  • Basic grasp of reaction stoichiometry
NEXT STEPS
  • Study the effects of volume changes on gaseous equilibria
  • Learn about the principles of chemical equilibrium
  • Explore examples of Le Chatelier's principle in various reactions
  • Investigate the role of solids and liquids in equilibrium systems
USEFUL FOR

Chemistry students, educators, and anyone studying chemical equilibrium and reaction dynamics will benefit from this discussion.

moxy
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Homework Statement



Which of the following equilibria is least affected by a change in the volume of the system?

A. 2 C (s) + O2 (g) = 2 CO (g)
B. 2 NO2 (g) = N2O4 (g)
C. H2 (g) + S (l) = H2S (g)
D. H2O (l) = H2O (g)
E. 2 NO (g) + Cl2 (g) = 2 NOCl (g)


Homework Equations



N/A

The Attempt at a Solution



I know the answer is C, but I don't know why. I thought maybe it had something to do with the number of moles on each side of the reaction. As volume decreases, pressure increases, and the reactions will favor the side with the least number of moles. I know this is true, but perhaps it doesn't apply? This is a sample exam problem, and any hints as to how I should approach it would be greatly appreciated.
 
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This deals with the Le Chatelier principle.

A. 2 C (s) + O2 (g) = 2 CO (g)
B. 2 NO2 (g) = N2O4 (g)
C. H2 (g) + S (l) = H2S (g)
D. H2O (l) = H2O (g)
E. 2 NO (g) + Cl2 (g) = 2 NOCl (g)

Take A for example. If you increase the decrease the volume, the reaction will shift left, because there are less moles of gas on that side. The change in volume usually only affects gases. So if you were to have an equal number moles of gases on both sides of the reaction, changing the volume wouldn't really affect the system. I think the only key thing you're missing is moles of "gases."
 
Ah, you save the day again. The gas part went over my head. This makes perfect sense now. Thanks...again!
 

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