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The following reaction is at equilibrium $$\ce{4HCl(aq) + O2(g) <=>2H2O(l) + 2Cl2(g)}\quad \Delta H>0$$Which one of the following changes will result in an increase in the concentration of $\ce{Cl2(g)}$

  1. adding some $\ce{H2O}$
  2. decreasing the temperature
  3. adding a catalyst
  4. decreasing the pressure
  5. removing some $\ce{HCl}$

Decreasing the pressure (4) is the answer but I don't understand why. I though decreasing pressure shifts the equilibrium to the side with the larger number of moles which is the left?

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    $\begingroup$ Welcom to Chemistry.SE! Your explanation would be correct if all the molecules would be gases. Are they? ;-) $\endgroup$ – Klaus-Dieter Warzecha Mar 10 '16 at 6:41
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Remember that $\ce{HCl}$ (aq) and $\ce{H2O}$(l) are liquids while the oxygen and chlorine are gases . As the number of moles of gases on the left hand side (one mole of oxygen) is smaller than that of the right hand side (two moles of chlorine), decreasing the pressure makes the equilibrium position move to the right hand side, therefore the concentration of chlorine increases.

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While checking for the effect of pressure on equilibrium, consider only the species which are in gaseous state. In above equilibrium only dioxygen and chlorine are in this state.

Ref:http://www.adichemistry.com/physical/equilibrium/le-chatelier/le-chatelier-principle.html

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