Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only












2














For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):



$$ce{A(s) + B(g) <=> C(g)},$$



if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).



And the opposite if pressure decreased.



E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?






physical-chemistry equilibrium pressure







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    2














    For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):



    $$ce{A(s) + B(g) <=> C(g)},$$



    if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).



    And the opposite if pressure decreased.



    E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?






    physical-chemistry equilibrium pressure







    share|improve this question









    New contributor




    frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      2












      2








      2







      For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):



      $$ce{A(s) + B(g) <=> C(g)},$$



      if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).



      And the opposite if pressure decreased.



      E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?






      physical-chemistry equilibrium pressure







      share|improve this question









      New contributor




      frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):



      $$ce{A(s) + B(g) <=> C(g)},$$



      if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).



      And the opposite if pressure decreased.



      E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?






      physical-chemistry equilibrium pressure




      physical-chemistry equilibrium pressure






      share|improve this question









      New contributor




      frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









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      edited 3 hours ago









      andselisk

      13.2k64698




      13.2k64698






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          2 Answers
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          There are many examples of such reactions, for instance:



          $$
          begin{align}
          ce{S (s) + O2 (g) &<=> SO2 (g)}\
          ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
          ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
          end{align}
          $$



          However, your assumption




          all that is considered is which side the solid/liquid is on, not the moles of gas present on either side




          is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.






          share|improve this answer































            0














            Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.






            share|improve this answer





















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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

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              active

              oldest

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              active

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              2














              There are many examples of such reactions, for instance:



              $$
              begin{align}
              ce{S (s) + O2 (g) &<=> SO2 (g)}\
              ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
              ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
              end{align}
              $$



              However, your assumption




              all that is considered is which side the solid/liquid is on, not the moles of gas present on either side




              is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.






              share|improve this answer




























                2














                There are many examples of such reactions, for instance:



                $$
                begin{align}
                ce{S (s) + O2 (g) &<=> SO2 (g)}\
                ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
                ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
                end{align}
                $$



                However, your assumption




                all that is considered is which side the solid/liquid is on, not the moles of gas present on either side




                is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.






                share|improve this answer


























                  2












                  2








                  2






                  There are many examples of such reactions, for instance:



                  $$
                  begin{align}
                  ce{S (s) + O2 (g) &<=> SO2 (g)}\
                  ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
                  ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
                  end{align}
                  $$



                  However, your assumption




                  all that is considered is which side the solid/liquid is on, not the moles of gas present on either side




                  is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.






                  share|improve this answer














                  There are many examples of such reactions, for instance:



                  $$
                  begin{align}
                  ce{S (s) + O2 (g) &<=> SO2 (g)}\
                  ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
                  ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
                  end{align}
                  $$



                  However, your assumption




                  all that is considered is which side the solid/liquid is on, not the moles of gas present on either side




                  is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 hours ago

























                  answered 3 hours ago









                  andselisk

                  13.2k64698




                  13.2k64698























                      0














                      Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.






                      share|improve this answer


























                        0














                        Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.






                        share|improve this answer
























                          0












                          0








                          0






                          Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.






                          share|improve this answer












                          Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 3 hours ago









                          harshit54

                          1108




                          1108






















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