Hydrogen and carbon dioxide react at a high temperature to give water and carbon

Question

Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide.
H2(g)+CO2(g)=H2O(g)+CO(g)
(a)Laboratory measurements at 986 degrees Celsius show that there are 0.11 mol each of CO2 at equilibrium in a 1.0L container. Calculate the equilibrium constant for the reaction at 986 degrees Celsius.
(b)Suppose 0.050 mol each of H2 and CO2 are placed in a 2.0L container. When equilibrium is achieved at 986 degrees Celsius, what amounts of CO(g) and H2O(g), in moles, would be present? Use the value Kc from part (a).

Explanation / Answer

            H2(g)+CO2(g)=H2O(g)+CO(g)

       [H2O]   = 0.11M

     [CO]   = 0.11M

      [H2] = 0.087M

      [CO2] = 0.087M

     Kc   = [H2O][CO]/[H2][CO2]

           = 0.11*0.11/0.087*0.087   = 1.59

b.      H2(g)+CO2(g)=H2O(g)+CO(g)

    I   0.05    0.05       0           0

C   -x         -x          +x         +x

E    0.05-x    0.05-x    +x        +x

     [H2] = no of moles/volume in L   = 0.05-x/2

   [CO2] = no of moles/volume in L   = 0.05-x/2

    [H2O] = no of moles/volume in L = x/2

   [CO] = no of moles/volume in L = x/2

      Kc = [H2O][CO]/[H2][CO2]

     1.59   = x*x/(0.05-x)(0.05-x)

      1.26    = x/0.05-x

    1.26(0.05-x) = x

      x =0.027

   [CO] = Molarity * volume in L

           = 0.027*2 = 0.054 moles

[H2O] = Molarity * volume in L

           = 0.027*2 = 0.054 moles

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