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CHEMICAL EQUILIBRIUM PROBLEMS WITH SOLUTIONS

1.  After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472oC it is found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3.  From these data calculate  the equilibrium constant Kp for this reaction.

Solution:


2.  An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25oC:   [CH3COOH] = 1.65x10-2 M, [H+] = 5.44x10-4 M, and [CH3COO-] = 5.44x10-4 M.  Calculate the equilibrium constant Kc, for the ionization of acetic acid at 25oC.  The reaction is 

                             CH3COOH(aq)    ⇄  H+(aq)   +   CH3COO-(aq)

Solution:



3.  Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions.  In the gas phase, it decomposes gaseous nitrogen dioxide

                                                   N2O4(g)   ⇄  2NO2(g)

Consider an experiment  in which gaseous  N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp = 0.133.  At equilibrium, the pressure of N2O4 was found to be 2.71 atm.  Calculate the equilibrium pressure of NO2.

Solution:


4.  Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:   CO(g)  +   2H2(g)  ⇄   CH3OH(g).   An equilibrium mixture in a 2.00 L vessel is found to contain 0.0406 mole CH3OH,  0.170 mol CO,  and 0.302 mol H2 at 500 K.  Calculate the Kc at this temperature.

Solution:

Calculation of concentrations of reactants and products:


 Calculating for Kc:




5.  Gaseous hydrogen iodide is placed in a closed container at 25oC, where it is partially decomposes to hydrogen and iodine:

                                            2HI(aq)   ⇄    H2(aq)    +    I2(aq)

At equilibrium it is found that [HI] = 3.53x10-3 M, [H2] = 4.79x10-4 M, and [I2] = 4.79x10-4 M.  What is the value of Kc at this temperature?

Solution:


6.  For the reaction:

                                   2NO(g)   +   2H2(g)   ⇄     N2(g)   +   2H2O(g)

it is determined that at equilibrium at a particular temperature, the concentrations are as follows:
[NO] = 8.1x10-3 M, [H2] = 4.1x10-5 M, [N2] = 5.3x10-2 M, and [H2O] = 2.9x10-3 M.  Calculate the value of Kc for the reaction of this temperature.

Solution:


7.  If Kc = 0.042 for PCl3(g)  +   Cl2(g)   ⇄    PCl5(g)  at 500K.  What is the value of Kp for this reaction at this temperature?

Solution:


8.  At 1000K, Kp = 1.85 for the reaction:  SO2(g)  +   ½O2(g)    ⇄     SO3(g)

a.  What is the value of Kp for the reaction SO3(g)  ⇄  SO2(g)  +   ½  O2?
b.  What is the value of Kp for the reaction  2SO2(g)  +  O2(g)    ⇄   2SO3(g)?
c.  What is the value of Kc for the reaction in b?

Solution:


9.  The following equilibria were obtained at 823K:
   
                  CoO(s)  +  H2(g)  ⇄   Co(s)  +  H2O(g),         Kc =  67
              CoO(s)   +  CO(g)    ⇄   Co(s)   +   CO2(g),        Kc = 490

Based of these equilibria , calculate the equilibrium constant for
           H2(g)   +   CO2(g)    ⇄   CO(g)   +   H2O(g)  at 823K.

Solution:  

Analyzing how the equations are manipulated:

           CoO(s)  +   H2(g)    ⇄   Co(s)   +  H2O(g)                Kc1 = 67
           CO2(g)  +  Co(s)     ⇄   CoO(s)  +  CO(g)                 Kc2 = 1÷ 490 = 2.04x10-3

After cancelling similar species from reactants and products we arrive the equation:
          H2(g)  +  CO2(g)   ⇄   CO(g)  +  H2O(g)                Kc3 = Kc1 x Kc2 = (67)(2.04x10-3) = 0.14
                                                                                               

10.  Consider the equilibrium:

                N2(g)  +  O2(g)   +   Br2(g)   ⇄    2NOBr(g)

Calculate the equilibrium constant Kp for this reaction, given the following information at 298 K.

              2NO(g)  +   Br2(g)   ⇄   2NOBr(g)        Kc = 2.0
              2NO(g)  ⇄  N2(g)  +  O2(g)                   Kc = 2.1x103

Solution: 

Analyzing how the equations are manipulated:

     2NO(g)  +   Br2(g)   ⇄   2NOBr (g)               Kc1 = 2.0
     N2(g)  +  O2(g)   ⇄   2NO(g)                         Kc2 = 1 ÷ 2.1x103 = 4.76x10-4

After cancelling similar species both in the reactant side and product side, the equation is:

    N2(g)  +  O2(g)  +  Br2(g)     ⇄    2NOBr(g) 

 Kc3 =  Kc1 x  Kc2 =(2.0)(4.76x10-4) = 9.52x10-4


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