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Consider the following equilibrium with a Kc = 55.6 at a temperature of 698 K. H2(g)...

Consider the following equilibrium with a Kc = 55.6 at a temperature of 698 K.

H2(g) + I2(g) <--> 2HI(g)              ΔH0 = + 26.5 kJ / mol

    1. If the initial concentrations were [H2] = 0.12 M; [I2] = 0.041 M; and [HI] = 2.6 M. Is the system at equilibrium, and if not, in which direction must it shift to establish equilibrium? Justify your answer.
    1. At the same 698 K, 0.50 mol of H2 and 0.88 mol of I2 are placed in a 3.0 L container. What are the concentrations of all three species at equilibrium?
    1. Based upon the information above, what is the numerical value of the Kc for the following reactions
      1. 2H2(g) + 2I2(g) « 4HI(g)
      2. 2HI(g)    « H2(g) + I2(g)
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Answer #1

Given equilibrium reaction is

H2(g) + I2(g) <--> 2HI(g)              ΔH0 = + 26.5 kJ / mol and Kc = 55.6 at a temperature of 698 K.

Reaction quotient, Qc = [HI]2/[H2][I2] A and initial concentrations arere [H2] = 0.12 M; [I2] = 0.041 M; and [HI] = 2.6 M

Qc = (2.6)^2/(0.12)(0.041) =1373.98

Now Qc > Kc, therefore the reaction is not at equilibrium and equilibrium will be shifted in backward reaction because Qc> Kc.

                     H2(g)            +         I2(g)      <--->         2HI(g)

initial            0.50 mole                0.88 mole                -

change           -x                              -x                        +2x

equilb     0.50-x                   0.88-x                         2x


    Kc = [HI]2/[H2][I2] = (2x/3)^2/((0.50-x/3)*(0.88-x/3)) = 55.6

x = 0.46 moles

At equilibrium , the concentration of H2, [H2]= 0.50-0.46/3 = 0.0133 M

                           the concentrations of I2, [I2]= 0.88-0.46/3 = 0.14 M

                          the concentrations of HI, [HI]= 2*0.46/3 = 0.306 M

----

1. The numerical value of the Kc for the reaction 2H2(g) + 2I2(g) <--> 2 4HI(g);             

                      K1c = (Kc)2 = (55.6)2 =3091.36

2.      The numerical value of the Kc for the reaction 2HI(g)  <--> H2(g) + I2(g)          

                      K1c = (1/Kc) = (1/55.6)= 0.0179

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