Question

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

H_2(g) + I_2(g) <--> 2HI_(g)              ΔH⁰ = + 26.5 kJ / mol

1. 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.
2. 2. 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?
3. 3. Based upon the information above, what is the numerical value of the K_c for the following reactions
      1. 2H_2(g) + 2I_2(g) « 4HI_(g)
      2. 2HI_(g)    « H_2(g) + I_2(g)

Answer 1

Given equilibrium reaction is

H_2(g) + I_2(g) <--> 2HI_(g)              ΔH⁰ = + 26.5 kJ / mol and K_c = 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 K_c for the reaction 2H_2(g) + 2I_2(g) <--> 2 4HI_(g);             

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

2.      The numerical value of the K_c for the reaction 2HI_(g)  <--> H_2(g) + I_2(g)          

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