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A reaction mixture that consisted of 0.4 mol H2 and 1.55 mol I2 was introduced into a 2 L flask and heated. At the equilibrium, 60% of the hydrogen gas had reacted. What is the equilibrium constant Kc for the reaction H2(g) + I2(g) ⇀↽ 2 HI(g) at this temperature?A reaction mixture that consisted of 0.4 mol H2 and 1.55 mol I2 was introduced into a 2 L flask and heated. At the equilibrium, 60% of the hydrogen gas had reacted. What is the equilibrium constant Kc for the reaction H2(g) + I2(g) ⇀↽ 2 HI(g) at this temperature?

Answer :

Edufirst

Answer:

  • Kc = 1.1

Explanation:

1) Equilibrium equation (given):

  • H₂ (g) + I₂ (g) ⇄ 2HI (g)

2) Equilibrium constant expression:

       [tex]Kc=\frac{[{HI]}^2}{[H][I]}[/tex]

3) Determine the conentrations from the stoichiometry:

a) Number of moles:

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

Start               0.4                                 1.55                         0

Change        - 0.4×0.6 = - 0.24        -  0.24                  + 0.24×2 = + 0.48

                   ---------------------------       -------------              --------------------------

End                0.16                               1.31                       0.48

b) Concentrations:

  • [ H₂ (g) ] = 0.16 mole / 2 liter = 0.08 M
  • [ I₂ (g) ] = 1.31 mole / 2 liter =   0.655 M
  • [HI (g) ] = 0.48 mole / 2 liter = 0.24 M

4) Compute Kc:

  • Kc = (0.24 M)² / (0.08 M × 0.655 M) = 1.099 ≈ 1.1 ← answer

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